# BMS CT-2017 Soln manual 06 05 2017

```60496
B.E/B.Tech. DEGREE EXAMINATION, NOV./DEC. 2016
Electrical and Electronics Engineering
EE2151 – CIRCUIT THEORY
1. In the circuit of Fig. 1 find the current I by mesh method.
15 Ω
5Ω
I
10 V
+
–
+
–
1.25 Ω
20 V
Fig. 1
The direction of mesh currents I1 and I2 are assumed as shown in Fig. 2.
15 Ω
5Ω
I
10 V
+
–
I1
1.25 Ω
I2
+
–
20 V
Fig. 2
By inspection, the mesh equations are written as,
6.25 −1.25
I1
10
=
−1.25 16.25
I2
−20
Solving the above matrix equation, we get
I1 = 1.375 A
I2 = −1.125 A
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Therefore, current I = I1 − I2 = 2.5 A
2. Write the nodal equations for the network of Fig. 3. Hence find the potential
difference between nodes 2 and 4.
3A
4A
1Ω
2Ω
4Ω
5
Ω
5
Ω
4A
Fig. 3
Taking node 4 as reference, by inspection the node equations are written as,


1
1 1
+
−
0
 5 1


 
1

 V1
1

1 
1 1 1
 −1
  V2  =  −1 
+ +
−

1
1 2 4
2 

 V3
0

1
1 1 
0
−
− +
2
5 2
Simplifying,


 

1
1.2
−1
0
V1
 −1 1.75 −0.5   V2  =  −1 
0 −0.5
0.7
0
V3
Solving the above matrix equation, we get
V1 = 0.5851 V
V2 = −0.2978 V
V3 = −0.2128 V
Hence, the potential difference between nodes 2 and 4, V24 = −0.2978 V
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3. Obtain the star connected equivalent for the delta connected circuit shown in
Fig. 4.
13 Ω
A
C
12 Ω
14 Ω
B
D
Fig. 4
The equivalent star elements for the given delta network elements are found as:
RAN =
=
RBN =
=
RCN =
=
RCA &times; RAB
RAB + RBC + RCA
13 &times; 12
=4Ω
12 + 14 + 13
RAB &times; RBC
RAB + RBC + RCA
12 &times; 14
= 4.3077 Ω
12 + 14 + 13
RBC &times; RCA
RAB + RBC + RCA
14 &times; 13
= 4.6667 Ω
12 + 14 + 13
The equivalent star network thus found is drawn as shown in Fig. 5.
A
4Ω
7Ω
07
3
N
.
4
4.6
66
7Ω
C
B
Fig. 5
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4. Determine the Thevenin’s equivalent circuit across AB for the given circuit
shown in Fig. 6.
10 Ω
5Ω
A
50 V
25 V
B
Fig. 6
Thevenin’s voltage
Let I be the current through the circuit and by applying KVL, we obtain
Therefore,
−50 + 10I + 5I + 25 = 0
I = 1.6667 A
Again applying KVL to the mesh involves 50 V, 10 Ω and VAB ,
Therefore,
−50 + 10(1.6667) + VAB = 0
VAB = Vth = 33.3333 V
Thevenin’s resistance
To compute Thevenin’s resistance the two voltage sources are short-circuited as
shown in Fig. 7.
10 Ω
5Ω
A
B
Fig. 7
With reference to Fig. 7, Rth is obtained as:
Rth = RAB =
10 &times; 5
= 3.3333 Ω
10 + 5
Thevenin’s equivalent circuit
Using the above found Vth and Rth values, the Thevenin’s equivalent circuit is
drawn as shown in Fig. 8.
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3.3333 Ω
A
33.33 V
B
Fig. 8
5. A series circuit with R = 10 Ω, L = 0.1 H and C = 50 &micro;F has an applied
voltage 50∠0◦ V with a variable frequency. Find (1) the resonant frequency, (2)
the value of frequency at which maximum voltage occurs across the inductor
(3) the value of frequency at which maximum voltage across capacitor and (4)
the quality factor.
Resonant frequency,
fr =
1
√
2π LC
1
2π 0.1 &times; 50 &times; 10−6
= 71.18 Hz.
=
√
The frequency at which maximum voltage occurs across the inductor is,
s
−1
R2 C
fL = (fr )
1−
2L
s
−1
102 &times; 50 &times; 10−6
= (71.18)
1−
2 &times; 0.1
= 72.09 Hz.
The frequency at which maximum voltage occurs across the capacitor is,
r
1
1
R2
fC =
−
2π r LC 2L
1
1
102
−
=
2π 0.1 &times; 50 &times; 10−6 2 &times; 0.1
= 71.09 Hz.
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r
1 L
Q=
R rC
0.1
1
=
= 4.47
10 50 &times; 10−6
Quality factor,
6. Obtain a conductively coupled equivalent circuit for the magnetically coupled
circuit shown below.
j6 Ω
j5 Ω
+
50∠0&deg; V
j10 Ω
3Ω
I2
I1
5Ω
–j4 Ω
–
Fig. 9
Applying KVL to mesh 1 in the given magnetically coupled circuit, we get
or
j5I1 − j6I2 + (3 − j4)(I1 − I2 ) = 50∠0◦
(3 + j1)I1 + (−3 − j2)I2 = 50∠0◦
(i)
Similarly, by applying KVL to mesh 2 in the given circuit, we get
or
j10I2 − j6I1 + 5I2 + (3 − j4)(I2 − I1 ) = 0
(−3 − j2)I1 + (8 + j6)I2 = 0
Equations (v) and (vi) are written in matrix form as,
(3 + j1) −(3 + j2)
I1
50∠0◦
=
−(3 + j2)
(8 + j6)
I2
0
(ii)
(iii)
The following points are noted carefully when constructing the conductively
coupled equivalent circuit.
If the two currents do not enter at dotted ends of coupled coils,
(a) Negative value of mutual reactance is connected in series with the individual elements of a particular mesh.
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(b) Positive value of mutual reactance is connected in series with the common
elements of two meshes.
Alternately, if the two currents enter at dotted ends of coupled coils,
(a) Positive value of mutual reactance is connected in series with the individual
elements of a particular mesh.
(b) Negative value of mutual reactance is connected in series with the common
elements of two meshes.
By above reasoning, the conductively coupled circuit is drawn as in Fig. 10.
50∠0&deg; V
j5 Ω
‒j6 Ω
j10 Ω
‒j6 Ω
j6 Ω
+
I1
‒j4 Ω
5Ω
I2
–
3Ω
Fig. 10
By inspection the following mesh equations are obtained.
(3 + j5 − j4 + j6 − j6)
−(3 − j4 + j6)
−(3 − j4 + j6) (8 + j10 − j4 + j6 − j6)
I1
I2
=
50∠0◦
0
(iv)
Simplifying, we obtain
(3 + j1) −(3 + j2)
−(3 + j2)
(8 + j6)
I1
I2
=
50∠0◦
0
(v)
Comparing Eqs. (vii) and (ix), it is observed that both are same.
7. In the circuit shown in Fig. 11, find the expression for i(t), vR (t) and vL (t) if
the switch is closed at t = 0.
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S
10 Ω
20 V
5H
i(t)
Fig. 11
After closing the switch S, application of KVL to the circuit gives
10i + 5
di
= 20
dt
Taking Laplace transformation on both sides
10I(s) + 5[sI(s) − I(0)] =
20
s
As there was no initial current through the inductor, I(0) = 0, then,
I(s)[10 + 5s] =
or
I(s) =
20
s
4
s(s + 2)
By applying partial fraction technique,
I(s) =
4
A
B
= +
s(s + 2)
s
s+2
(vi)
Solving for constants A and B,
A=
4
(s + 2)
= 2 and B =
s=0
4
s
= −2
s=−2
Substituting the values of A and B into Eq. (vi), we get
I(s) =
2
2
−
s s+2
Taking inverse Laplace transformation on both sides, the current equation is
obtained as:
i(t) = 2 − 2e−2t A
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Voltage across the resistor
VR = Ri(t)
= 10 &times; 2(1 − e−2t )
VR = 20(1 − e−2t ) V
Voltage across the inductor
di
dt
d
= 5 (2 − 2e−2t )
dt
VL = 20e−2t V
VL = L
8. In the circuit shown in Fig. 12, find the value of current i at t = 50 &micro;S if the
switch is closed at t = 0 and vC (t = 0) = 0.
S
10 Ω
50 V
i(t)
10 &micro;F
Fig. 12
After closing the switch S, application of KVL to the circuit gives
Ri + vC = V
Using, i = C
dv
, Eq. (i) is rewritten as
dt
RC
dvC
+ vC = V
dt
Substituting the corresponding R and C values
dvC
+ vC = 50
dt
dvC
0.0001
+ vC = 50
dt
10 &times; 10 &times; 10−6
or
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(i)
Applying Laplace transformation on both sides, we get
0.0001[svC (s) − vC (0)] + vC (s) =
50
s
Given that vC (t = 0) = 0. That is, there was no initial voltage across the
capacitor, So, vc (0) = 0. Substituting this in the above equation,
50
s
50
vC (s)[1 + 0.0001s] =
s
0.0001svC (s) + vC (s) =
50
s(1 + 0.0001s)
500000
vC (s) =
s(s + 10000)
vC (s) =
or
By applying partial fraction technique
vC (s) =
500000
A
B
= +
s(s + 10000)
s
s + 10000
(ii)
Solving for constants A and B,
A=
500000
(s + 10000)
= 50 and B =
s=0
500000
s
= −50
s=−10000
Substituting the values of A and B into Eq. (ii), we get
vC (s) =
50
50
−
s
s + 10000
Taking inverse Laplace transformation on both sides, the voltage across the
capacitor is obtained as
vC (t) = 50 − 50e−10000t V
From which, equation for current is obtained as
i(t) = C
dvC
dt
= (10 &times; 10−6 )
d
(50 − 50e−10000t )
dt
Hence, i(t) = 5e−10000t A. Then, at t = 50 &micro;S,
−6 )
i(t) = 5e−(10000&times;50&times;10
= 3.0327 A
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9. In the circuit shown in Fig. 13, the applied voltage is 150 sin (314t + φ) volts.
If the switch is closed when φ = 40◦ , determine the expression for i(t).
S
5Ω
0.5 H
100 &micro;F
i(t)
Fig. 13
Determination of initial conditions
(a) The switch is closed when φ = 40◦ . Therefore, at φ &lt; 40◦ , the current through the inductor was zero, i.e., i(0− ) = 0, which is also equal
to i(0+ ), since the current through the inductor cannot change instantaneously. Hence, we can write
i(0− ) = 0 = i(0+ )
(i)
(b) Initial voltage on the capacitor is zero, which results in
i(0− ) = 0 = i(0+ )
(ii)
Applying KVL after closing the switch (at φ = 40◦ )
Z
1
di
5i + 0.5 +
idt = 150 sin (314t + 40◦ )
−6
dt 100 &times; 10
= 150 [sin 314t cos 40◦ + cos 314t sin 40◦ ]
= 114.9 sin 314t + 96.5 cos 314t
Taking Laplace transform on both sides, we get
5I(s) + 0.5[sI(s) − I(0)] +
314
s
I(s)
1
= (114.9) 2
+ (96.5) 2
−6
2
100 &times; 10
s
s + 314
s + 3142
From Eq. (i), i(0) = 0, and in s-domain, I(0) = 0. Therefore,
I(s)
36078.6 + 96.5s
=
s
s2 + 3142
10000
36078.6 + 96.5s
I(s) 5 + 0.5s +
=
s
s2 + 3142
5I(s) + 0.5sI(s) + 10000
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36078.6 + 96.5s
5s + 0.5s2 + 10000
=
I(s)
s
s2 + 3142
36078.6 + 96.5s
s
I(s) =
s2 + 3142
5s + 0.5s2 + 10000
=
96.5s2 + 36078.6s
(0.5s2 + 5s + 10000)(s2 + 3142 )
=
193s2 + 72157.2s
(s2 + 10s + 20000)(s2 + 3142 )
By applying partial fraction technique
I(s) =
193s2 + 72157.2s
As + B
Cs + D
= 2
+ 2
2
2
2
(s + 10s + 20000)(s + 314 )
s + 10s + 20000 s + 3142
(iii)
Expanding
193s2 + 72157.2s = (As + B)(s2 + 3142 ) + (Cs + D)(s2 + 10s + 20000)
= As3 + 3142 As + Bs2 + 3142 B + Cs3 + 10Cs2 + 20000Cs
+ Ds2 + 10Ds + 20000D
= (A + C)s3 + (B + 10C + D)s2 + (3142 A + 20000C + 10D)s
+ (3142 B + 20000D)
Comparing the coefficients of s3
0=A+C
(iv)
193 = B + 10C + D
(v)
Comparing the coefficients of s2
Comparing the coefficients of s
72157.2 = 3142 A + 20000C + 10D
(vi)
Comparing the constant terms
0 = 3142 B + 20000D
(vii)
Solving Eqs. (iv) to (vii), we obtain
A = 0.8859
B = −51.3669
C = −0.8859
D = 253.2247
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Substituting the above found constants into Eqn. (iii), we get
0.8859s − 51.3669 −0.8859s + 253.2247
+
s2 + 10s + 20000
s2 + 3142
0.8859s − 51.3669 −0.8859s + 253.2247
+
=
(s + 5)2 + 19975
s2 + 3142
I(s) =
=
0.8859s − 51.3669 −0.8859s + 253.2247
+
(s + 5)2 + 141.332
s2 + 3142
0.8859s
51.3669
0.8859s
253.2247
−
− 2
+ 2
2
2
2
2
2
(s + 5) + 141.33
(s + 5) + 141.33
s + 314
s + 3142
141.33
0.8859s
51.3669
=
−
(s + 5)2 + 141.332
141.33 (s + 5)2 + 141.332
0.8859s
253.2247
314
− 2
+
2
2
s + 314
314
s + 3142
=
Taking inverse Laplace transformation
i(t) = 0.8859 e−5t cos 141.33t − 0.3635 e−5t sin 141.33t − 0.8859 cos 314t + 0.8064 sin 314t
i(t) = [e−5t (0.8859 cos 141.33t − 0.3635 sin 141.33t) − 0.8859 cos 314t + 0.8064 sin 314t] A
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80365
B.E/B.Tech. DEGREE EXAMINATION, NOV./DEC. 2016
Electrical and Electronics Engineering
EE 6201 – CIRCUIT THEORY
1. Calculate the node voltages of given circuit in Fig. 1.
5A
4Ω
2Ω
6Ω
10 A
Fig. 1
Applying KCL at node 1, we get
V1 V1 − V2
+
−5=0
2
4
or
0.75V1 − 0.25V2 = 5
(i)
Applying KCL at node 2, we get
V2 − V1 V2
+
+ 5 − 10 = 0
4
6
or
−0.25V1 + 0.4167V2 = 5
Solving Eqs. (i) and (ii), we get
V1 = 13.33 V
V2 = 20 V
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(ii)
2. Determine the current I0 for the given circuit in Fig. 2, when Vs = 12 V.
2Ω
8Ω
+ Vx
I0
4Ω
4Ω
i2
i1
6Ω
+
Vs
3 Vx
+
Fig. 2
Applying KVL in mesh 1,
6i1 + 2i1 + 4(i1 − i2 ) + Vs = 0
Given that, Vs = 12. Therefore, the above equation becomes
12i1 − 4i2 = −12
(i)
Applying KVL in mesh 2,
−Vs − 4(i1 − i2 ) + 8i2 + 4i2 − 3Vx = 0
Using Vs = 12 and Vx = 2i1 , the above equation becomes
−10i1 + 16i2 = 12
(ii)
Solving Eqs. (i) and (ii), we get
i1 = −0.9474 A
i2 = 0.1579 A
Hence, I0 = i2 = 0.1579 A.
3. Using mesh analysis for the given circuit in Fig. 3, find the current I2 and
voltage drop across 1 Ω resistor.
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XL1 = 2 Ω
R1
XC
8Ω
R2
4Ω
I2
1Ω
XL2
+
6Ω
–
E1 = 8 V∠20&deg;
E2 = 10 V∠0&deg;
–
+
Fig. 3
Assuming the current delivered by the voltage source E1 is I1 , and by applying
KVL to the two meshes, we obtain
or
and
or
I1 + j2I1 − j8(I1 − I2 ) + 4(I1 − I2 ) − 10∠0◦ − 8∠20◦ = 0
(5 − j6)I1 + (−4 + j8)I2 = 17.73∠8.88◦ (i)
j6I2 + 10∠0◦ − 4(I1 − I2 ) + j8(I1 − I2 ) = 0
(−4 + j8)I1 + (4 − j2)I2 = −10
(ii)
Solving Eqs. (i) and (ii), we get
I1 = (0.9 + j0.5) A
I2 = (−0.67 + j3.74) A
So, the voltage drop across 1 Ω resistor is
I1 &times; 1 = (0.9 + j0.5) V
4. Find the equivalent capacitance Ceq between terminals a and b of Fig. 4.
5 &micro;F
60 &micro;F
a
Ceq
20 &micro;F
6 &micro;F
20 &micro;F
b
Fig. 4
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Capacitors 20 &micro;F and 5 &micro;F are in series. Therefore, the equivalent capacitance
obtained from them is
20 &times; 5
=4
20 + 5
Now the capacitors 4 &micro;F, 6 &micro;F and 20 &micro;F are in parallel. Therefore, the
equivalent capacitance is
4 + 6 + 20 = 30
Finally, 30 &micro;F and 60 &micro;F are in series. Therefore, the equivalent capacitance
between the terminals a and b is
30 &times; 60
= 20 &micro;F
30 + 60
5. Obtain the equivalent resistance Rab of the circuit given in Fig. 5 and calculate
the total current i.
i
a
13 Ω
24 Ω
10 Ω
20 Ω
100 V
30 Ω
50 Ω
b
Fig. 5
The star connected network comprising 24 Ω, 20 Ω and 30 Ω resistors are
converted into equivalent delta connected network as shown below.
c
c
24 Ω
20 Ω
n
d
⇒
60 Ω
90 Ω
30 Ω
e
e
d
75 Ω
Fig. 6
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Rcd =
(24 &times; 40) + (20 &times; 30) + (30 &times; 24)
= 60 Ω
30
Rde =
(24 &times; 40) + (20 &times; 30) + (30 &times; 24)
= 75 Ω
24
(24 &times; 40) + (20 &times; 30) + (30 &times; 24)
= 90 Ω
20
Replacing the cde star into delta, the circuit is redrawn as shown in Fig. 7.
Rec =
i
13 Ω
a
60
Ω
10 Ω
90 Ω
Ω
100 V
75
50 Ω
b
Fig. 7
From Fig. 7,
Rab = 13 + 90||
60 &times; 10
60 + 10
+
75 &times; 50
75 + 50
= 40 Ω
100
= 2.5 A.
40
6. Find the value of RL in Fig. 8 for maximum power to RL and calculate the
maximum power.
Therefore, the total current, i =
I
6A
R2
R1
E1
3Ω
+ 68 V ‒
RL
10 Ω
R3
2Ω
Fig. 8
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According to maximum power transfer theorem, maximum power is transferred
to RL , when
RS = RL
R1
3Ω
R2
⇐ RL
10 Ω
R3
2Ω
Fig. 9
From Fig. 9, RS is obtained as 3 + 10 + 2 = 15 Ω.
R2
R1
E1
3Ω
+ 68 V ‒
10 Ω
RL = 15 Ω
60 V
R3
2Ω
Fig. 10
By source transformation technique, the 6 A current source of given circuit in
Fig. 8 is converted into an equivalent voltage source as shown in Fig. 10 and
the current through RL is obtained as
IL =
128
= 4.2667 A
30
Therefore, the maximum power that is transferred to the load resistance RL ,
Pmax = 4.26672 &times; 15 = 273 W
7. Apply superposition theorem to determine the current i through 3 Ω resistor
for the given circuit in Fig. 11.
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24 V
8Ω
+
4Ω
4Ω
i
+
12 V
3Ω
3A
Fig. 11
Response due to 12 V source only
The current through 3 Ω (I 0 ) due to 12 V source only is obtained by keeping
only that source and killing other sources as shown in Fig. 12 (a).
24 V
8Ω
8Ω
+
I2
I4
4Ω
4Ω
4Ω
4Ω
I′
12 V
+
I1
I′′
I3
3Ω
3Ω
(a)
(b)
Fig. 12
Referring to Fig. 12 (a), the mesh equations are written as
7 −4
I1
12
=
−4 16
I2
0
Solving, I 0 (I1 ) is obtained as 2 A.
Response due to 24 V source only
The current through 3 Ω (I 00 ) due to 24 V source only is obtained by keeping
only the 24 V source in the circuit and killing other two sources as shown in
Fig. 12 (a).
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With reference to Fig. 12 (b), the mesh equations are written as
16 −4
I3
−24
=
−4
7
I4
0
Solving, I 00 (I3 ) is obtained as −1 A.
Response due to 3 A source only
The current through 3 Ω (I 0 ) due to 3 A source only is obtained by keeping
only that source and killing other sources as shown in Fig. 13.
8Ω
4Ω
4Ω
I′′′
3Ω
3A
Reference node
Fig. 13
Referring to Fig. 13, the nodal equations are written as follows:
Applying KCL at node 1, we obtain
V1 − V2 V1
+
−3=0
4
8
or
0.375V1 − 0.25V2 = 3
(i)
Similarly, applying KCL at node 2, we obtain
V2 − V1 V2 V2
+
+
=0
4
4
3
or
−0.25V1 + 0.83V2 = 0
(ii)
Solving Eqs. (i) and (ii), we get V2 = 3 V. Therefore, I 000 is obtained as
V2
3
= =1
3
3
Thus, by the use of superposition principle i = I 0 + I 00 + I 000 = 2 − 1 + 1 = 2 A.
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8. For the series resonant of Fig. 14, find I, VR , VL and VC at resonance. Also,
if resonant frequency is 5000 Hz, determine bandwidth, Q factor, half power
frequencies, and power dissipated in the circuit at resonance and at the half
power frequencies. Derive the expression for resonant frequency.
I
+ VR –
+ VL –
R=2Ω
XL = 10 Ω
+
XC = 10 Ω
E = 10 V∠0&deg;
–
+
VC
–
Fig. 14
At resonant condition, the sum of reactances is zero, therefore,
Current at resonance, I =
V
10
=
=5A
R
2
Voltage drop across resistor, VR = IR = 5 &times; 2 = 10 V
Voltage drop across inductor, VL = IXL = 5 &times; 10 = 50 V
Voltage drop across capacitor, VC = IXC = 5 &times; 10 = 50 V
10
XL
=
=5
R
2
Resonant frequency
5000
Bandwidth, β =
=
= 1000 Hz
Q-factor
5
Q-factor, Q =
Lower half power frequency, f1 = fr −
β
1000
= 5000 −
= 4500 Hz
2
2
Upper half power frequency, f2 = fr +
β
1000
= 5000 +
= 5500 Hz
2
2
Power at resonant frequency = I 2 R = 52 &times; 2 = 50 W
Power at half power frequencies = (0.707I)2 R = (0.707 &times; 5)2 &times; 2 = 25 W
At resonance,
XL = X C
1
or
2πfr L =
2πfr C
1
.
Hence, resonant frequency,
fr = √
2π LC
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9. Obtain the conductively coupled equivalent circuit for the given circuit in Fig.
15 and find the voltage drop across 12 Ω resistor.
j3 Ω
‒j4 Ω
+
I1
12 ∠0&deg; V
j5 Ω
j6 Ω
I2
12 Ω
–
Fig. 15
For the given circuit the mesh current I1 enters the dot whereas I2 leaves the
dot. So, using the dot rule the conductively coupled circuit is drawn as shown
in Fig. 16.
‒j4 Ω ‒(j3) Ω
‒(j3) Ω
j3 Ω
+
12∠0&deg; V
I2
I1
–
12 Ω
j5 Ω
Fig. 16
By applying mesh analysis, the mesh equations are written as
j1
−j8
I1
12∠0◦
=
−j8 12 + j5
I2
0
Using Cramer’s rule
j1 12∠0◦
−j8
0
(j8)(12∠0◦ )
∆I2
=
=
I2 =
∆
(j1)(12 + j5) − (−j8)2
j1
−j8
−j8 12 + j5
= 1.5945∠78.5◦ A
Therefore, the voltage drop across 12 Ω resistor
= 12 &times; 1.5945∠78.5◦ = 19.134∠78.5◦ V
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10. The number of turns in two coupled coils are 500 turns and 1500 turns respectively. When 5 A current flows in coil 1, the total flux in this coil is
0.6 &times; 10−3 Wb and the flux linking in second coil is 0.3 &times; 10−3 Wb. Determine L1 , L2 , M and K.
Given that N1 = 500; N2 = 1500; I1 = 5 A; φ1 = 0.6&times;10−3 and φ12 = 0.3&times;10−3
K=
φ12
0.3
=
=5
φ1
0.6
N1 φ1
500 &times; 0.6 &times; 10−3
=
= 0.06 H
I1
5
r
L1
N1
=
N2
L2
L1 =
From which,
L2 = L1
0.06
2 = 500 2 = 0.54 H
N1
N2
M =K
1500
p
p
L1 L2 = 0.5 (0.06 &times; 0.54) = 0.09 H
11. A series RL circuit with R = 50 Ω and L = 30 H has a constant voltage V =
50 volts applied at t = 0 as shown in Fig. 17. Determine the current i, voltage
across inductor. Derive the necessary expression and plot the respective curves.
2Ω
R
t=0
i(t)
+
VS
+
L
v(t)
–
Fig. 17
Applying KVL to the circuit in Fig. 17, after closing the switch
di
=V
dt
di
L = V − Ri
dt
di
V − Ri
=
dt
L
Ri + L
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Separating the variables,
di
dt
=
V − Ri
L
Multiplying by ‘−R’ on both sides, we get
−R
−R
di =
dt
V − Ri
L
Integrating both sides
Z −R
V − Ri
Z di =
ln(V − Ri) =
−R
L
dt
−R
t+K
L
(i)
where K is a constant of integration and is evaluated by applying initial conditions. The initial condition is that at t = 0, i = 0.
Substituting this into Eq. (i), we get
ln V = K
Therefore, Eq. (i) becomes
−R
t + ln V
L
−R
t
ln(V − Ri) − ln V =
L
V − Ri
−R
ln
t
=
V
L
ln(V − Ri) =
Taking antilog on both sides
R
V − Ri
= e− L t
V
R
Ri
1−
= e− L t
V
R
V i=
1 − e− L t
R
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(ii)
Thus using Eq. (ii),
V −R
t
L
i=
1−e
R
50
50 1 − e− 30 t
=
20
= 2.5(1 − e−1.6667t ) A
Voltage across inductor
di(t)
dt
d
= 30 [2.5 − 2.5e−1.6667t ]
dt
VL = L
= 30[0 − (2.5)(−1.6667)e−1.6667t ]
= 125e−1.6667t V
12. Determine the impedance parameter (z ) of the given two port network in Fig.
18.
4I1
I1
5Ω
10 Ω
+
V1
I2
+
+
V2
20 Ω
–
–
Fig. 18
The defining equations of z -parameters are:
z11 I1 + z12 I2 = V1
z21 I1 + z22 I2 = V2
(i)
(ii)
Applying KVL to the mesh of input-port side,
or
5I1 + 20(I1 + I2 ) = V1
25I1 + 20I2 = V1
(iii)
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Similarly, by applying KVL to the mesh of output-port side,
4I1 + 10I2 + 20(I1 + I2 ) = V2
24I1 + 30I2 = V2
or
(iv)
Upon comparing Eqs. (iii) and (iv) with Eqs. (i) and (ii) respectively, we get
the z -parameters as
z11 z12
25 20
[z] =
=
z21 z22
24 30
13. Find the hybrid parameter (h) of the two port network in Fig. 19.
2Ω
3Ω
6Ω
Fig. 19
The defining equations of h-parameters are:
V1 = h11 I1 + h12 V2
I2 = h21 I1 + h22 V2
(i)
(ii)
The currents and voltages for the respective ports are labelled as shown in Fig.
Fig. 20.
I1
2Ω
3Ω
I2
I1 + I2
V1
6Ω
V2
–
Fig. 20
Applying KVL to the mesh of input-port side,
or
2I1 + 6(I1 + I2 ) = V1
8I1 + 6I2 = V1
(iii)
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Similarly, by applying KVL to the mesh of output-port side,
or
3I2 + 6(I1 + I2 ) = V2
6I1 + 9I2 = V2
(iv)
V2 − 6I1
9
I2 = −0.6666I1 + 0.1111V2
(v)
From Eq. (iv),
I2 =
or
Putting the value of I2 into Eq. (iii), we obtain
V1 = 8I1 + 6 (0.1111V2 − 0.6667I1 )
= 8I1 + 0.6666V2 − 4I1
V1 = 12I1 + 0.6666V2
(vi)
Upon comparing Eqs. (vi) and (v) with Eqs. (i) and (ii) respectively, we get
the h-parameters as
12
0.6666
h11 h12
=
[h] =
h21 h22
−0.6666 0.1111
14. For the ∆ − ∆ system shown in Fig. 21, find the phase angles θ2 and θ3 for the
specified phase sequence. Also find the phase current and line current in each
IAa
A
a
Iab
5Ω
EAB = 120 V ∠0&deg;
3-phase
∆-connected ac
ECA = 120 V ∠θ3
generator
Ica
Phase sequence: ACB
c
IBb
B
5Ω
5Ω
5Ω
5Ω
5Ω
Ibc
b
ICc
C
EBC = 120 V ∠θ2
Fig. 21
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The phasor diagram of the line voltages for the given phase sequence is shown
in Fig. 22.
ECB
EBA
θ3
θ2
EAC
Fig. 22
It is seen from Fig. 22 that, for the given ACB phase sequence, θ2 = 240◦ and
θ3 = 120◦ . Thus, the line voltages are as follows:
EBA = 120∠0◦ V
EAC = 120∠120◦ V
ECB = 120∠240◦ V
Impedance in each branch,
Zph =
(5)(−j5)
= (2.5 − j2.5) Ω
(5 − j5)
We recall that in ∆-connected networks, line voltage and phase voltage are
same. Therefore, the phase currents are:
Iab =
120∠0◦
EBA
=
= (24 + j24) = 33.94∠45◦ A
Zph
(2.5 − j2.5)
Ibc =
ECB
120∠240◦
=
= (8.7846 − j32.7846) = 33.94∠ − 75◦ A
Zph
(2.5 − j2.5)
Ica =
EAC
120∠120◦
=
= (−32.7846 + j8.7846) = 33.94∠165◦ A
Zph
(2.5 − j2.5)
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Applying KCL at node ‘a’, we obtain
Line current, IAa = Iab − Ica
= (24 + j24) − (−32.7846 + j8.7846)
= 56.7846 + j15.2154 = 58.7877∠15◦ A
Similarly, other two line currents are obtained as
Line current, IBb = Ibc − Iab
= (8.7846 − j32.7846) − (24 + j24)
= −15.2154 − j56.7846 = 58.7877∠ − 105◦ A
Line current, ICc = Ica − Ibc
= (−32.7846 + j8.7846) − (8.7846 − j32.7846)
= −41.5692 + j41.5692 = 58.7877∠135◦ A
15. A 3-phase, 400 V supply is given to balanced star connected load of impedance
(8+j6) Ω in each branch. Determine line current, power factor and total power.
√
Given that load impedance, Z = 8 + j6 = 82 + 62 = 10 Ω
Phase voltage, Vph =
Line voltage
√
3
400
= √ = 230.94 V
3
Vph
Z
230.94
= 23.094 A
=
10
Phase current, Iph =
In star-connected networks, magnitude of line current and phase current are
same. Therefore, line current
IL = 23.094 A
Power factor,
cos φ =
R
8
=
= 0.8
Z
10
The total power,
√
√
P = 3VL IL cos φ = 3 &times; 400 &times; 23.094 &times; 0.8 = 12.8 kW
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16. The two wattmeter method produces wattmeter readings P1 = 1560 W and
P2 = 2100 W when connected to a delta connected load. If the line voltage
is 220 V, calculate (i) the per phase average power, (ii) total reactive power,
(iii) power factor and (iv) the phase impedance. Is the impedance inductive or
capacitive? Justify.
By two wattmeter method of measuring power,
Total real power, P = P1 + P2 = 1560 + 2100 = 3660 W
∴ Per phase average power = 3660
= 1220 W
3
√
√
Total reactive power, Q = 3(P2 − P1 ) = 3(2100 − 1560) = 935.31 var
√ P2 − P1
−1
Power factor, cos φ = cos tan
3
P1 + P2
√ 2100 − 1560
−1
= cos tan
3
1560 + 2100
= 0.97
Line current, IL = √
P
3660
=√
= 9.9 A
3VL cos φ
3 &times; 220 &times; 0.97
So,
9.9
IL
phase current, Iph = √ = √ = 5.72 A
3
3
Then,
phase impedance, Zph =
Vph
220
=
= 38.44 Ω
Iph
5.72
and in complex form,
Zph = 38.44∠14.33◦ Ω
Since the total reactive power Q is found to be positive, meaning there is an
absorption of reactive power by the load. Hence, the impedance is inductive.
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57307
B.E/B.Tech. DEGREE EXAMINATION, MAY/JUNE 2016
Electrical and Electronics Engineering
EE 6201 – CIRCUIT THEORY
1. Determine the current IL in the circuit shown below.
4V
3Ω
3Ω
3Ω
IL
8V
6V
5Ω
1Ω
1Ω
Fig. 1
The mesh currents are assumed in the directions as indicated in Fig. 2.
4V
3Ω
I3
3Ω
3Ω
IL
I1
8V
5Ω
1Ω
6V
I2
1Ω
Fig. 2
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By mesh analysis method, the mesh equations are written as


  
9 5 −3
I1
8
 5 9




3
I2 = 6 
−3 3
9
I3
4
Using Cramer’s rule, I1 is found as, I1 =
∆I1 =
∆I1
.
∆
8 5 −3
6 9
3
4 3
9
= 8(81 − 9) − 5(54 − 12) − 3(18 − 36) = 420
∆=
9 5 −3
5 9
3
−3 3
9
= 9(81 − 9) − 5(45 + 9) − 3(15 + 27) = 252
∆I1
420
Therefore, I1 =
=
= 1.6667
∆
252
Similarly, I2 is found as, I2 =
∆I2 =
∆I2
.
∆
9 8 −3
5 6
3
−3 4
9
= 9(54 − 12) − 5(54 − 12) − 3(20 + 18) = −168
Therefore,
I2 =
∆I2
−168
=
= −0.6667
∆
252
Hence, the current IL = I1 + I2 = (1.6667 − 0.6667) = 1 A.
2. Calculate the voltage across A and B in the circuit shown in Fig. 3?
12 V
6Ω
12
V
A
6V
4Ω
4Ω
10 Ω
B
Fig. 3
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6
= 0.6 A
10
= 0.6 &times; 4
= 2.4 V (A being negative)
12
I2 =
= 0.86 A
14
= 0.86 &times; 4
= 3.44 V (B being positive)
The current through 6 Ω and 4 Ω circuit,
I1 =
Therefore, voltage across 4 Ω resistor
The current through 4 Ω and 10 Ω circuit,
Therefore, voltage across 4 Ω resistor
Replacing the resistances connected between A and B with their respective
voltage drops, the given circuit is redrawn as shown in Fig. 4.
12 V
A
12
V
6Ω
6V
2.4 V
3.44 V
10 Ω
B
Fig. 4
From Fig. 4 it is found that,
VAB = −2.4 + 12 + 3.44 = 13.04 V
3. Three loads A, B and C are connected in parallel to a 240 V source. Load
A takes 9.6 kW, Load B takes 60 A and Load C has a resistance of 4.8 Ω.
Calculate (1) RA and RB (2) the total current (3) the total power, and (4)
equivalent resistance.
(1) Resistances RA and RB
IA =
P
V
9.6 &times; 103
= 40 A
240
V
RA =
IA
=
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240
=6Ω
40
V
RB =
IB
=
=
240
=4Ω
60
(2) Total current
IC =
=
V
RC
240
= 50 A
4.8
Therefore, the total current, IT = IA + IB + IC
= 40 + 60 + 50 = 150 A
(3) Total power
Power, P = V IT = 240 &times; 150 = 36000 W
(4) Equivalent resistance
1
1 1
1
= + +
Req
6 4 4.8
= 0.625 Ω
1
Therefore, the equivalent resistance, Req =
= 1.6 Ω
0.625
4. For the circuit shown in figure below, determine the total current IT , phase
angle and power factor.
100 &micro;F
10 Ω
IT
50 V,
100 Hz
30 Ω
0.1 H
Fig. 5
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Capacitive reactance of 100 &micro;F,
XC =
1
2πf C
1
= 15.92 Ω
2π(100)(100 &times; 10−6 )
Inductive reactance of 0.1 H,
XL = 2πf L
= 2π(100)(0.1) = 62.83 Ω
Equivalent impedance of the circuit
= (10 − j15.92) + (30)||(j62.83)
= (10 − j15.92) + (24.43 + j11.66)
= (34.43 − j4.26) Ω
50
Total current,
IT =
34.43 − j4.26
= (1.43 + j0.17) = 1.44∠7.05◦
So, phase angle,
φ = 7.05◦
and power factor,
cos φ = cos 7.05◦ = 0.99.
=
5. Find the current in 4 Ω resistor in the circuit shown in Fig. 6.
3Ω
57 V
5Ω
4Ω
7Ω
6Ω
42 V
4V
25 V
70 V
Fig. 6
Mesh currents are assumed with their direction as shown in Fig. 7.
3Ω
57 V
5Ω
4Ω
42 V
I1
6Ω
I2
25 V
7Ω
I3
4V
70 V
Fig. 7
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By mesh analysis

7
 −4
0
method, the mesh equations are written as

 

−4
0
I1
42 + 25 = 67
15 −6   I2  =  −25 − 57 − 70 = −152 
−6 13
I3
70 + 4 = 74
Solving the above equations, we get
I1 = 5 A
I2 = −8 A
I3 = 2 A
Therefore, current through 4 Ω resistor, I1 − I2 = 5 − (−8) = 13 A
6. Obtain the Norton’s model and find the maximum power that can be transferred
to the 100 Ω ohm load resistance in the circuit shown in Fig. 8.
220 Ω
10 V
RL = 100 Ω
470 Ω
200 Ω
380 Ω
5V
Fig. 8
To find Norton’s surrent, IN
To obtain the Norton’s model the circuit in Fig. 8 is first redrawn by shortcircuiting the 100 Ω resistance as shown in Fig. 9 and finding the Norton’s
current, IN . Mesh analysis is used and the directions of mesh currents are as
indicated in Fig. 9.
220 Ω
10 V
I1 470 Ω
A
B
IN
I2
380 Ω
200 Ω
I3
5V
Fig. 9
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By inspection the mesh equations are written as


 

690 −470
0
I1
10
 −470
850 −380   I2  =  0 
0 −380
580
I3
−5
Solving the above equations, we get
I2 = IN = 12.5876 mA
To find Thevenin’s resistance, Rth
Rth
220 Ω
470 Ω
200 Ω
B
⇒
A
380 Ω
Fig. 10
As seen in Fig. 10, Rth appearing across the terminals A and B is
220 &times; 470
200 &times; 380
Rth =
+
220 + 470
200 + 380
= 280.88 Ω
Using the values of IN and Rth found above the Norton’s model is drawn as
shown in Fig. 11.
12.5876 mA
280.88 Ω
A
RL = 100 Ω
B
Fig. 11
By current division rule, current through the load resistance
= 12.5876 &times; 10−3 &times;
280.88
= 9.2827 mA.
280.88 + 100
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Therefore, the maximum power that can be transferred to 100 Ω resistance
= (9.2827 &times; 10−3 )2 &times; 100 = 8.6169 mW.
7. Determine the resonant frequency, bandwidth and quality factor of the coil for
the series resonant circuit consisting of R = 10 Ω, L = 0.1 H and C = 10 &micro; F.
Derive the formula used for bandwidth.
1
Resonant frequency,
fr = √
2π LC
1
2π 0.1 &times; 10 &times; 10−6
= 159.15 Hz.
r
1 L
Q=
R rC
1
0.1
=
= 10
10 10 &times; 10−6
fr
BW =
Q
=
Quality factor,
Bandwidth,
=
√
159.15
= 15.915
10
[For derivation of the formula for bandwidth, please refer Section 10.1.5 in the book mentioned at footer ]
8. Write the mesh equations and obtain the conductively coupled equivalent circuit
for the magnetically coupled circuit shown in Fig. 12.
j6 Ω
j5 Ω
+
50∠0&deg; V
j10 Ω
3Ω
I2
I1
5Ω
–j4 Ω
–
Fig. 12
Applying KVL to mesh 1 in the given magnetically coupled circuit, we get
or
j5I1 − j6I2 + (3 − j4)(I1 − I2 ) = 50∠0◦
(3 + j1)I1 + (−3 − j2)I2 = 50∠0◦
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(i)
Similarly, by applying KVL to mesh 2 in the given circuit, we get
j10I2 − j6I1 + 5I2 + (3 − j4)(I2 − I1 ) = 0
(−3 − j2)I1 + (8 + j6)I2 = 0
or
Equations (i) and (ii) are written in matrix form as
(3 + j1) −(3 + j2)
I1
50∠0◦
=
−(3 + j2)
(8 + j6)
I2
0
(ii)
(iii)
The following points are noted carefully when constructing the conductively
coupled equivalent circuit.
If the two currents do not enter at dotted ends of coupled coils,
(i) Negative value of mutual reactance is connected in series with the individual elements of a particular mesh.
(ii) Positive value of mutual reactance is connected in series with the common
elements of two meshes.
Alternately, if the two currents enter at dotted ends of coupled coils,
(i) Positive value of mutual reactance is connected in series with the individual elements of a particular mesh.
(ii) Negative value of mutual reactance is connected in series with the common
elements of two meshes.
By the above reasoning, the conductively coupled circuit is drawn as shown in
Fig. 13.
50∠0&deg; V
j5 Ω
‒j6 Ω
j10 Ω
‒j6 Ω
j6 Ω
+
I1
‒j4 Ω
I2
5Ω
–
3Ω
Fig. 13
By inspection the following mesh equations are obtained.
(3 + j5 − j4 + j6 − j6)
−(3 − j4 + j6)
I1
50∠0◦
=
−(3 − j4 + j6)
(8 + j10 − j4 + j6 − j6)
I2
0
(iv)
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Simplifying, we obtain
(3 + j1) −(3 + j2)
I1
50∠0◦
=
−(3 + j2)
(8 + j6)
I2
0
(v)
Comparing Eqs. (iii) and (v), it is observed that both are same.
9. A sinusoidal voltage of 10 sin 100t is connected in series with a switch and
R = 10 Ω and L = 0.1 H. If the switch is closed at t = 0 determine the transient
current i(t).
Applying KVL to the circuit after closing the switch, we get
di
= 10 sin 100t
dt
di
100i +
= 100 sin 100t
dt
Taking Laplace transformation on both sides,
10i + 0.1
100
+ 1002
10000
I(s)(100 + s) =
s2 + 1002
10000
1
I(s) =
s + 100
s2 + 1002
100I(s) + sI(s) = 100
I(s) =
s2
10000
A
Bs + C
=
+ 2
2
2
(s + 100)(s + 100 )
(s + 100) (s + 1002 )
(vi)
Solving for the constant,
A=
10000
+ 1002
s2
= 0.5
s=−100
On cross-multiplying Eq. (vi), we get
10000 = A(s2 + 1002 ) + (Bs + C)(s + 100)
= As2 + 10000A + Bs2 + 100Bs + Cs + 100C
= s2 (A + B) + s(100B + C) + (10000A + 100C)
(vii)
On equating the coefficients of s2 of Eq. (vii), we get
or
A+B=0
B = −A = −0.5
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On equating the coefficients of s of Eq. (vii), we get
100B + C = 0
C = −100B = 50
Hence,
Substituting the values of A, B and C into Eq. (vi), we get
0.5
−0.5s + 50
+ 2
(s + 100) (s + 1002 )
0.5
0.5s
50
=
− 2
+ 2
2
(s + 100) (s + 100 ) (s + 1002 )
0.5
0.5s
50
100
+
=
− 2
2
2
(s + 100) (s + 100 )
100 (s + 1002 )
I(s) =
Taking inverse Laplace transformation on both sides, we get
i(t) = 0.5e−100t − 0.5 cos 100t + 0.5 sin 100t
(viii)
0.
5
=
0.5
+
2
0.
5
2
0.
70
7
A right-angled triangle is constructed using 0.5 and 0.5 as two sides as
shown in Fig. 14.
φ
0.5
Fig. 14
Referring to Fig. 14,
−1
φ = tan
Also,
or
0.5
0.5
= 45◦
cos φ =
0.5
0.707
0.5 = 0.707 cos 45◦
Similarly,
0.5 = 0.707 sin 45◦
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Therefore, Eq. (viii) becomes,
i(t) = 0.5e−100t − 0.707 sin 45◦ cos 100t + 0.707 cos 45◦ sin 100t
= 0.5e−100t + 0.707(cos 45◦ sin 100t − sin 45◦ cos 100t)
i(t) = 0.5e−100t + 0.707 sin(100t − 45◦ ) A
10. In the circuit shown below, find the transient current after switch is closed at
time t = 0, given that an initial charge of 100 &micro;C is stored in the capacitor.
t=0
15 Ω
50 V
200 &micro;F
i(t)
Fig. 15
Initial voltage on the capacitor,
V =
Q
100 &times; 10−6
=
= 0.5 V
C
200 &times; 10−6
Applying KVL after closing the switch,
Ri + vc = V
dvc
+ vc = V
dt
dvc
15 &times; 200 &times; 10−6
+ vc = 50
dt
dvc
0.003
+ vc = 50
dt
RC
Taking Laplace transformation on both sides, we get
0.003[sVC (s) − VC (0)] + VC (s) =
50
s
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At t = 0, VC (0) = 0.5 V. Therefore, the above equation becomes
50
s
50
+ 0.0015
VC (s)[1 + 0.003s] =
s
0.003[sVC (s) − 0.5] + VC (s) =
( 50
) + 0.0015
VC (s) = s
(1 + 0.003s)
=
50 + 0.0015s
s(1 + 0.003s)
=
16666.67 + 0.5s
50 + 0.0015s
=
0.003s(s + 333.33)
s(s + 333.33)
By applying partial fraction technique,
I(s) =
16666.67 + 0.5s
A
B
= +
s(s + 333.33)
s
s + 333.33
(ix)
Solving for constants A and B,
A=
16666.67 + 0.5s
(s + 333.33)
= 50 and B =
s=0
16666.67 + 0.5s
s
= −49.5
s=−333.33
Substituting the values of A and B into Eq. (vii), we get
VC (s) =
50
49.5
−
s
s + 333.33
Taking inverse Laplace transformation on both sides, the voltage across the
capacitor is obtained as
vC (t) = 50 − 49.5e−333.33t V
The transient current is given by
i(t) = C
dvC
dt
d
(50 − 49.5e−333.33t )
dt
A
= (200 &times; 10−6 )
Hence,
i(t) = 3.3e−333.33t
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11. Obtain the readings of two wattmeters connected to a three phase, 3 wire,
120 V system feeding a balanced ∆ connected load with a load impedance of
12∠30◦ Ω. Assume RYB phase sequence. Determine the phase power and
compare the total power to the sum of wattmeter readings.
Phase current,
IRY =
VRY
ZRY
120∠0◦
=
12∠30◦
= 10∠ − 30◦ A
Similarly, the other phase currents
and
Power factor,
IYB = 10∠ − 150◦ A
IBR = 10∠ − 270◦ A
cos φ = cos 30◦ = 0.866
Line current (IL ) in R-line is calculated by using KCL as follows:
So, total power
Also, total power
Per phase power
IL = IRY − IBR
= (10∠ − 30◦ ) − (10∠ − 270◦ )
= 17.32∠ − 60◦
√
P = 3VL IL cos φ
√
= 3 &times; 120 &times; 17.32 &times; 0.866 = 3117.6 W
P = W1 + W2
W1 = VL IL cos (30◦ + φ)
= 120 &times; 17.32 &times; cos (30◦ + 30◦ )
W1 = 1039.2 W
(x)
W2 = VL IL cos (30◦ − φ)
= 120 &times; 17.32 &times; cos (30◦ − 30◦ )
W2 = 2078.4 W
(xi)
= Vph Iph cos φ
= 120 &times; 10 &times; 0.866
= 1039.2 W
(xii)
Upon comparing Eqs. (x) and (xii), it is observed that when the power factor is
0.866, one of the wattmeters reads per phase power, i.e., one-third of the total
power, while the other wattmeter reads two-third of the total power.
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12. If W1 and W2 are the readings of the two wattmeters which measures power in
W1
the three phase balanced system and if
= a, show that the power factor of
W2
a+1
the circuit is given by cos φ = √
.
2 a2 − a + 1
By employing two wattmeter method of measuring three phase power, we have
√ W2 − W1
tan φ = 3
W1 + W2
W1
√ W2 1 − W2
= 3
W1
W2 1 +
W2
=
√ (1 − a)
3
(1 + a)
Squaring both sides,
(1 − a)2
(1 + a)2
(1 − a)2
φ = 1 + (3)
(1 + a)2
(1 + a)2 + 3(1 − a)2
φ=
(1 + a)2
(1 + a2 + 2a) + 3(1 + a2 − 2a)
=
φ
(1 + a)2
(1 + a2 + 2a + 3 + 3a2 − 6a)
=
φ
(1 + a)2
(4a2 − 4a + 4)
=
(1 + a)2
4(a2 − a + 1)
=
(1 + a)2
(1 + a)2
φ=
4(a2 − a + 1)
tan2 φ = (3)
1 + tan2
sec2
1
cos2
1
cos2
cos2
Therefore,
1+a
cos φ = √
2 a2 − a + 1
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13. A symmetrical three-phase, three-wire 440 V, ABC system feeds a balanced
Y-connected load with ZA = ZB = ZC = 10∠30◦ Ω. Obtain the line currents.
Phase voltage, Vph
440∠0◦
= √
= 254.03∠ − 30◦
3
[It is to be noted that in star-connected networks, the phase voltage lags behind the line voltage
by 30◦ ]
So, phase current in A-phase,
IAN =
=
Phase current in B-phase,
Phase current in C-phase,
IBN
ICN
Vph
Zph
254.03∠ − 30◦
10∠30◦
= 25.403∠ − 60◦ A
= 25.403∠ − 180◦ A
= 25.403∠ − 300◦ A
In star-connected networks, IL = Iph . Therefore,
Line current in A-phase,
Line current in B-phase,
Line current in C-phase,
IA = 25.403∠ − 60◦ A
IB = 25.403∠ − 180◦ A
IC = 25.403∠ − 300◦ A
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77122
B.E/B.Tech. DEGREE EXAMINATION, APRIL/MAY 2015
Electrical and Electronics Engineering
EE6201 – CIRCUIT THEORY
1. Use branch currents in the network shown below to find the current supplied
by the 60 V source. Solve the circuit by the mesh current method.
7Ω
I1
60 V
+
I2
I3
I4
12 Ω
6Ω
12 Ω
Fig. 1
The various branch currents are marked as shown in Fig. 2.
A
7Ω
I1
60 V
+
H
I1 ‒ I2
B
I1 ‒ I2 ‒ I3
C
D
I2
I3
I4
12 Ω
6Ω
12 Ω
G
F
E
Fig. 2
Applying Kirchhoff’s voltage law in mesh ABGHA, we get
or
7I1 + 12I2 − 60 = 0
7I1 + 12I2 = 60
(i)
Applying Kirchhoff’s voltage law in mesh BCFGB, we get
6I3 − 12I2 = 0
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(ii)
Applying Kirchhoff’s voltage law in mesh CDEFC, we get
12I4 − 6I3 = 0
(iii)
Also, from Fig. 2,
I1 − I2 − I3 = I4
Therefore, Eq. (iii) becomes
12I1 − 12I2 − 18I3 = 0
(iv)
Solving Eqs. (i), (ii) and (iv), we get
The current supplied by the 60 V source = I1 = 6 A.
2. Solve the network given below by the node voltage method.
10 Ω
1
2
5Ω
2Ω
25 V
+
2Ω
4Ω
+
50 V
ref
Fig. 3
Applying KCL at node 1, we get
or
V1 V1 − 25 V1 − V2
+
+
=0
2
5
10
0.8V1 − 0.1V2 = 5
(v)
Applying KCL at node 2, we get
or
V2 − V1 V2 V2 + 50
+
+
=0
10
4
2
−0.1V1 + 0.85V2 = −25
(vi)
Solving Eqs. (v) and (vi), we get
V1 = 2.6119 V
V2 = −29.1045 V
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3. Compute the current in the 23 Ω resistor of the figure shown below by applying
superposition principle.
4Ω
I23Ω
47 Ω
27 Ω
20 A
+
‒
200 V
23 Ω
Fig. 4
Response due to 200 V source only
The current through 23 Ω due to 200 V source only is obtained by keeping only
that source and open-circuiting 20 A current source as shown in Fig. 5.
4Ω
I′23Ω
47 Ω
I1
I2
‒
200 V
23 Ω
+
27 Ω
Fig. 5
Mesh analysis applied on the circuit shown in Fig. 5 and by inspection the mesh
equations are written as:
74 47
47 74
I1
I2
=
200
200
0
Solving the above equation, we obtain I2 = I23Ω
= 1.65 A
Response due to 20 A source only
The current through 23 Ω due to 20 A source only is obtained by keeping only
that source and short-circuiting 200 V voltage source as shown in Fig. 6(a).
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4Ω
4Ω
I′′23Ω
47 Ω
27 Ω
20 A
I′′23Ω
17.15 Ω
23 Ω
(a)
20 A
23 Ω
(b)
Fig. 6
As the two resistors 27 Ω and 47 Ω are found parallel it is simplified and the
simplified circuit is shown in Fig. 6(b).
By current division technique,
00
I23Ω
= 20 &times;
17.15 + 4
= 9.581 A
17.15 + 4 + 23
Hence, by applying superposition principle
0
00
I23Ω = I23Ω
+ I23Ω
= 1.653 + 9.581 = 11.234 A
4. Obtain the Thevenin and Norton equivalent circuits for the active network
shown below.
3Ω
3Ω
6Ω
+
‒
20 V
a
10 V
‒
+
b
Fig. 7
Thevenin’s voltage, Vth
Let I be the current flowing through the only loop. Applying KVL to it,
−20 + 3I + 6I − 10 = 0
30
or
I=
= 3.3333 A
9
So, voltage across 6 Ω resistor
= 6 &times; 3.3333 = 20 V
Hence, voltage across a-b
Vth = 20 − 10 = 10 V
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Thevenin’s resistance, Rth
Figure 8 shows the circuit used to calculate the Thevenin’s resistance, Rth .
With reference to Fig. 8, the two resistors 3 Ω and 6 Ω are connected in
parallel. Hence, Thevenin’s resistance,
Rth = Rab = 3 +
6&times;3
=5Ω
6+3
3Ω
3Ω
a
6Ω
b
Fig. 8
Norton’s current, IN
The two terminals a and b are short-circuited as shown in Fig. 9, and the
Norton’s current is computed using mesh analysis.
3Ω
3Ω
6Ω
I1
+
IN
I2
10 V
‒
20 V
a
‒
+
b
Fig. 9
9 −6
−6
9
I1
I2
=
30
−10
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Applying Cramer’s rule, Norton’s current
IN = I2 =
=
=
∆I2
∆
9
30
−6 −10
9 −6
−6
9
−90 + 180
=2A
81 − 36
Using the above found values of Vth , IN and Rth , the Thevenin and Norton
equivalent circuits are obtained as shown in Fig. 10.
Rth = 5 Ω
a
Vth = 10 V
+
a
IN = 2 A
b
Rth = 5 Ω
b
(a) Thevenin’s equivalent circuit
(b) Norton’s equivalent circuit
Fig. 10
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71496
B.E/B.Tech. DEGREE EXAMINATION, APRIL/MAY 2015
Electrical and Electronics Engineering
EE2151 – CIRCUIT THEORY
1. Determine the current supplied by each battery in the circuit shown below using
mesh analysis.
5Ω
4 Ω E3 = 5 V
3Ω
2Ω
E1 = 20 V
E5 = 30 V
E2 = 5 V
E4 = 5 V
Fig. 1
Mesh currents as indicated in Fig. 2 are assumed and by inspection the mesh
equations are written as:
5Ω
E1 = 20 V
I1
4 Ω E3 = 5 V
3Ω
E2 = 5 V
I2
2Ω
I3
E5 = 30 V
E4 = 5 V
Fig. 2


 

8 −3
0
I1
15
 −3
9 −2   I2  =  15 
0 −2
2
I3
−35
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(i)
Applying Cramer’s rule to Eq. (i), the mesh currents are obtained as follows:
∆=
8 −3
0
−3
9 −2
0 −2
2
= 8(18 − 4) + 3(−6) = 94
∆I1 =
15 −3
0
15
9 −2
−35 −2
2
= 15(18 − 4) + 3(30 − 70) = 90
∆I2 =
8
15
0
−3
15 −2
0 −35
2
= 8(30 − 70) − 15(−6) = −230
∆I3 =
8 −3
15
−3
9
15
0 −2 −35
= 8(−315 + 30) + 3(105) + 15(6) = −1875
Therefore,
∆I1
90
=
= 0.9574 A
∆
94
∆I2
−230
I2 =
=
= −2.4468 A
∆
94
∆I3
−1875
I3 =
=
= −19.9468 A
∆
94
I1 =
Hence,
Current
Current
Current
Current
Current
supplied
supplied
supplied
supplied
supplied
by
by
by
by
by
battery
battery
battery
battery
battery
E1
E2
E3
E4
E5
= I1 = 0.9574 A
= I2 − I1 = −2.4468 − 0.9574 = −3.4042 A
= I2 = −2.4468 A
= I3 − I2 = −19.9468 + 2.4468 = −17.5 A
= −I3 = 19.9468 A
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2. Use nodal voltage method to find the voltages of nodes ‘m’ and ‘n’ and currents
through j2 Ω and −j2 Ω reactances in the network of Fig. 3.
5Ω
500&deg; V
4Ω
m
n
2Ω
‒j2 Ω
j2 Ω
5090&deg; V
Fig. 3
Applying KCL at node ‘m’, we get
Vm − 50∠0◦ Vm Vm − Vn
+
+
=0
5
j2
4
0.2Vm − 10∠0◦ − j0.5Vm + 0.25Vm − 0.25Vn = 0
(0.45 − j0.5)Vm − 0.25Vn = 10∠0◦
(ii)
Applying KCL at node ‘n’, we get
Vn − Vm Vn − 50∠90◦
Vn
+
+
=0
−j2
4
2
j0.5Vn + 0.25Vn − 0.25Vm + 0.5Vn − 25∠0◦ = 0
−0.25Vm + (0.75 + j0.5)Vn = 25∠90◦
(iii)
Putting Eqs. (ii) and (iii) in matrix form, we obtain
0.45 − j0.5
−0.25
Vm
10∠0◦
=
−0.25 0.75 + j0.5
Vn
25∠90◦
Applying Cramer’s rule, the node voltages Vm and Vn are obtained as follows:
∆=
0.45 − j0.5
−0.25
−0.25 0.75 + j0.5
= (0.5875 − j0.15) − (0.0625)
= 0.525 − j0.15
∆Vm =
10∠0◦
−0.25
◦
25∠90 0.75 + j0.5
= (7.5 + j5) + (j6.25)
= 7.5 + j11.25
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∆Vn =
0.45 − j0.5 10∠0◦
−0.25 25∠90◦
= (12.5 + j11.25) − (−2.5)
= 15 + j11.25
Therefore, the node voltages
Vm =
∆Vm
7.5 + j11.25
=
= 5.625 + j4.7813 V
∆
0.525 − j0.15
Vn =
∆Vn
15 + j11.25
=
= 20.7547 + j27.3585 V
∆
0.525 − j0.15
Hence,the current flowing through
j2 Ω reactance =
Vm
5.625 + j4.7813
=
j2
j2
= (2.39 − j2.8125) = 3.6913∠ − 49.64◦ A
−j2 Ω reactance =
20.7547 + j27.3585
Vn
=
−j2
−j2
= (−13.6793 + j10.3774) = 17.17∠ − 142.82◦ A
3. Determine the current through 2 Ω resistor in the following network using
Thevenin’s theorem.
1Ω
12 V
+
1Ω
2Ω
+
6V
Fig. 4
The 2 Ω resistor is removed as shown in Fig. 5(a) and the voltage across the
two terminals (Thevenin’s voltage Vth ) is obtained using Kirchhoff’s voltage law
as,
−12 + 2I + 6 = 0
6
Therefore, I = = 3 A. Hence, Vth = 12 − 3 &times; 1 = 9 V.
2
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1Ω
12 V
+
1Ω
1Ω
+
Vth
1Ω
+ 6V
Rth
‒
(a)
(b)
Fig. 5
As seen in Fig. 5(b), the Thevenin’s resistance, Rth = 1||1 =
1&times;1
= 0.5 Ω.
1+1
Rth = 0.5 Ω
Vth = 9 V
+
2Ω
Fig. 6
Using the above found values of Vth and Rth , the Thevenin’s equivalent circuit
is drawn as shown in Fig. 6. Thus, the current through 2 Ω resistor is,
9
Vth
=
= 3.6 A
2 + Rth
2 + 0.5
4. In the wheatstone bridge circuit of Fig. 7, find (i) the effective resistance between P and Q. Find the current supplied by a 10 V battery connected to PQ.
0.25 Ω
A
2
5
Ω
Ω
P
3Ω
B
10 V
1.
4
0.
5
Ω
Q
C
Ω
D
Fig. 7
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Equivalent resistance between P and Q
Using star-delta transformation technique, delta ABC is converted into star
[Fig. 8] and its values are obtained as,
5&times;2
=1Ω
5+3+2
3&times;5
= 1.5 Ω
=
5+3+2
2&times;3
=
= 0.6 Ω
5+3+2
RAN =
RBN
RCN
A
A
5Ω
2Ω
1Ω

1 .5
B
C
3Ω
N 0
.6
Ω
B
Ω
C
Fig. 8
Replacing the ABC delta by its equivalent star, the circuit is redrawn as shown
in Fig. 9.
0.25 Ω
A
P
1Ω
N
5
1.
10 V
0.
Ω
6
Ω
B
C
1.
5
4
0.
Ω
Q
Ω
D
Fig. 9
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Both 1.5 Ω resistances are in series. Also in series are 0.6 Ω and 0.4 Ω. These
two series combinations together form one parallel connection and its equivalent
value is,
RDN = (1.5 + 1.5)||(0.6 + 0.4)
3&times;1
=
= 0.75
3+1
Therefore, the effective resistance between P and Q, RPQ = 0.25+1+0.75 = 2 Ω
Total current
I=
VPQ
10
=
=5A
RPQ
2
5. For the circuit shown, determine Norton’s equivalent circuit between the output
terminals A and B.
3Ω
j4 Ω
A
+
4Ω
250&ordm; V
–
–j5 Ω
B
Fig. 10
The terminals A and B are short-circuited [Fig. 11] and the current through it
(Norton’s current) is calculated as,
25∠0◦
Norton’s current, IN =
= 3 − j4 = 5∠ − 53.13◦ A
3 + j4
3Ω
j4 Ω
A
+
4Ω
IN
250&ordm; V
–
–j5 Ω
B
Fig. 11
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Thevenin’s impedance, Zth is calculated by killing the 25∠0◦ voltage source as
shown in Fig. 12 and looking back the circuit through the terminals A and B.
j4 Ω
3Ω
A
4Ω
Zth
–j5 Ω
B
Fig. 12
From Fig. 12,
Zth = (3 + j4)||(4 − j5)
(3 + j4)(4 − j5)
= 3.5 + j0.5
=
(3 + j4) + (4 − j5)
Using the above found values of IN and Zth the Norton’s equivalent circuit is
drawn as shown in Fig. 13.
IN = 5‒53.13&ordm; A
A
3.5 Ω
j0. 5 Ω
B
Fig. 13
6. Verify superposition theorem for the (2 + j5) Ω impedance.
j4 Ω
–j3 Ω
2Ω
2∠30&ordm; A
50∠0&ordm; V
j5 Ω
Fig. 14
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50∠0◦ V source is acting alone
With reference to Fig. 15(a), the current I 0 flowing through (2+j5) Ω impedance
due to 50∠0◦ V source is obtained as:
50∠0◦
= (1.1765 − j5.2941) A
I0 =
2 + j9
j4 Ω
j4 Ω
–j3 Ω
2Ω
500&ordm;
–j3 Ω
2Ω
20&ordm; A
V
j5 Ω
j5 Ω
(a)
(b)
Fig. 15
2∠30◦ A source is acting alone
As shown in Fig. 15(b), only 2∠30◦ A is made active, while 50∠0◦ V source
is short-circuited. By current division rule, the current I 00 flowing through
(2 + j5) Ω impedance due to 2∠30◦ A source is:
I 00 = 2∠30◦ &times;
j4
= (0.6395 + j0.5865) A
j4 + 2 + j5
By the principle of superpositon, current through (2 + j5) Ω impedance due to
both sources,
I = I 0 + I 00
= (1.1765 − j5.2941) + (0.6395 + j0.5865)
= (1.8160 − j4.7076) A
j4 Ω
V1
(iv)
–j3 Ω
2Ω
2∠30&ordm; A
50∠0&ordm; V
j5 Ω
Fig. 16
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To verify the result obtained above, node analysis method is used. As the given
circuit contains only one node, a voltage of V1 as shown in Fig. 16 is assumed.
Applying KCL to the only node,
V1 − 50
V1
+
= 2∠30◦
j4
2 + j5
V1 (0.0689 − j0.4224) = 2∠30◦ − j12.5
2∠30◦ − j12.5
V1 =
0.0689 − j0.4224
= 27.1713 − j0.3317 V
Therefore, current through (2 + j5) Ω impedance
I=
V1
= (1.8167 − j4.7076) A
2 + j5
(v)
Comparing Eqs. (iv) and (v), it is observed that both results are same. Thus,
superposition theorem is verified.
7. For the circuit shown below, determine the frequency at which the circuit resonates. Also find the voltage across the inductor at resonance and the Q factor
of the circuit.
10 Ω
50 μF
0.1 H
+
‒
100 Vrms
Fig. 17
1
√
2π LC
Resonant frequency,
fr =
Q-factor,
1
p
= 71.18 Hz.
2π (0.1)(50 &times; 10−6 )
r
1 L
Q=
R rC
1
0.1
=
= 4.47
10 50 &times; 10−6
=
At resonance, voltage across the inductor
VL = QV = 4.47 &times; 100 = 447 V
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8. The number of turns in two coupled coils are 500 turns and 1500 turns respectively. When 5 A current flows in coil 1, the total flux in this coil is 0.6 &times; 10−3
Wb and the flux linking the second coil is 0.3 &times; 10−3 Wb. Determine L1 , L2 , M
and k.
N1 φ1
Self-inductance of coil 1,
L1 =
I1
500 &times; 0.6 &times; 10−3
=
5
= 60 mH
r
L1
N1
Turns ratio,
=
N2
L2
So,
L2 =
=
∴ Self-inductance of coil 2,
Coefficient of coupling,
Mutual inductance
N22
L1
N12
15002
&times; 60 &times; 10−3
5002
L2 = 540 mH
0.3
φ12
=
=5
k=
φ1
0.6
p
p
M = k L1 L2 = 0.5 (0.06 &times; 0.54) = 0.09 H
9. A coil having an inductance of 100 mH is magnetically coupled to another coil
having an inductance of 900 mH. The coefficient of coupling between the coils
is 0.45. Calculate the equivalent inductance if the two coils are connected in
(a)
(b)
(c)
(d)
Series aiding
Series opposing
Parallel aiding
Parallel opposing
p
L1 L2
√
= 0.45 100 &times; 10−3 &times; 900 &times; 10−3 = 135 mH
Mutual inductance, M = k
Eqt. inductance if series aiding, Leq = L1 + L2 + 2M
= (100 &times; 10−3 ) + (900 &times; 10−3 ) + (2 &times; 135 &times; 10−3 )
= 1270 mH
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Eqt. inductance if series opposing, Leq = L1 + L2 − 2M
= (100 &times; 10−3 ) + (900 &times; 10−3 ) − (2 &times; 135 &times; 10−3 )
= 730 mH
Eqt. inductance if parallel aiding, Leq =
L1 L2 − M 2
L1 + L2 − 2M
(0.1 &times; 0.9) − 0.1352
0.1 + 0.9 − (2 &times; 0.135)
= 98.3219 mH
=
Eqt. inductance if parallel opposing, Leq =
L1 L2 − M 2
L1 + L2 + 2M
(0.1 &times; 0.9) − 0.1352
0.1 + 0.9 + (2 &times; 0.135)
= 56.5157 mH
=
10. A series RLC circuit with R = 100 Ω, L = 0.1 H and C = 100 &micro;F has a DC
voltage of 200 V applied to it at t = 0 through a switch. Find the expression
for the transient current. Assume initially relaxed circuit condition.
t=0
0.1 H
100 Ω
100 μF
i(t)
200 V
Fig. 18
Determination of initial conditions
(a) The switch is closed at t = 0. Therefore, at t = 0− , the current through
the inductor was zero, i.e., i(0− ) = 0, which is also equal to i(0+ ), since
the current through the inductor cannot change instantaneously. Hence,
we can write
i(0− ) = 0 = i(0+ )
(vi)
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(b) Initial voltage on the capacitor is zero, which results in
i(0− ) = 0 = i(0+ )
(vii)
Applying KVL after closing the switch
di
1
100i + 0.1 +
dt 100 &times; 10−6
Z
idt = 200
Taking Laplace transform on both sides, we get
100I(s) + 0.1[sI(s) − I(0)] +
1
I(s)
200
=
−6
100 &times; 10
s
s
From Eq. (vi), i(0) = 0, and in s-domain, I(0) = 0. Therefore,
I(s)
1
=
−6
100 &times; 10
s
1
1
I(s) 100 + 0.1s +
=
100 &times; 10−6 s
0.01s + 10 &times; 10−6 s2 + 1
I(s)
=
100 &times; 10−6 s
100I(s) + 0.1sI(s) +
200
s
200
s
200
s
200
100 &times; 10−6 s
I(s) =
s
0.01s + 10 &times; 10−6 s2 + 1
0.02
0.01s + 10 &times; 10−6 s2 + 1
0.02
=
10 &times; 10−6 (s2 + 1000s + 100000)
=
2000
+ 1000s + 100000
2000
=
(s + 112.7)(s + 887.3)
=
s2
By applying partial fraction technique,
I(s) =
2000
A
B
=
+
(s + 112.7)(s + 887.3)
s + 112.7 s + 887.3
(viii)
Solving for constants A and B,
A=
2000
(s + 887.3)
= 2.582 and B =
s=−112.7
2000
s + 112.7
= −2.582
s=−887.3
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Substituting the values of A and B into Eq. (viii), we get
I(s) =
2.582
2.582
−
s + 112.7 s + 887.3
Taking inverse Laplace transform on both sides, the current equation is obtained
as
i(t) = 2.582e−112.7t − 2.582e−887.3t A
11. In the circuit shown in Fig. 19, find the time when the voltage across the
capacitor becomes 25 V, after the switch is closed at t = 0.
t=0
1 μF
20 Ω
100 V
Fig. 19
Applying KVL after closing the switch,
Ri + vc = V
dvc
+ vc = V
dt
dvc
20 &times; 1 &times; 10−6
+ vc = 100
dt
dvc
20 &times; 10−6
+ vc = 100
dt
RC
Taking Laplace transformation on both sides, we get
20 &times; 10−6 [sVC (s) − VC (0)] + VC (s) =
100
s
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At t = 0, VC (0) = 0 V. Therefore, the above equation becomes
100
s
100
VC (s)[1 + 20 &times; 10−6 s] =
s
20 &times; 10−6 sVC (s) + VC (s) =
VC (s) =
)
( 100
s
(1 + 20 &times; 10−6 s)
=
100
s(1 + 20 &times; 10−6 s)
=
5 &times; 106
100
=
20 &times; 10−6 s(s + 50000)
s(s + 50000)
By applying partial fraction technique,
VC (s) =
5 &times; 106
A
B
= +
s(s + 50000)
s
s + 50000
(ix)
Solving for constants A and B,
5 &times; 106
A=
(s + 50000)
s=0
5 &times; 106
= 100 and B =
s
= −100
s=−50000
Substituting the values of A and B into Eq. (ix), we get
VC (s) =
100
100
−
s
s + 50000
Taking inverse Laplace transformation on both sides, the voltage across the
capacitor is obtained as
vC (t) = 100 − 100e−50000t V
(x)
The time at which the voltage across the capacitor becomes 25 V is found by
equating Eq. (x) to 25. So,
or
Therefore,
100 − 100e−50000t = 25
e−50000t = 0.75
−50000t = −0.2877
t = 5.754 &micro;s.
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27205
B.E/B.Tech. DEGREE EXAMINATION, NOV./DEC. 2015
Electrical and Electronics Engineering
EE6201 – CIRCUIT THEORY
1. Determine the magnitude and direction of the current in the 2 V battery in the
circuit shown in Fig. 1.
2Ω
4V
3Ω
2V
1.5 Ω
3V
Fig. 1
The various currents through the different branches are assumed as shown in
Fig. 2.
A
4V
C
+
‒
B
3Ω
I1
2V
E
2Ω
I1 + I2
‒
+
D
1.5 Ω
I2
3V
‒
+
F
Fig. 2
Writing Kirchhoff’s voltage equation to the loop ABDCA
or
−4 + 2(I1 + I2 ) + 3I1 − 2 = 0
5I1 + 2I2 = 6
(i)
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Writing Kirchhoff’s voltage equation to the loop CDFEC
2 − 3I1 + 1.5I2 − 3 = 0
−3I1 + 1.5I2 = 1
or
(ii)
Solving Eqs. (i) and (ii), we obtain
Current through 2 V battery = I1 = 0.5185 A
2. Determine the power dissipation in the 4 Ω resistor of the given circuit shown
in Fig. 3.
5Ω
50 V
2Ω
3Ω
6Ω
4Ω
10 V
Fig. 3
Mesh analysis is applied to find the current through 4 Ω resistor first. Figure 4
is used in which the assumed mesh currents I1 , I2 and I3 are shown.
5Ω
50 V
I1
2Ω
3Ω
I2
6Ω
4Ω
I3
10 V
Fig. 4
Mesh equations by inspection are written as:


 

8 −3
0
I1
50
 −3
9 −4   I2  = 
0 
0 −4 10
I3
−10
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The mesh currents I2 and I3 are found using Cramer’s rule as follows.


8 −3
0
9 −4 
∆ =  −3
0 −4 10
= 8(90 − 16) + 3(−30) = 502


8
50
0
0 −4 
∆I2 =  −3
0 −10 10
(iii)
= 8(−40) − 50(−30) = 1180


8 −3
50
9
0 
∆I3 =  −3
0 −4 −10
(iv)
= 8(−90) + 3(30) + 50(12) = −30
(v)
Using Eqs. (iii) to (v), we get
1180
∆I2
=
= 2.3506 A
∆
502
−30
∆I3
=
= −0.0598 A
I3 =
∆
502
I2 =
Therefore, current through 4 Ω resistor = I2 − I3 = 2.3506 + 0.0598 = 2.4104 A
Hence, power dissipation in the 4 Ω resistor = 2.41042 &times; 4 = 23.24 W
3. Using node analysis, find the voltage Vx for the circuit shown in Fig. 5.
+
2
Vx
4Ω
Ω
Ω
2
4A
2A
‒
1Ω
1Ω
5Vx
+ ‒
Fig. 5
Nodal voltages are assumed as shown in Fig. 6.
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Reference
+
2
Vx
Ω
Ω
2
4A
4Ω
2A
‒
V1
V3
1Ω
V2
1Ω
5Vx
+ ‒
Fig. 6
Since the nodes 1 and 3 are having a voltage source of magnitude 5Vx supernode
analysis is used.
By inspection, we can write
V1 − V3 = 5Vx
V2 = −Vx
V1 + 5V2 − V3 = 0
Also,
Therefore,
(vi)
Applying KCL to the supernode 1&amp;3, we get
V1 − V2 V1
V3 − V2 V3
+
+ 4+
+
+2=0
1
2
1
2
|
{z
} |
{z
}
w.r.t node 1
or
w.r.t node 3
1.5V1 − 2V2 + 1.5V3 = −6
(vii)
Applying KCL to the simple node 2, we get
or
V2 − V1 V2 V2 − V3
+
+
=0
1
4
1
−V1 + 2.25V2 − V3 = 0
(viii)
Solving Eqs. (vi), (vii) and (viii), we get
V2 = −4.3636
Therefore,
Vx = −V2 = 4.3636 V
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4. Find the Thevenin’s equivalent circuit of the network shown in Fig. 7
3Ω
1Ω
A
2Ω
+
10 V
5A
5A
2Ω
+
10 V
B
Fig. 7
Thevenin’s voltage
To find Thevenin’s voltage (Vth ), the two 5 A current sources are converted into
voltage sources as shown in Fig. 8. Also shown are the mesh currents I1 and I2 .
5V
3Ω
1Ω
+
A
2Ω
10 V
+
2Ω
I1
I2
10 V
+
10 V
+
B
Fig. 8
Mesh equations by inspection are written as
5 −2
I1
0
=
I2
25
−2
5
(ix)
Solving Eq. (ix), I2 is obtained as 5.9524 A. Therefore, voltage across A and B
VAB = Vth = 2 &times; 5.9524 − 10 = 1.9048 V
Thevenin’s resistance
To find Thevenin’s resistance (Rth ) the circuit shown in Fig. 10 is used where
all the sources are killed and the circuit is looked back through the terminals A
and B.
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3Ω
1Ω
A
2Ω
Rth
2Ω
B
Fig. 9
From Fig. 10, Rth = RAB is obtained as
Rth = [(3||2) + 1]||2
3&times;2
=
+ 1 ||2
3+2
2.2 &times; 2
= 2.2||2 =
2.2 + 2
Rth = 1.0476 Ω
Thevenin’s equivalent circuit
Vth = 1.9048 V
Rth = 1.0476 Ω
A
+
B
Fig. 10
5. Determine the value of resistance that may be connected across A and B so that
maximum power is transferred from the circuit to the resistance. Also, estimate
the maximum power transferred to the resistance shown in Fig. 11.
2Ω
5V
4Ω
+
20 V
+
8Ω
A
10 Ω
B
Fig. 11
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By maximum power transfer theorem, maximum power is transferred to R, if
the value of R is equal to Rth . Hence, we have to find the value of Thevenins
resistance, Rth . To calculate Rth , the two voltage sources are replaced by a
short-circuits as shown in Fig. 12 and the circuit is looked back from the
terminals A and B.
Value of load resistance for maximum power transfer
2Ω
4Ω
A
8Ω
Rth
10 Ω
B
Fig. 12
From Fig. 12, Rth = RAB is obtained as
= [(2||8) + 4]||10
2&times;8
=
+ 4 ||10
2+8
5.6 &times; 10
= 5.6||10 =
5.6 + 10
Rth = 3.59 Ω
Therefore, maximum power is transferred to the resistance across the terminals
A and B when it is equal to 3.59 Ω.
Maximum power
To find the maximum power transfer to the load resistance, current through it
is found first. Mesh analysis is applied and the circuit shown in Fig. 13 is used
for the calculation of current.
2Ω
4Ω
5V
+
20 V
+
I1
8Ω
I2 10 Ω
I3
A
3.59 Ω
B
Fig. 13
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

 

10 −8
0
I1
20
 −8
22 −10   I2  =  0 
0 −10 13.59
I3
5
Using Cramer’s rule
I3 =
∆I3
∆


10 −8 20
 −8
22 0 
0 −10 5

=
10 −8
0
 −8
22 −10 
0 −10 13.59
2380
= 2.1249 A
=
1120.04
Therefore, the maximum power transferred to load resistance
= I32 &times; R = 2.12492 &times; 3.59 = 16.21 W
6. For the circuit shown in Fig. 14, determine the frequency at which the circuit
resonates. Also find the quality factor, voltage across inductance an voltage
across capacitance at resonance.
0.03 H 100 μF
5Ω
+
20 V
‒
Fig. 14
Resonant frequency,
fr =
1
√
2π LC
1
p
2π (0.03)(100 &times; 10−6 )
= 91.89 Hz.
=
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r
1 L
Q=
Rs C
Q-factor,
=
(0.03)
(100 &times; 10−6 )
1
5
= 3.46
At resonance, voltages across inductance and capacitance are equal. Therefore,
VL = VC = QV = 3.46 &times; 20 = 69.2 V
7. Find the mutual reactance Xm in the coupled coils shown in Fig. 15, if the
average power in 8 Ω resistance is 100 W.
Xm
5Ω
+
j5 Ω
1000&deg; V
j12 Ω
8Ω
–
Fig. 15
Two mesh currents I1 and I2 are assumed in the direction as shown in Fig.
16(a). Current I1 is entering the dot whereas I2 is leaving the dot, hence the
value of mutual inductance M is negative. Accordingly, the electrical equivalent
circuit of the coupled coils drawn and is shown in Fig. 16(b).
Xm
5 Ω (j5 ‒ jXm) Ω
+
I1
j5 Ω
j12 Ω
–
I2
8Ω
1000&deg; V
1000&deg; V
5Ω
(a)
(j12 ‒ jXm) Ω
+
I1
jXm Ω
I2
8Ω
–
(b)
Fig. 16
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The mesh equations for the circuit shown in Fig. 16(b) is written as:
5 + j5 − jXm + jXm
−jXm
−jXm jXm + j12 − jXm + 8
5 + j5 −jXm
−jXm 8 + j12
I1
I2
I1
I2
100∠0◦
0
100∠0◦
0
=
=
Using Cramer’s rule,
I2 =
∆I2
=
∆
=
=
5 + j5 100∠0◦
−jXm
0
5 + j5 −jXm
−jXm 8 + j12
j100Xm
(5 + j5)(8 + j12) − (jXm )2
2
Xm
j100Xm
− 20 + j100
It is given that power in 8 Ω resistance is 100 W.
|I2 |2 &times; 8 = 100
or
2
j100Xm
&times; 8 = 100
2
Xm − 20 + j100
!2
p
2
02 + 1002 Xm
p
&times; 8 = 100
2 − 20)2 + 1002
(Xm
2
1002 Xm
100
=
4
2
2
2
Xm + 20 − 40Xm + 100
8
4
2
Xm
− 840Xm
+ 10400 = 0
2
By substituting x = Xm
into the above equation, it becomes
x2 − 840x + 10400 = 0
Solving,
or
x = 827.43
Xm = 28.77 Ω
and
and
12.57
3.55 Ω
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8. In the series RL circuit shown in Fig. 17, the switch is closed on position 1 at
t = 0. At t = 100 ms, the switch is moved to position 2. Find i(t) and sketch
the transient.
0.2 H
2Ω


5V
i(t)
20 V
Fig. 17
At t = 0, the switch is connected to position 1 and the differential equation is
2i + 0.2
di
= −5
dt
Taking Laplace transformation on both sides, we get
2I(s) + 0.2[sI(s) − I(0)] =
−5
s
(x)
At t = 0, i(t) = 0, and in s-domain, I(0) = 0. Therefore, Eq. (x) becomes
−5
s
−5
I(s)[2 + 0.2s] =
s
2I(s) + 0.2sI(s) =
−5
s(0.2s + 2)
−5
−25
=
=
0.2s(s + 10)
s(s + 10)
I(s) =
By applying partial fraction technique,
=
A
B
+
s
(s + 10)
(xi)
Solving for the constants A and B,
A=
−250
s + 10
= −2.5
s=0
and
B=
−25
s
= 2.5
s=−10
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Substituting the values of A and B into Eq. (xi), we get
I(s) =
−2.5
2.5
+
s
(s + 10)
Taking inverse Laplace transformation on both sides, the current equation is
obtained as:
i(t) = −2.5 + 2.5e−10t A
(xii)
The switch is kept at this position for 100 ms. Therefore, at t = 0.1 s, the
current through the circuit is determined by putting t = 0.1 in Eq. (xii), i.e. at
t = 0.1 s.
i(t) = −2.5 + 2.5e−10(0.1) = −1.5803 A
This value becomes the initial current for the circuit when the switch is put
in position 2. Then, the switch is moved to position 2 at t = 0.1 s. Due to this
action, 20 V source is connected in the circuit and 5 V source is disconnected.
This moment (t = 0.1 s) can be redefined as t0 = 0.
Applying KVL,
di
= 20
dt0
Taking Laplace transformation on both sides, we get
2i + 0.2
2I(s) + 0.2[sI(s) − I(0)] =
20
s
(xiii)
At t0 = 0, I(0) = −1.5803 A. Therefore, Eq. (xiii) becomes
20
s
20
I(s)[2 + 0.2s] =
− 0.31606
s
20 − 0.31606s
I(s) =
s(0.2s + 2)
2I(s) + 0.2sI(s) + 0.31606 =
=
20 − 0.31606s
0.2s(s + 10)
=
100 − 1.5803s
s(s + 10)
By applying partial fraction technique,
=
A
B
+
s
(s + 10)
(xiv)
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Solving for the constants A and B,
A=
100 − 1.5803s
s + 10
= 10
and
B=
s=0
100 − 1.5803s
s
= −11.5803
s=−10
Substituting the values of A and B into Eq. (xiv), we get
10
11.5803
I(s) =
−
s
(s + 10)
Taking inverse Laplace transformation on both sides, the current equation is
obtained as:
0
i(t) = 10 − 11.5803e−10t A
or
i(t) = 10 − 11.5803e−10(t−0.1) A
(xv)
Therefore,
i(t) = −2.5 + 2.5e−10t A
= 10 − 11.5803e−10(t−0.1) A
for 0 &lt; t &lt; 0.1
for t &gt; 0.1
i
10 A
0
1
.1
47
s
t
0
–1.5803 A
0.1 s
Fig. 18
With reference to Eq. (xv), the time t at which i(t) becomes zero is obtained
as:
11.5803e−10(t−0.1) = 10
Therefore,
e−10(t−0.1) = 0.8635
−10(t − 0.1) = ln 0.8635
t = 0.1147 s
The transient current is shown in Fig. 18.
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9. Determine the driving point impedance of the network shown in Fig. 19.
2H
2H
1
1Ω
2Ω
1′
Fig. 19
The transformed network of the given network is shown in Fig. 20.
2s
2s
1
1
2
1′
Fig. 20
The driving point impedance, Z(s) of the network is the equivalent impedance
across the input port 1 − 10 . By network reduction technique the given network
is successively reduced as shown in Figs. 21(a) through 21(c).
Z(s) = 1||[2s + (2||2s)]
4s
= 1|| 2s +
2 + 2s
Therefore, driving point impedance, Z(s) =
4s2 + 8s
4s2 + 10s + 2
2s
1
1
1
4s
2  2s
1
1′
1
1′
(a)
4 s 2  8s
2  2s
2
4 s  8s
Zeq 
2
4 s  10 s  2
1′
(b)
(c)
Fig. 21
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10. Determine the h-parameters of the two-port network shown in Fig. 22.
10 Ω
1
2
20 Ω
1′
2′
Fig. 22
The performance equations for h-parameters are given as:
V1 = h11 I1 + h12 V2
I2 = h21 I1 + h22 V2
(xvi)
(xvii)
By definition,
h11 =
h12 =
h21 =
h22 =
V1
I1
V2 =0
V1
V2
I1 =0
I2
I1
V2 =0
I2
V2
I1 =0
To determine the above h-parameters, mesh analysis may be used. Accordingly,
the two mesh currents are assumed as shown in Fig. 23.
10 Ω
1
V1
2
I1
20 Ω
1′
I2
V2
2′
Fig. 23
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By inspection ,the mesh equations are written as:
30 20
I1
V1
=
20 20
I2
V2
Expanding the above matrix equation,
30I1 + 20I2 = V1
20I1 + 20I2 = V2
(xviii)
(xix)
From Eq. (xix),
I2 =
−20I1 + V2
20
I2 = −I1 + 0.05V2
or
(xx)
Substituting Eq. (xx) into Eq. (xviii), we get
V1 = 10I1 + V2
(xxi)
Upon comparison of Eqs. (xxi) and (xx) respectively with Eqs. (xvi) and (xvii),
we get
h11 = 10;
h21 = −1;
h12 = 1;
h22 = 0.05
11. Three equal impedances, each of (8 + j10) Ω are connected in star. This is
further connected to a 440 V, 50 Hz, three phase supply. Calculate the active
and reactive power and line and phase currents.
Phase voltage
Vph =
440∠0◦
√
= 254.03∠ − 30◦
3
[It is to be noted that in star-connected networks, the phase voltage lags behind the line voltage
by 30◦ ]
Vph
Zph
254.03∠ − 30◦
=
(8 + j10)
= (2.99 − j19.61) A
= 19.84∠ − 81.34◦ A
= 19.84∠ − 201.34◦ A
= 19.84∠ − 321.34◦ A
So, phase current in R − phase, IRN =
Phase current in Y − phase,
Phase current in B − phase,
IYN
IBN
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In star-connected networks, IL = Iph . Therefore,
IR = 19.84∠ − 81.34◦ A
IY = 19.84∠ − 201.34◦ A
IB = 19.84∠ − 321.34◦ A
10
−1
φ = tan
8
◦
= 51.34
√
P = 3 VL IL cos φ
√
= 3 (440)(19.84)(cos 51.34◦ )
= 9445.5 W
√
Q = 3 VL IL sin φ
√
= 3 (440)(19.84)(sin 51.34◦ )
= 11806.8 var
Line current in R − phase,
Line current in Y − phase,
Line current in B − phase,
Phase angle,
Total active power,
Total reactive power,
12. Two wattmeters connected to measure the input to a balanced three phase
circuit indicate 2000 W and 500 W respectively. Find the power factor of the
circuit,
(a) When both readings are positive and
(b) When the latter is obtained after reversing the connections to the current
coil of one instrument.
Let
W1 = 500 W
W2 = 2000 W
So, total three-phase power input = W1 + W2 = 500 + 2000 = 2500 W
√
W2 − W1
tan φ = 3
W + W2
1
√ 2000 − 500
−1
φ = tan
3
= 46.1◦
2000 + 500
So, power factor,
cos φ = 0.69
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(b)When the reading of 500 W is obtained after reversing the connections to the
current coil
Let
W1 = −500 W
W2 = 2000 W
So, total three-phase power input = W1 + W2 = −500 + 2000 = 1500 W
√
W2 − W1
tan φ = 3
W + W2
1
√ 2000 + 500
−1
3
= 70.89◦
φ = tan
2000 − 500
So, power factor,
cos φ = 0.33
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21496
B.E/B.Tech. DEGREE EXAMINATION, NOV./DEC. 2015
Electrical and Electronics Engineering
EE2151 – CIRCUIT THEORY
1. Find the current in 10 Ω resistance, V1 and source voltage VS in the circuit
shown in Fig. 1.
A
5Ω
30 V
C
I5 = 1 A
I6 = 4 A
10 Ω
VS
6Ω
V1
B
Fig. 1
Given that,
Current through 5 Ω resistor,
Current through 6 Ω resistor,
I5 = 1 A
I6 = 4 A
Therefore by applying KCL at node C, we get
Current through 10 Ω resistor,
I10 = I5 + I6 = 1 + 4 = 5 A
Applying KVL around the loop comprising 6 Ω, 10 Ω and V1
(I6 &times; 6) + (I10 &times; 10) − V1 = 0
V1 = (4 &times; 6) + (5 &times; 10) = 74 V
Applying KVL around the loop comprising ACBA, we obtain
−VS + (I5 &times; 5) − 30 + (I10 &times; 10) − V1 = 0
(1 &times; 5) − 30 + (5 &times; 10) − 74 = VS
VS = −49 V
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2. Write the node voltage equations and determine the currents in each branch for
the network shown in Fig. 2.
3Ω
V1
5A
1Ω
V2
10 Ω
5Ω
+ 10 V
Fig. 2
Applying KCL at node 1
V1 V1 − V2
+
=5
10
3
0.4333V1 − 0.3333V2 = 5
or
(i)
Similarly, applying KCL at node 2
V2 − V1 V2 V2 − 10
+
+
=0
3
5
1
−0.3333V1 − 1.5333V2 = 0
or
(ii)
Solving Eqs. (i) and (ii), we get
V1 = 19.88 V
V2 = 10.84 V
Therefore,
V1
19.88
=
= 1.988 A
10
10
19.88 − 10.84
V1 − V2
Current through 3 Ω branch =
=
= 3.013 A
3
3
V2
10.84
Current through 5 Ω branch =
=
= 2.168 A
5
5
V2 − 10
19.88 − 10
Current through 1 Ω branch =
=
= 0.84 A
1
1
Current through 10 Ω branch =
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3. Obtain the star connected equivalent for the delta connected circuit shown in
Fig. 3
13 Ω
A
C
12 Ω
14 Ω
B
D
Fig. 3
The equivalent star elements for the given delta network elements are found as:
RAN =
=
RBN =
=
RCN =
=
RCA &times; RAB
RAB + RBC + RCA
13 &times; 12
=4Ω
12 + 14 + 13
RAB &times; RBC
RAB + RBC + RCA
12 &times; 14
= 4.3077 Ω
12 + 14 + 13
RBC &times; RCA
RAB + RBC + RCA
14 &times; 13
= 4.6667 Ω
12 + 14 + 13
The equivalent star network thus found is drawn as shown in Fig. 4.
A
4Ω
7Ω
07
3
N
.
4
4.6
66
7Ω
C
B
Fig. 4
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4. Determine the Thevenin’s equivalent circuit across AB for the given circuit
shown in Fig. 5.
10 Ω
5Ω
A
50 V
25 V
B
Fig. 5
Thevenin’s voltage
Let I be the current through the circuit and by applying KVL, we obtain
Therefore,
−50 + 10I + 5I + 25 = 0
I = 1.6667 A
Again applying KVL to the mesh involves 50 V, 10 Ω and VAB ,
Therefore,
−50 + 10(1.6667) + VAB = 0
VAB = Vth = 33.3333 V
Thevenin’s resistance
To compute Thevenin’s resistance the two voltage sources are short-circuited as
shown in Fig. 6.
10 Ω
5Ω
A
B
Fig. 6
With reference to Fig. 6, Rth is obtained as:
Rth = RAB =
10 &times; 5
= 3.3333 Ω
10 + 5
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Thevenin’s equivalent circuit
Using the above found Vth and Rth values, the Thevenin’s equivalent circuit is
drawn as shown in Fig. 7.
3.3333 Ω
A
33.33 V
B
Fig. 7
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57018
B.E/B.Tech. DEGREE EXAMINATION, MAY/JUNE 2014
Electrical and Electronics Engineering
EE6201 – CIRCUIT THEORY
1. Find the current I and voltage across 30 Ω resistor of the circuit shown in Fig.
1.
8Ω
2Ω
40 V
100 V
30 Ω
Fig. 1
Let the current be I amperes leaving the positive terminal of 100 V battery.
Applying KVL around the loop,
−100 + 8I + 40 + 30I + 2I = 0
60
or
I=
= 1.5 A
40
Therefore, voltage across 30 Ω resistor = 1.5 &times; 30 = 45 V
2. Determine the current in all the resistors of the circuit shown in Fig. 2.
A
i1
50 A
2Ω
i2
1Ω
i3
5Ω
B
Fig. 2
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Applying KCL at node A, we get
i1 + i2 + i3 = 50
VAB VAB VAB
+
+
= 50
2
1
5
or
Solving, we get
VAB = 29.41 V
Therefore, i1 =
5.882 A
29.41
= 14.705 A;
2
i2 =
29.41
= 29.41 A;
1
i3 =
29.41
=
5
3. Determine the current through each resistor in the circuit shown in Fig. 3.
12 A
VS
+
4Ω
4Ω
I1
4Ω
I2
I3
Fig. 3
According to current division rule, the total current 12 A divides equally among
all the three resistors since they are having of equal value. Hence, the currents
12
= 3 A.
I1 = I2 = I3 =
4
4. When a DC voltage is applied to a capacitor, voltage across its terminals is
found to build up in accordance with vc = 50(1 − e−100t ). After 0.01 s, the
current flow is equal to 2 mA.
(a) Find the value of capacitance in farad.
(b) How much energy is stored in the electric field?
(a)Value of capacitance
dvc
dt
d
= C (50 − 50e−100t )
dt
Ic = 5000Ce−100t
Current through the capacitor, Ic = C
(i)
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Given that, Ic = 2 mA after t = 0.01 s. Substituting the values of Ic and t into
Eq. (i), we get
0.002 = 5000Ce−100&times;0.01
Therefore,
C = 1.0873 &micro;F
(b) Energy stored
1
Energy stored = Cvc2
2
1
= (1.0873 &times; 10−6 )(50 − 50e−100&times;0.01 )2
2
= 590.49 &times; 10−12 J
5. Determine the current in the 5 Ω resistor in the network shown in Fig. 4.
10 Ω
2
2A
50 V
5Ω
1
Fig. 4
Since there exits a current source on the common branch of the two adjacent
meshes, supermesh analysis is used. Mesh currents and consequent voltage
drops across each resistor are indicated in Fig. 5.
+
10(I1 ‒ I2)
‒
50 V
+
I2
2A
2I2
‒
I1
+
+
5(I1 ‒ I3)
I3
‒
1I3
‒
Fig. 5
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Applying KVL to the simple mesh, we get
−50 + 10(I1 − I2 ) + 5(I1 − I3 ) = 0
15I1 − 10I2 − 5I3 = 50
or
(ii)
Applying KVL to the supermesh, we get
−5(I1 − I3 ) − 10(I1 − I2 ) + 2I2 + I3 = 0
−15I1 + 12I2 + 6I3 = 0
I2 − I3 = 2
or
Also,
(iii)
(iv)
Solving Eqs. (ii) to (iv), we obtain
I1 = 20
I2 = 17.33
I3 = 15.33
Hence, the current through 5 Ω resistor = I1 − I3 = 20 − 15.33 = 4.67 A
6. Find out the current in each branch of the circuit shown in Fig. 6.
3
5A
10 Ω
1
10 V
5Ω
Fig. 6
Nodal analysis is used. Let the voltages of the two nodes in the circuit shown
in Fig. 7 be V1 and V2 respectively.

5A
10 Ω
3

5Ω
1
10 V
Fig. 7
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Writing KCL for node 1, we get
V1 V1 − V2
+
−5=0
10
3
0.4333V1 − 0.3333V2 = 5
or
(v)
Similarly, writing KCL for node 1, we get
V2 − V1 V2 V2 − 10
+
+
=0
3
5
1
−0.3333V1 + 1.5333V2 = 10
or
(vi)
Solving Eqs. (v) and (vi), we obtain
V1 = 19.88 V
V2 = 10.84 V
Therefore,
V1
19.88
=
= 1.988 A
10
10
V1 − V2
19.88 − 10.84
Current through 3 Ω resistor =
=
= 3.0133 A
3
3
10.84
V2
=
= 2.168 A
Current through 5 Ω resistor =
5
5
V2 − 10
10.84 − 10
Current through 1 Ω resistor =
=
= 0.84 A
1
1
Current through 10 Ω resistor =
7. Determine current in each mesh of the circuit shown in Fig. 8.
10 V
10 A
3Ω
1Ω
2Ω
Fig. 8
By source transformation technique, the 10 A source is converted into its equivalent voltage source as shown in Fig. 9. The various branch currents are also
marked on the modified circuit.
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Ix + Iy
Iy
Ix
10 V
30 V
1Ω
3Ω
2Ω
Fig. 9
By writing KVL equations for the two meshes, we get
3(Ix + Iy ) − 30 + 10 + 2Ix
5Ix + 3Iy
−2Ix − 10 + Iy
−2Ix + Iy
or
and
=0
= 20
=0
= 10
(vii)
(viii)
Solving Eqs. (vii) and (viii), we get
Ix = −0.9091
Iy = 8.1818
To find the mesh currents the numeric values of Ix and Iy are marked on the
given circuit as shown in Fig. 10.
8.1818 A
0.9091 A
10 V
10 A
I1 3 Ω
I2
I3
1Ω
2Ω
Fig. 10
It is apparent from Fig. 10 that, the mesh currents
I1 = 10 A
I2 = 8.1818 − 0.9091 = 7.2727 A
I3 = 8.1818 A
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8. Determine the voltages at each node of the circuit shown in Fig. 11.
3Ω
10 Ω
3Ω
5Ω
10 V
2Ω
5A
6Ω
1Ω
Fig. 11
The voltage source is converted into current source as shown in Fig. 12 and
nodal analysis is applied.
3Ω
3Ω
V1
1A
5Ω
10 Ω
V2
2Ω
5A
V3
1Ω
6Ω
Fig. 12
By inspection, the node equations are written as:

1 1 1
1 1
1
+ + +
− −
0
 10 5 3 3
3 3


1 1
1 1 1
1

− −
+ +
−

3 3
2 3 3
2


1 1 1
1
0
−
+ +
2
2 1 6


  
 V1
1

  V2  =  5 

 V3
0

Simplifying,

0.9667 −0.6667

 −0.6667
0
0

  
V1
1

1.1667
−0.5   V2  =  5 
V3
0
−0.5 1.6667
(ix)
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Solving Eq. (ix), we obtain
V1 = 8.0805 V
V2 = 10.2166 V
V3 = 3.0649 V
9. For the circuit shown in Fig. 13, determine the impedance at resonance frequency, 10 Hz above resonant frequency and 10 Hz below resonant frequency.
10 Ω
10 μF
0.1 H
VS
Fig. 13
Resonant frequency,
fr =
1
√
2π LC
1
√
2π 0.1 &times; 10 &times; 10−6
= 159.1549 Hz
=
Impedance at resonant frequency (Zr )
At resonant frequency, XL = XC . Therefore, Zr = R = 10 Ω.
Impedance at 10 Hz above resonant frequency (Z1 )
10 Hz above resonant frequency,
Therefore, inductive reactance,
Similarly, capacitive reactance,
Impedance,
f1 = 169.1549 Hz
XL1 = 2πf1 L
= 2π(169.1549)(0.1)
= 106.28 Ω
1
XC1 =
2πf1 C
1
=
2π(169.1549)(10 &times; 10−6 )
= 94.09 Ω
p
Z1 = R2 + (XL1 − XC1 )2
p
= 102 + (106.28 − 94.09)2
= 15.77 Ω
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Impedance at 10 Hz below resonant frequency (Z2 )
10 Hz above resonant frequency,
Therefore, inductive reactance,
f2 = 149.1549 Hz
XL2 = 2πf2 L
= 2π(149.1549)(0.1)
= 93.72 Ω
1
XC1 =
2πf1 C
1
=
2π(149.1549)(10 &times; 10−6 )
= 106.7 Ω
p
Z2 = R2 + (XL2 − XC2 )2
p
= 102 + (93.72 − 106.7)2
= 16.38 Ω
Similarly, capacitive reactance,
Impedance,
10. A series RL circuit with R = 30 Ω and L = 15 H has a constant voltage V =
60 V applied at t = 0 as shown in Fig. 14. Determine the current i, the voltage
across resistor and the voltage across the inductor.
S
60 V
30 Ω
15 H
i(t)
Fig. 14
After closing the switch S, application of KVL to the circuit gives
30i + 15
di
= 60
dt
Taking Laplace transformation on both sides
30I(s) + 15[sI(s) − I(0)] =
60
s
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As there was no initial current through the inductor, I(0) = 0, therefore,
I(s)[30 + 15s] =
or
I(s) =
60
s
4
s(s + 2)
By applying partial fraction technique,
I(s) =
4
A
B
= +
s(s + 2)
s
s+2
(x)
Solving for constants A and B,
A=
4
(s + 2)
= 2 and B =
s=0
4
s
= −2
s=−2
Substituting the values of A and B into Eq. (x), we get
I(s) =
2
2
−
s s+2
Taking inverse Laplace transformation on both sides, the current equation is
obtained as:
i(t) = 2 − 2e−2t A
Voltage across the resistor
VR = Ri(t)
= 30 &times; 2(1 − e−2t )
VR = 60(1 − e−2t ) V
Voltage across the inductor
di
dt
d
= 15 (2 − 2e−2t )
dt
VL = 60e−2t V
VL = L
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11. The circuit shown in Fig. 15 consists of resistance, inductance and capacitance
in series with 100 V DC when the switch is closed at t = 0. Find the transient
current.
S
0.05 H
20 Ω
L
R
20 μF
C
100 V
Fig. 15
After closing the switch S, application of KVL to the circuit gives
Z
di
1
20i + 0.05 +
idt = 100
dt 20 &times; 10−6
Taking Laplace transformation on both sides
20I(s) + 0.05[sI(s) − I(0)] +
1
I(s)
100
=
−6
20 &times; 10
s
s
As there was no initial current through the inductor, I(0) = 0, then,
I(s)
=
20I(s) + 0.05[sI(s)] + 50000
s
50000
I(s) 20 + 0.05s +
=
s
20s + 0.05s2 + 50000
I(s)
=
s
100
s
100
s
100
s
Therefore,
I(s) =
or
100
s
s
0.05s2 + 20s + 50000
2000
s2 + 400s + 1000000
2000
=
(s + 200)2 + 960000
I(s) =
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2000
(s + 200)2 + (979.79)2
979.79
2000
=
979.79 (s + 200)2 + (979.79)2
=
Taking inverse Laplace transformation on both sides
i(t) = 2.0412e−200t sin 979.79t A
12. A symmetrical three-phase, three-wire 440 V supply to a star connected load.
The impedance in each branch are ZR = (2 + j3) Ω, ZY = (1 − j2) Ω and ZB =
(3 + j4) Ω. Find its equivalent delta connected load.
The equivalent delta element values are obtained as follows:
ZRY =
=
ZR ZY + ZY ZB + ZB ZR
ZB
(2 + j3)(1 − j2) + (1 − j2)(3 + j4) + (3 + j4)(2 + j3)
(3 + j4)
ZRY = (3.8 − j0.4) Ω
ZYB =
=
ZR ZY + ZY ZB + ZB ZR
ZR
(2 + j3)(1 − j2) + (1 − j2)(3 + j4) + (3 + j4)(2 + j3)
(2 + j3)
ZYB = (5.23 − j0.85) Ω
ZBR =
=
ZR ZY + ZY ZB + ZB ZR
ZY
(2 + j3)(1 − j2) + (1 − j2)(3 + j4) + (3 + j4)(2 + j3)
(1 − j2)
ZBR = (−3 + j8) Ω
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13. A three phase, balanced delta-connected load of (4 + j8) Ω is connected across
a 400 V, 3 φ balanced supply. Determine the phase currents and line currents.
Phase sequence is RYB.
Phase currents
Line voltage VRY is taken as reference. For a phase sequence of RYB, the phase
currents are:
VRY
ZRY
400∠0◦
=
4 + j8
= 44.72∠ − 63.43◦ A
IRY =
Since the delta-connected load is a balanced one,
IYB = 44.72∠ − 183.43◦ A
IBR = 44.72∠ − 303.43◦ A
Line currents
By applying KCL the line currents are obtained as follows:
IR = IRY − IYB
= (44.72∠ − 63.43◦ ) − (44.72∠ − 183.43◦ )
= 77.46∠ − 33.43◦ A
IY = IYB − IBR
= (44.72∠ − 183.43◦ ) − (44.72∠ − 303.43◦ )
= 77.46∠ − 153.43◦ A
IB = IBR − IRY
= (44.72∠ − 303.43◦ ) − (44.72∠ − 63.43◦ )
= 77.46∠86.57◦ A
14. A symmetrical three-phase, three-wire 400 V supply is connected to a deltaconnected load. Impedances in each branch are ZRY = 10∠30◦ Ω, ZYB =
10∠45◦ Ω and ZBR = 2.5∠60◦ Ω. Find its equivalent star-connected load.
The equivalent star element values are obtained as follows:
ZR =
ZRY ZBR
ZRY + ZYB + ZBR
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=
(10∠30◦ )(2.5∠60◦ )
10∠30◦ + 10∠45◦ + 2.5∠60◦
= 1.13∠50◦ Ω
ZRY ZYB
ZY =
ZRY + ZYB + ZBR
=
(10∠30◦ )(10∠45◦ )
10∠30◦ + 10∠45◦ + 2.5∠60◦
= 4.51∠35◦ Ω
ZYB ZBR
ZB =
ZRY + ZYB + ZBR
=
(10∠45◦ )(2.5∠60◦ )
10∠30◦ + 10∠45◦ + 2.5∠60◦
= 1.13∠65◦ Ω
15. A balanced star connected load having an impedance of (15 + j20) Ω per phase
is connected to 3φ, 440 V, 50 Hz. Find the line current and power absorbed by
Line voltage VRY is taken as reference. For a phase sequence of RYB, the phase
currents are:
VRY = 440∠0◦
440∠0◦ − 30◦
√
VRN =
3
= 254.03∠ − 30◦ V
Line voltage,
Therefore, phase voltage
[In Y-connected networks, phase voltage lags behind the line voltage by 30◦ ]
Since, the star-connected load is a balanced one,
VYN = 254.03∠ − 150◦ V
VBN = 254.03∠ − 270◦ V
In Y-connected networks line current and phase current is same. Therefore,
IR =
=
VRN
Zph
254.03∠ − 30◦
15 + j20
= 10.16∠ − 83.13◦ A
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So,
and
IY = 10.16∠ − 203.13◦ A
IB = 10.16∠ − 323.13◦ A
√
P = 3VL IL cos φ
√
−1 20
= 3 &times; 440 &times; 10.16 &times; cos tan
15
= 4645.78 W
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51430
B.E/B.Tech. DEGREE EXAMINATION, MAY/JUNE 2014
Electrical and Electronics Engineering
EE2151 – CIRCUIT THEORY
1. Determine the currents through the resistances in the bridge network shown in
Fig. 1 using Kirchhoff’s laws.
8
Ω
Ω
4
15 Ω
2
Ω
5
Ω
+–
6V
Fig. 1
The currents through various elements in the given circuit are marked along
with their direction in Fig. 2.
B
Ω
Ω
8
4
I1 ‒ I3
I3
I1
A
C
15 Ω
I2
2
Ω
5
Ω
I1 + I2
I2 + I3
D
+–
6V
Fig. 2
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Applying Kirchhoff’s voltage law around loop ABDA, we get
4I1 + 15I3 − 2I2 = 0
4I1 − 2I2 + 15I3 = 0
Rearranging,
(i)
Similarly, applying Kirchhoff’s voltage law around loop BCDB, we get
8(I1 − I3 ) − 5(I2 + I3 ) − 15I3 = 0
8I1 − 5I2 − 28I3 = 0
or
(ii)
Also, Kirchhoff’s voltage law around loop ADCA gives
2I2 + 5(I2 + I3 ) − 6 = 0
7I2 + 5I3 = 6
or
(iii)
Solving Eqs. (i), (ii) and (iii), we obtain
I1 = 0.49 A
I2 = 0.86 A
I3 = −0.015 A
and
Hence, the current through various elements are obtained by applying Kirchhoff’s current law as follows:
Current
Current
Current
Current
Current
Current
through
through
through
through
through
through
4 Ω resistor,
2 Ω resistor,
15 Ω resistor,
8 Ω resistor,
5 Ω resistor,
6 V source,
I1
I2
I3
I1 − I3
I2 + I3
I1 + I2
= 0.49 A
= 0.86 A
= −0.015 A
= 0.505 A
= 0.845 A
= 1.35 A
2. For the circuit shown in Fig. 3,
(i) Determine the currents in all the branches.
(ii) Calculate the power and power factor of the source.
(iii) Show that power delivered by the source is equal to power consumed by
2 Ω resistor.
j2.5 Ω
+
2Ω
1000&deg; V
– j1 Ω
–
Fig. 3
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Branch currents
Equivalent impedance, Zeq = j2.5 + (2)||(−j1)
= j2.5 +
(2)(−j1)
(2 − j1)
= (0.4 + j1.7) Ω
Therefore, total current from the source,
I=
100∠0◦
V
=
Zeq
0.4 + j1.7
= (13.1148 − j55.7377) = 57.26∠ − 76.76◦ A
By current division rule,
(−j1)
(2 − j1)
= −19.6721 − j16.3935 = 25.6074∠ − 140.19◦ A
(2)
Current through − j1 Ω reactance = (13.1148 − j55.7377) &times;
(2 − j1)
= 32.7869 − j39.3442 = 51.2148∠ − 50.19◦ A
Current through 2 Ω resistor = (13.1148 − j55.7377) &times;
Power &amp; Power factor
Equivalent impedance,
Therefore,
Zeq = (0.4 + j1.7) Ω
1.7
X
=
R
0.4
1.7
−1
φ = tan
= 76.76◦
0.4
tan φ =
or
Therefore,
Hence, power,
cos φ = cos 76.76◦ = 0.229
P = V I cos φ
P = 100 &times; 57.26 &times; 0.229 = 1311.25 W
(iv)
Power consumed by 2 Ω resistor,
P = I 2R
P = 25.60742 &times; 2 = 1311.47 W
(v)
Comparing Eqs. (iv) and (iv) it is observed that power delivered by the source
is equal to power consumed by 2 Ω resistor.
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3. For the circuit shown in Fig. 4, using Thevenin’s theorem, find the current in
the 10 Ω resistor.
10 Ω
10 V
+
+ 4V
3Ω
3Ω
2Ω
1Ω
Fig. 4
The circuit is redrawn by opening the 10 Ω resistance as shown in Fig. 5. Vth
is the voltage between the terminals A and B.
B
A
+
10 V
Vth
+
+ 4V
3Ω
3Ω
2Ω
1Ω
Fig. 5
From Fig. 5,
I1 =
10
=2A
3+2
and
I2 =
4
=1A
3+1
Therefore, voltage across 3 Ω resistor (LHS)
=2&times;3=6 V
Similarly, voltage across 3 Ω resistor (RHS)
=1&times;3=3 V
Hence, Thevenin’s voltage,
Vth = 6 − 3 = 3 V
Thevenin’s resistance, Rth is found by looking back the circuit shown in Fig. 5,
through the terminals A and B after short-circuiting the two voltage sources.
So,
Rth = (2||3) + (3||1)
3&times;1
2&times;3
+
=
2+3
3+1
= 1.95 Ω
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Using the above found Vth and Rth , Thevenin’s equivalent circuit is drawn as
shown in Fig. 6.
1.95 Ω
3V
A
+
10 Ω
B
Fig. 6
From Fig. 6, hence, the current through 10 Ω is obtained as,
Vth
3
=
= 0.251 A
Rth + 10
1.95 + 10
4. Determine the current through 20 V source in the circuit of Fig. 7.
20 V
+
10 Ω
1A
Fig. 7
Using source transformation technique, the current source in the given circuit
is converted into a voltage source as shown in Fig. 8.
10 Ω
20 V
+
+
10 V
Fig. 8
Hence, the current through 20 V source is obtained as,
20 − 10
=1A
10
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5. Calculate the current in the 4 Ω of Fig. 9 using superposition theorem.
3Ω
2Ω
+
10 V
4Ω
2Ω
1A
Fig. 9
10 V source is acting alone
The given circuit is redrawn by open-circuiting 1 A source as shown in Fig. 10.
3Ω
10 V
+
I1
2Ω
4Ω
I2
2Ω
Fig. 10
By mesh analysis method, the mesh equations are written as
7 −4
I1
10
=
−4
8
I2
0
Solving, we get
I1 = 2 A
I2 = 1 A
Therefore, current through 4 Ω resistor due to 10 V source is I 0 = I1 − I2 =
2 − 1 = 1 A.
1 A source is acting alone
The circuit is then redrawn by short-circuiting 10 V source as shown in Fig.
11(a). By source transformation technique, the current source is converted into
voltage source [Fig. 11(b)].
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3Ω
2Ω
4Ω
3Ω
2Ω
2Ω
I3 4 Ω
1A
(a)
2Ω
I4
2V
+
(b)
Fig. 11
By mesh analysis method, the mesh equations are written as
7 −4
I3
0
=
−4
8
I4
2
Solving, we get
I3 = 0.2 A
I4 = 0.35 A
Therefore, current through 4 Ω resistor due to 1 A source is, I 00 = I4 − I3 =
0.35 − 0.2 = 0.15 A.
By the principle of superposition, current through 4 Ω resistor,
I = I 0 + I 00 = 1 + 0.15 = 1.15 A
6. In the circuit of Fig. 12, find the value of R for maximum power transfer. Also,
calculate the maximum power.
15 Ω
12 V
+
10 Ω
R
1A
Fig. 12
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By maximum power transfer theorem, maximum power is transferred to R, if
the value of R is equal to Rth . To find Rth , adjustable resistance R is removed
and the two sources are killed as shown in Fig. 13.
15 Ω
10 Ω
Rth
Fig. 13
From Fig. 13, Rth = 15 Ω. Therefore, the value of R for maximum power
transfer is 15 Ω.
To find the maximum power, Thevenin’s theorem is used here. To find
Thevenin’s voltage, Vth , the circuit shown in Fig. 14 is used.
15 Ω
12 V
+
10 Ω
Vth
1A
Fig. 14
As the circuit carries only 1 A current, this current develops a voltage drop of
15 V across 15 Ω resistor. So, Vth = 12 + 15 = 17 V. Using the above found Vth
and Rth , Thevenin’s equivalent circuit is drawn as shown in Fig. 15.
15 Ω
27 V
+
R =15 Ω
Fig. 15
From Fig. 15, hence, the current through 10 Ω is obtained as,
Vth
27
=
= 0.9 A
Rth + 10
15 + 15
Thus, maxmum power transferred to R = I 2 R = 0.92 &times; 15 = 12.15 W
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7. A RLC series circuit consists of R = 16 Ω, L = 5 mH and C = 2 &micro;F. Calculate
the quality factor at resonance, bandwidth and half-power frequencies.
Quality factor
r
1 L
Q=
R rC
1 5 &times; 10−3
=
= 3.125
16 2 &times; 10−6
Quality factor,
Bandwidth
Bandwidth,
BW =
=
R
2πL
16
= 509.2958 Hz.
2π &times; 5 &times; 10−3
Half-power frequencies
Resonant frequency,
fr =
1
√
2π LC
1
√
2π 5 &times; 10−3 &times; 2 &times; 10−6
= 1591.55 Hz.
R
f1 = fr −
4πL
16
= 1591.55 −
4π &times; 5 &times; 10−3
= 1336.9 Hz.
R
f2 = fr +
4πL
16
= 1591.55 +
4π &times; 5 &times; 10−3
= 1846.2 Hz.
=
Lower half-power frequency,
Upper half-power frequency,
8. Determine the value of RL for resonance in the network shown in Fig. 16.
RL
20 Ω
V
j20 Ω
–j10 Ω
Fig. 16
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The total impedance,
Y = Y1 + Y2
1
1
=
+
RL + j20 20 − j10
RL − j20
1
20 + j10
1
&times;
+
&times;
=
RL + j20 RL − j20
20 − j10 20 + j10
=
RL − j20
20 + j10
+ 2
2
2
RL + 20
20 + 102
RL
20
20
10
−j 2
+ 2
+j 2
2
2
2
+ 20
RL + 20
20 + 10
20 + 102
10
20
RL
20
+j
−
=
+
RL2 + 202 500
500 RL2 + 202
=
RL2
At resonance, the imaginary part of admittance is zero. Hence,
10
20
− 2
=0
500 RL + 202
Solving,
RL = 24.49 Ω
9. A coil having an inductance of 100 mH is magnetically coupled to another coil
having an inductance of 900 mH. The coefficient of coupling between the coils is
0.45. Calculate the equivalent inductance if the coils are connected in (1) series
opposing and (2) parallel opposing.
Mutual inductance,
Eqt. inductance for series opposing,
p
L1 L2
p
= 0.45 (100 &times; 10−3 )(900 &times; 10−3 )
= 0.135
= L1 + L2 − 2M
= (100 &times; 10−3 ) + (900 &times; 10−3 ) − (2 &times; 0.135)
= 0.73 H
L1 L2 − M 2
=
L1 + L2 + 2M
(100 &times; 10−3 )(900 &times; 10−3 ) − 0.1352
=
(100 &times; 10−3 ) + (900 &times; 10−3 ) + (2 &times; 0.135)
= 56.5157 mH
M =k
Leq
Eqt. inductance for parallel opposing, Leq
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10. For the circuit shown in Fig. 17, determine the voltage ratio V1 /V2 , which will
make the current I1 equal to zero.
j2 Ω
5Ω
2Ω


V1
I1
j8 Ω
–
ω=
50
sec
j4 Ω
I2
Fig. 17
V2
–
ω=
50
sec
Current I1 is entering the dot and I2 is leaving the dot, hence the polarities of
mutually induced voltage and self-induced voltage are opposite.
Applying KVL to the first mesh,
5I1 + j8I1 − j2I2 = V1
(5 + j8)I1 − j2I2 = V1
(vi)
Applying KVL to the second mesh,
2I2 + V2 + j4I2 − j2I1 = 0
−j2I1 + (2 + j4)I2 = −V2
Writing Eqs. (vi) and (vii) in matrix form,
5 + j8
−j2
I1
V1
=
−j2 2 + j4
I2
−V2
(vii)
(viii)
Using Cramer’s law, I1 is calculated as:
∆I1
= 0 (given)
∆
∆I1 = 0
V1
−j2
∆I1 =
=0
−V2 2 + j4
I1 =
or
From Eq. (viii),
or
Hence,
(2 + j4)V1 = j2V2
V1
j2
=
V2
2 + j4
= 0.4 + j0.2 = 0.4472∠26.57◦
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11. In the circuit of Fig. 18, the switch is closed at t = 0. Determine the mesh
currents i1 (t) and i2 (t).
t=0
100 Ω
i1(t)
i2(t)
+
50 V
100 Ω
2 μF
Fig. 18
After closing the switch at t = 0, and by applying KVL to the inner loop, we
get
or
or
−50 + 100i1 + 100(i1 − i2 ) = 0
200i1 − 100i2 = 50
i2 = 2i1 − 0.5
(ix)
(x)
Similarly by applying KVL to the outer loop, we get
1
−50 + 100i1 +
2 &times; 10−6
Z
i2 (t)dt = 0
Z
100i1 + 500000
i2 (t)dt = 50
Differentiating on both sides, we get
100
di1
+ 500000i2 = 0
dt
Substituting the value of i2 from Eq. (x), we get
100
or
di1
+ 500000(2i1 − 0.5) = 0
dt
di1
+ 10000i1 = 2500
dt
(xi)
Taking Laplace transformation on both sides, we get
2500
s
2500
I1 (s)[s + 10000] =
s
2500
I1 (s) =
s(s + 10000)
sI1 (s) + 10000I1 (s) =
So,
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By applying partial fraction technique,
I1 (s) =
2500
A
B
= +
s(s + 10000)
s
s + 10000
(xii)
Solving for constants A and B,
A=
2500
s + 10000
= 0.25
and
B=
s=0
2500
s
= −0.25
s=−10000
Substituting the value of A and B into Eq. (xii), we get
I1 (s) =
0.25
0.25
−
s
s + 10000
Taking inverse Laplace transformation on both sides, the current i1 (t) is obtained as:
i1 (t) = 0.25 − 0.25e−10000t A
Substituting the value of i1 (t) into Eq. (x), we obtain
i2 (t) = 2i1 (t) − 0.5
= 2(0.25 − 0.25e−10000t ) − 0.5
i2 (t) = −0.5e−10000t A
12. An RL circuit excited by a sinusoidal source e(t) = 10 sin 100t volts, by closing
the switch at t = 0. Take R = 10 Ω and L = 0.1 H. Determine the current i(t)
flowing through the RL circuit.
Applying KVL to the circuit after closing the switch, we get
di
= 10 sin 100t
dt
di
= 100 sin 100t
100i +
dt
10i + 0.1
Taking Laplace transformation on both sides,
100
s2 + 1002
10000
I(s)(100 + s) =
s2 + 1002
1
10000
I(s) =
s + 100
s2 + 1002
100I(s) + sI(s) = 100
I(s) =
10000
A
Bs + C
=
+ 2
2
2
(s + 100)(s + 100 )
(s + 100) (s + 1002 )
(xiii)
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Solving for the constant,
10000
+ 1002
A=
s2
= 0.5
s=−100
On cross-multiplying Eq. (xiii), we get
10000 = A(s2 + 1002 ) + (Bs + C)(s + 100)
= As2 + 10000A + Bs2 + 100Bs + Cs + 100C
= s2 (A + B) + s(100B + C) + (10000A + 100C)
(xiv)
On equating the coefficients of s2 of Eq. (xiv), we get
A+B=0
B = −A = −0.5
or
On equating the coefficients of s of Eq. (xiv), we get
100B + C = 0
C = −100B = 50
Hence,
Substituting the values of A, B and C into Eq. (xiii), we get
0.5
−0.5s + 50
+ 2
(s + 100) (s + 1002 )
0.5
0.5s
50
− 2
+ 2
=
2
(s + 100) (s + 100 ) (s + 1002 )
0.5
0.5s
50
100
=
− 2
+
(s + 100) (s + 1002 )
100 (s2 + 1002 )
I(s) =
Taking inverse Laplace transformation on both sides, we get
i(t) = 0.5e−100t − 0.5 cos 100t + 0.5 sin 100t
(xv)
0.5
0.
52

0.
52

0.
70
7
A right-angled triangle is constructed using 0.5 and 0.5 as two sides as
shown in Fig. 19.

0.5
Fig. 19
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Referring to Fig. 19,
−1
φ = tan
Also,
0.5
0.5
= 45◦
cos φ =
0.5
0.707
0.5 = 0.707 cos 45◦
or
Similarly,
0.5 = 0.707 sin 45◦
Therefore, Eq. (xv) becomes,
i(t) = 0.5e−100t − 0.707 sin 45◦ cos 100t + 0.707 cos 45◦ sin 100t
= 0.5e−100t + 0.707(cos 45◦ sin 100t − sin 45◦ cos 100t)
i(t) = 0.5e−100t + 0.707 sin(100t − 45◦ ) A
13. A symmetrical 3-phase, 100 V, 3-wire supply feeds an unbalanced star connected
load with impedances of the load has ZR = 5∠0◦ Ω, ZY = 2∠90◦ Ω and ZB =
4∠ − 90◦ Ω. Find the line currents, voltage across the impedances and draw the
phasor diagram. Also calculate the power consumed by the load.
The given values of load impedances are converted into rectangular form.
ZR = 5∠0◦ = 5 + j0
ZY = 2∠90◦ = 0 + j2
ZB = 4∠ − 90◦ = 0 − j4
IR
I1
(5 + j0) Ω
1000&deg; V
IY
j
0+
2)
(0
Ω
(
‒j
4)
Ω
I2
100‒120&deg; V
IB
Fig. 20
Figure 20 shows the arrangement in which rectangular form of impedance values
are marked on it. To calculate the line currents in an unbalanced star connected
load, mesh analysis is used here and the mesh currents are chosen as shown in
Fig. 20.
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By inspection, mesh equations are written as,
5 + j2 −j2
I1
100∠0◦
=
−j2 −j2
I2
100∠ − 120◦
Using Cramer’s rule, I1 and I2 are obtained as follows:
I1 =
100∠0◦
−j2
100∠ − 120◦ −j2
5 + j2 −j2
−j2 −j2
∆I1
=
∆
=
173.21 − j300
8 − j10
I1 = 26.74 − j4.07 = 27.05∠ − 8.66◦ A
I2 =
5 + j2
100∠0◦
−j2
100∠ − 120◦
5 + j2 −j2
−j2 −j2
∆I2
=
∆
=
−76.79 − j333
8 − j10
I2 = 16.56 − j20.93 = 26.69∠ − 51.65◦ A
So, the line currents are obtained as follows:
IR = I1
= 27.05∠ − 8.66◦ A
IY = I2 − I1
= (26.69∠ − 51.65◦ ) − (27.05∠ − 8.66◦ )
= 19.69∠ − 121.13◦ A
IB = −I2
= (−1)26.69∠ − 51.65◦
= 26.69∠128.35◦ A
Therefore,
Voltage across 5∠0◦ Ω impedance = IR &times; 5∠0◦ = 135.25∠ − 8.66◦ V.
Voltage across 2∠90◦ Ω impedance = IY &times; 2∠90◦ = 39.28∠ − 31.13◦ V.
Voltage across 4∠ − 90◦ Ω impedance = IB &times; 4∠ − 90◦ = 106.72∠38.35◦ V.
P = |IR |2 RR + |IY |2 RY + |IB |2 RB
= (27.052 &times; 5) + 0 + 0
= 3658.51 W
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Phasor diagram
The phasor diagram indicating all line currents (= phase currents), line voltages
and phase voltages has been drawn as shown in Fig. 21. It is to be noted that
line voltage, VRY has been taken as reference.
VBR
VB
IBR = IB
128.35&deg;
8.66&deg;
VRY
IRY = IR
121.13&deg;
VR
IYB = IY
VY
VYB
Fig. 21
14. The two wattmeter method is used to measure power in a three-phase delta
connected load. The delta connected load consists of ZRY = (10+j10) Ω, ZYB =
(15 − j15) Ω, and ZBR = (0 + j10) Ω and it is connected to a 400 V, threephase supply of phase sequence RYB. Calculate the reading of wattmeters with
current coil in line R and B.
Phase currents
IRY =
400∠0◦
VRY
= 28.28∠ − 45◦ A
=
ZRY
10 + j10
IYB
VYB
400∠ − 120◦
=
=
= 18.86∠ − 75◦ A
ZYB
15 − j15
IBR
VBR
400∠ − 240◦
=
=
= 17.89∠93.43◦ A
ZBR
20 + j10
Power
Total real power, P = |IRY |2 RRY + |IYB |2 RYB + |IBR |2 RBR
= (28.282 &times; 10) + (18.862 &times; 15) + (17.892 &times; 20)
= 19734.11 W
√
Also,
P = 3VL IL cos φ = W2 + W1
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Total reactive power,
Also,
Therefore,
and
Q = |IRY |2 XRY + |IYB |2 XYB + |IBR |2 XBR
= (28.282 &times; 10) − (18.862 &times; 15) + (17.892 &times; 10)
= 5862.61 var
√
√
Q = 3VL IL sin φ = 3(W2 − W1 )
W2 + W1 = 19734.11
(xvi)
5862.61
W2 − W1 = √
= 3384.78
(xvii)
3
Solving, Eqs. (xvi) and (xvii), we get the reading of two wattmeters as:
W1 = 8174.66 W
W2 = 11559.45 W
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97063
B.E/B.Tech. DEGREE EXAMINATION, NOV./DEC. 2014
Electrical and Electronics Engineering
EE6201 – CIRCUIT THEORY
1. Using node analysis, find the node voltages and the currents through all the
resistors for the circuit shown in Fig. 1.
10 Ω
V1
20 A
25 A
25 A
I1
4Ω
15 A
20 A
1Ω
V2
I2
2Ω
V3 8 A
2A
10 Ω
20 V
Fig. 1
Voltage at node 1,
Voltage at node 2,
Voltage at node 3,
Current through 10 Ω,
Current through 1 Ω,
V1
V2
V3
I1
I2
= 20 &times; 4 = 80 V
= 15 &times; 2 = 30 V
= 2 &times; 10 = 20 V
= 25 − 20 = 5 A
= 25 − 15 = 10 A
2. Find the equivalent resistance between the terminals a and b for the network
shown in Fig. 2.
Using star-delta transformation technique, delta acd is converted into star [Fig.
2] and its values are obtained as,
2&times;4
= 0.6667 Ω
4+6+2
4&times;6
=
=2Ω
4+6+2
Ran =
Rcn
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Rdn =
6&times;2
=1Ω
4+6+2
a
Ω
2
4
Ω
6Ω
c
RT
d
3
Ω
3
Ω
b
Fig. 2
a
a
4Ω
2Ω
0.6667 Ω

n
2Ω
c
d
6Ω
c
1Ω
d
Fig. 3
Replacing the acd delta by its equivalent star, the circuit is redrawn as shown
in Fig. 4.
a
0.6667 Ω
n
2
RT
Ω
1
Ω
c
d
3
Ω
3
Ω
b
Fig. 4
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With reference to Fig. 4, the equal resistance between a and b,
RT = 0.6667 + [(2 + 3)||(1 + 3)] = 2.8889 Ω
3. For the circuit shown in Fig. 5, find the (i) currents in different branches, (ii)
current supplied by the battery and (iii) potential difference between A and B.
Ω
6
2
Ω
10 V
A
B
4Ω
3
Ω
8
Ω
Fig. 5
Mesh analysis is used and the currents are assumed as shown in Fig. vi.
Ω
6
2
Ω
10 V
I1
A
I2
B
4Ω
3
Ω
8
Ω
Fig. 6
Mesh equations by inspection are written as
9 4
I1
10
=
4 18
I2
10
(i)
Solving Eq. (i), we get
I1 = 0.9589 A
I2 = 0.3425 A
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Therefore, current through
2
3
4
6
8
Ω
Ω
Ω
Ω
Ω
resistor
resistor
resistor
resistor
resistor
= I1 = 0.9589 A
= I1 = 0.9589 A
= I1 + I2 = 1.3014 A
= I2 = 0.3425 A
= I2 = 0.3425 A
Current supplied by the battery
= I1 + I2 = 1.3014 A
Potential difference between A and B
A fictitious battery of VAB is connected between A and B as shown in Fig. 7.
I1
I2
Ω
6
2
Ω
A
B
VAB
Fig. 7
With reference to Fig. 7,
−VAB + 2I1 + 6I2 = 0
VAB = 2(0.9589) + 6(0.3425)
= 3.9728 V
4. Find the current I, through the 20 Ω resistor shown in Fig. 8 using Thevenin’s
theorem.
20 V
10 Ω
1Ω
9V
50 V
20 Ω
5Ω
10 V
2Ω
Fig. 8
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Thevenin’s voltage (Vth )
20 Ω resistor is disconnected first from the circuit as shown in Fig. 9 and
the voltage across the open circuit(Vth ) is found by applying KVL to the loop
ABCDA through 50 V and Vth .
Vth − 50 + voltage across 10 Ω + voltage across 5 Ω = 0
Vth − 50 + 20 − 10 = 0
So,
Vth = 40 V
1Ω
9V
A
B
50 V
20 V
10 Ω
‒ Vth +
5Ω
10 V
2Ω
C
D
Fig. 9
Thevenin’s resistance (Rth )
Figure 10(a) is used to compute Thevenin’s resistance, Rth , in which all voltage
sources are short-circuited. Further simplification of Fig. 10(a) leads to Fig.
10(b).
1Ω
1Ω
Rth
10 Ω
5Ω
2Ω
Rth
0Ω
(b)
(a)
Fig. 10
With reference to Fig. 10(b),
Rth =
1&times;0
=0Ω
1+0
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Thevenin’s equivalent circuit
The above found Vth and Rth values are used to draw the Thevenin’s equivalent
circuit as shown in Fig. 11.
Rth= 0 Ω
20 Ω
Vth = 40 V
Fig. 11
From the Thevenin’s equivalent circuit shown in Fig. 11,
I=
Vth
40
=
=2A
Rth + 20
20
5. Find the current through 5 Ω resistor using superposition theorem in the circuit
shown in Fig. 12.
4Ω
I1
32 V
9A
5Ω
2Ω
I2
10 Ω
4A
Fig. 12
32 V source is only active
To apply superposition theorem any one source should be active, and 32 V
source is accordingly kept active first as shown in Fig. 13.
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4Ω
I1
32 V
5Ω
2Ω
I2
10 Ω
Fig. 13
Mesh analysis is used to find the current I 0 through 5 Ω due to 32 V source by
writing mesh equations by inspection as follows:
6 2
I1
32
=
(ii)
2 17
I2
32
Solving Eq. (ii), we obtain
I 0 = I2 = 1.3061 A
9 A source is only active
Fig. 14(a) shows the circuit for the condition that only 9 A current source is
active whereas other two sources are inactive. As 4 Ω and 2 Ω are found parallel
to each other they are simplified to get 1.3333 Ω and is replaced in the circuit
as shown in Fig. 14(b).
4Ω
1.3333 Ω
2Ω

9A
5Ω
9A
5Ω
10 Ω
10 Ω
(b)
(a)
Fig. 14
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Referring to Fig. 14(b) and by applying current division rule, current I 00 through
5 Ω due to 9 A source is
I 00 = 9 &times;
(1.3333 + 10)
= 6.2449 A
(5 + 1.3333 + 10)
4 A source is only active
Fig. 15(a) shows the circuit for the condition that only 4 A current source is
active whereas other two sources are inactive. Again 4 Ω and 2 Ω are found
parallel to each other and the connection is simplified to get 1.3333 Ω which is
replaced in the circuit as shown in Fig. 15(b).
4Ω
1.3333 Ω
2Ω

5Ω
10 Ω
5Ω
10 Ω
4A
4A
(b)
(a)
Fig. 15
Referring to Fig. 15(b) and by applying current division rule, current I 000
through 5 Ω due to 4 A source is
I 000 = 4 &times;
(10)
= 2.449 A
(5 + 1.3333 + 10)
By superposition principle, current through 5 Ω resistor is
I = I 0 + I 00 + I 000 = 1.3061 + 6.2449 + 2.449 = 10 A
6. Impedances Z1 and Z2 are parallel and this combination is in series with an
impedance Z3 , connected to a 100 V, 50 Hz AC supply. Z1 = (5−jXC ) Ω, Z2 =
(5 + j0) Ω, Z3 = (6.25 + j1.25)Ω . Determine the value of capacitance such that
the total current of the circuit will be in phase with the total voltage. Find the
circuit current and power.
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Capacitance at resonance
Total current and total voltage of an AC circuit is in-phase at resonance only.
Therefore, equivalent impedance of the circuit is
Zeq =
5(5 − jXC )
+ (6.25 + j1.25)
5 + 5 − jXC
=
25 − j5XC
+ (6.25 + j1.25)
10 − jXC
=
87.5 + 1.25XC + j(12.5 − 11.25XC )
10 − jXC
=
87.5 + 1.25XC + j(12.5 − 11.25XC ) 10 + jXC
&times;
10 − jXC
10 − jXC
(875 + 11.25XC2 ) + j(125 − 25XC + 1.25XC2 )
=
102 + XC2
=
(875 + 11.25XC2 )
(125 − 25XC + 1.25XC2 )
+
j
102 + XC2
102 + XC2
(iii)
At resonance, the reactive component of the circuit becomes zero, i.e.,
125 − 25XC + 1.25XC2 = 0
Solving the above quadratic equation, we get
XC = 10 Ω
From which, capacitance is found as below:
Capacitive reactance,
XC =
1
2πf C
1
1
=
2πf XC
2π &times; 50 &times; 10
So,
C=
Hence,
C = 318.31 &micro;F
Current at resonance
Impedance at resonance (Zr ) is given by the real part of Zeq and by substituting
the value of XC into the real part of Eq. (iii), we obtain
Zr = 10 Ω
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Therefore, current at resonance
Ir =
V
100
=
= 10 A
Zr
10
Also, power at resonance
Pr = V Ir = 100 &times; 10 = 1000 W
7. The switch in the circuit shown in Fig. 16 is moved from position 1 to 2 at
t = 0. Find the expression for voltage across resistance and capacitor, energy
in the capacitor for t &gt; 0.

S

100 V
i(t)
50 V
5k Ω
+
1 μF
‒
Fig. 16
Determination of initial conditions
Before t = 0, the switch was kept at position 1 for a long time, and by this time
the circuit should have attained steady state conditions. So, the voltage across
the capacitor would be 100 V with the polarity indicated in Fig. 16.
At t = 0 the switch is moved to position 2, and the Kirchhoff’s voltage equation
is
Ri + vC = V
(iv)
[Note: vC is opposite to the polarity mentioned in Fig. 16].
Eq. (iv) is rewritten as
dvC
+ vC = V
dt
Substituting the corresponding values
RC
dvC
+ vC = 50
dt
dvC
0.005
+ vC = 50
dt
5000 &times; 1 &times; 10−6
or
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Applying Laplace transformation on both sides, we get
0.005[svC (s) − vC (0)] + vC (s) =
50
s
as vC (0) being the initial voltage across the capacitor, i.e., 100 V (opposite
polarity with vC (s))
So,
50
s
50
vC (s)[1 + 0.005s] =
− 0.5
s
10000 − 100s
vC (s) =
s(s + 200)
0.005[svC (s) + 100] + vC (s) =
Therefore,
By applying partial fraction technique
10000 − 100s
A
B
= +
s(s + 200)
s
s + 200
vC (s) =
(v)
Solving for constants A and B,
A=
10000 − 100s
(s + 200)
= 50 and B =
s=0
10000 − 100s
s
= −150
s=−200
Substituting the values of A and B into Eq. (v), we get
50
150
−
s
s + 200
Taking inverse Laplace transformation on both sides, the voltage across the
capacitor is obtained as
VC (s) =
vC (t) = 50 − 150e−200t V
From which, equation for current will be
i(t) = C
dvC
dt
d
(50 − 150e−200t )
dt
i(t) = 0.03e−200t A
= (1 &times; 10−6 )
Hence,
Also, voltage across the resistor is
vR (t) = Ri(t)
= 5000 &times; 0.03e−200t
So,
vR (t) = 150e−200t A
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8. For the coupled circuit shown in Fig. 17, find the value of V2 so that the current
I1 = 0.
j2 Ω
5Ω
2Ω
+
1000&deg;V
+
I1
j2 Ω I2
j8 Ω
V2
–
–
Fig. 17
Applying KVL to mesh 1 gives
5i1 + j8i1 + j2i2 = 10∠90◦
(5 + j8)i1 + j2i2 = 10∠90◦
(vi)
Applying KVL to mesh 2 gives
2i2 + j2i2 + j2i1 = V2
j2i1 + (2 + j2)i2 = V2
(vii)
Writing Eqs. (vi) and (vii) in matrix, we have
5 + j8
j2
i1
10∠90◦
=
V2
j2 2 + j2
i2
By Cramer’s rule
I1 =
∆I1
∆
(viii)
Given that I1 = 0. So, from Eq. (viii), ∆I1 = 0.
10∠90◦
j2
=0
∆I1 =
V2
2 + j2
= −20 + j20 − j2V2 = 0
Solving, V2 = (10 + j10) = 14.14∠45◦ V
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9. A 400 V (line to line) is applied to three star connected identical impedances
each consisting of a 4 Ω resistance in series with 3 Ω inductive reactance. Find
(1) line current and (2) total power supplied.
RYB phase sequence is assumed. Phase voltage of R phase,
VRN =
400∠0◦
√
= 230.94∠ − 30◦ V
3
[It is to be noted that in star-connected networks, the phase voltage lags behind the line voltage
by 30◦ ]
In star connected networks, line current and phase current is same. So, line
current in R phase is
IR =
=
VRN
ZRN
230.94∠ − 30◦
4 + j3
= 18.14 − j42.48 = 46.19∠ − 66.87◦ A
IY = 46.19∠ − 186.87◦ A
IB = 46.19∠ − 306.87◦ A
Power factor
−1 3
−1 X
= cos tan
= cos 36.87◦ = 0.8.
cos φ = cos tan
R
4
Total power supplied,
P =
√
√
3VL IL cos φ
= 3 &times; 400 &times; 46.19 &times; 0.8
P = 25601 W
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91433
B.E/B.Tech. DEGREE EXAMINATION, NOV./DEC. 2014
Electrical and Electronics Engineering
EE2151 – CIRCUIT THEORY
1. Two 50 Ω resistors are connected in series. When a resistor R is connected
across one of them, the total circuit resistance is 60 Ω. Calculate the value of
R. If the supply voltage across the above circuit is 60 V, find the current passing
through individual resistance.
50 Ω
50 Ω
R
60 V
Fig. 1
Equivalent resistance of the circuit
Req = 60 = 50 + (50||R)
50 &times; R
10 =
50 + R
500 + 10R = 50R
Hence,
R = 12.5 Ω
Total current supplied by the voltage source =
V
60
=
=1A
Req
60
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Using current division rule,
50
= 0.8 A
50 + 12.5
12.5
=1&times;
= 0.2 A
50 + 12.5
=1&times;
Current through the resistor R(12.5 Ω)
Current through 50 Ω resistor (across R)
Current through 50 Ω (series with 60 V battery)
= total current = 1 A
2. In the circuit given below, obtain the load current.
4Ω
120 V
+
6Ω
12 Ω
3Ω
15 Ω
9Ω
Fig. 2
Mesh analysis method is used to find the current through the load. Mesh
currents I1 , I2 and I3 are assumed as shown in Fig. 3.
4Ω
120 V
6Ω
I1 12 Ω
+
I2 9 Ω
3Ω
I3
15 Ω
Fig. 3


 

16 −12
0
I1
120
 −12
27 −9   I2  =  0 
0 −9 27
I3
0
Solving the above equation, we get
I3 = 2 A
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3. In the circuit given below, obtain the equivalent resistance at AD.
Ω
30
60
Ω
A
90 Ω
B
C
75
Ω
15
Ω
D
Fig. 4
Using star-delta transformation technique, delta ABC is converted into star
[Fig. 5] and its values are obtained as,
60 &times; 30
= 10 Ω
30 + 60 + 90
30 &times; 90
= 15 Ω
=
30 + 60 + 90
90 &times; 60
=
= 30 Ω
30 + 60 + 90
RAN =
RBN
RCN
A
10 Ω

Ω
30
60
Ω
A
15
B
90 Ω
C
Ω
B
N 3
0
Ω
C
Fig. 5
Replacing the ABC delta by its equivalent star, the circuit is redrawn as shown
in Fig. 6.
RDN = (15 + 75)||(30 + 15)
90 &times; 45
=
= 30
90 + 45
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A
10 Ω
N
15
30
Ω
Ω
B
C
75
15
Ω
B
Ω
D
Fig. 6
Therefore, the equal resistance between A and B, RAB = 30 + 10 = 40 Ω
4. Find the load current IL , the load voltage VL and load power PL by the principle
of superposition theorem in the following circuit.
2.1 Ω
11/3 Ω
IL
225 V
RB = 0.9 Ω
6Ω
RL
220 V
RB = 1 Ω
Fig. 7
225 V source is acting alone
Fig. 8(a) is used to find the current I 0 due to 225 V voltage source only.
By network reduction technique, the equivalent resistance (Req1 ) across 225 V
source is:
11
Req1 = (0.9 + 2.1) + 6||
+1
3
6 &times; 14
3
= (3) +
6 + 14
3
= 5.625 Ω
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Therefore, current supplied by 225 V battery =
2.1 Ω
11/3 Ω
11/3 Ω
IL
RL
6Ω
IL
1Ω
0.9 Ω
(a)
RL
6Ω
220 V
RB = 1 Ω
225 V
RB = 0.9 Ω
2.1 Ω
V
225
=
= 40 A.
Req1
5.625
(b)
Fig. 8
Using current division rule, current through 6 Ω due to 225 V voltage source I 0
is
11
+1
0
3
I = 40 &times;
6 + 11
+1
3
= 17.5 A
Fig. 8(b) is used to find the current I 00 due to 220 V voltage source only.
By network reduction technique, the equivalent resistance (Req2 ) across 220 V
source is:
11
+ [6||(2.1 + 0.9)]
Req2 = 1 +
3
14
6&times;3
=
+
3
6+3
20
Ω
=
3
Current supplied by 220 V battery =
V
220
= 20 = 33 A.
Req2
(3)
Using current division rule, current through 6 Ω due to 220 V voltage source I 00
is
0.9 + 2.1
00
I = 33 &times;
0.9 + 2.1 + 6
= 11 A
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By superposition principle
Total current through 6 Ω resistor,
IL = I 0 + I 00
= 17.5 + 11 = 28.5 A
VL = IL &times; RL
= 28.5 &times; 6 = 171 V
PL = IL2 &times; RL
= 4873.5 W
5. A series RL circuit with R = 100 Ω and L = 20 H has a DC voltage of 200
V applied through a switch at t = 0. Find the equation for the current and
voltage across the different elements.
After closing the switch, applying KVL to the circuit
100i + 20
di
= 200
dt
Applying Laplace transformation on both sides, we get
100I(s) + 20[sI(s) − I(0)] =
200
s
As there was no initial current through the inductor, I(0) = 0, therefore,
200
s
200
I(s)[100 + 20s] =
s
10
I(s) =
s(s + 5)
100I(s) + 20[sI(s)] =
or
By applying partial fraction technique
I(s) =
10
A
B
= +
s(s + 5)
s
s+5
(i)
Solving for constants A and B
A=
10
(s + 5)
= 5 and B =
s=0
10
s
= −2
s=−5
Substituting the values of A and B into Eq. (i), we get
I(s) =
2
2
−
s s+5
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Taking inverse Laplace transformation on both sides, the current equation is
obtained as:
i(t) = 2 − 2e−5t A
Voltage across the resistor
VR = Ri(t)
= 100 &times; 2(1 − e−5t )
VR = 200(1 − e−5t ) V
Voltage across the inductor
di
dt
d
= 20 (2 − 2e−5t )
dt
VL = 200e−5t V
VL = L
6. A series RL circuit with R = 100 Ω and L = 1 H has a sinusoidal voltage source
200 sin (500t + φ) applied at a time when φ = 0.
(i) Find the expression for the current.
(ii) At what value of angle φ must the switch be closed so that the current
Expression for the current
After closing the switch, applying KVL to the circuit
100i +
di
= 200 sin 500t
dt
Applying Laplace transformation on both sides, we get
100I(s) + [sI(s) − I(0)] = 200
s2
500
+ 5002
As there was no initial current through the inductor, I(0) = 0, therefore,
100000
+ 5002
100000
I(s) =
(s + 100)(s2 + 5002 )
I(s)[s + 100] =
s2
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By applying partial fraction technique
I(s) =
100000
A
Bs + C
=
+ 2
2
2
(s + 100)(s + 500 )
s + 100 s + 5002
(ii)
Solving for constant A
A=
100000
+ 5002 )
(s2
= 0.3846
s=−100
On cross-multiplying Eq. (ii), we get
100000 = A(s2 + 5002 ) + (Bs + C)(s + 100)
= As2 + A5002 + Bs2 + Cs + 100Bs + 100C
= s2 (A + B) + s(100B + C) + A5002 + 100C
(iii)
On equating the coefficients of s2 of Eq. (iii), we get
A+B=0
B = −A = −0.3846
or
On equating the coefficients of s of Eq. (iii), we get
100B + C = 0
C = 38.46
Hence,
Substituting the values of A, B and C into Eq. (ii), we have
0.3846
−0.3846s + 38.46
+
s + 100
s2 + 5002
0.3846
0.3846s
38.46
=
− 2
+ 2
2
s + 100 s + 500
s + 5002
0.3846
0.3846s
38.46 500
=
− 2
+
2
s + 100 s + 500
500 s2 + 5002
I(s) =
Taking inverse Laplace transformation on both sides, we get
i(t) = 0.3846e−100t − 0.3846 cos 500t + 0.07692 sin 500t
A right-angled triangle is constructed using 0.07692 and 0.3846 as two sides as
seen in Fig. 9.
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22
39
0.

2
46
38
69
22

0.
0.3846
0.
07

0.07692
Fig. 9
Referring to Fig. 9,
−1
φ = tan
Also,
cos φ =
0.3846
0.07692
= 78.69◦
0.07692
0.3922
0.07692 = 0.3922 cos φ = 0.3922 cos 78.69◦
or
Similarly,
0.3846 = 0.3922 sin φ = 0.3922 sin 78.69◦
Hence,
i(t) = 0.3846e−100t − 0.3942 sin 78.69◦ cos 500t + 0.3942 cos 78.69◦ sin 500t
= 0.3846e−100t + 0.3942 (sin 500t cos 78.69◦ − cos 500t sin 78.69◦ )
−100t
− 78.69◦ ) A
i(t) = 0.3846e
{z
} + |0.3942 sin (500t
|
{z
}
Transient part
The above result is also obtained by directly substituting the necessary data
into the complete solution for i(t) as follows:
&quot;
i(t) = p
−Vm
R2 + (ωL)2
sin
φ − tan
−1
ωL
R
#
#
−1 ωL
+ p
sin ωt + φ − tan
R
R2 + (ωL)2
&quot;
−R
Lt
e
Vm
Given that,
Vm = 200;
R = 100 Ω;
L = 1 H;
ω = 500;
φ=0
Substituting these given values into the above equation, we obtain
&quot;
&quot;
#
#
200
500
−1 500
− 100
t
−1
i(t) = p
sin − tan
e 1 + p
sin 500t − tan
100
100
1002 + (500)2
1002 + (500)2
−200
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Thus,
i(t) = 0.3846e−100t + 0.3942 sin (500t − 78.69◦ ) A
as found earlier.
The angle φ at which the switch is closed so that the current directly enters the
The current directly enters the steady state means the transient part has become
zero. That is,
&quot;
#
R
ωL
−Vm
sin φ − tan−1
i(t) = p
e− L t = 0
2
2
R
R + (ωL)
−1 ωL
=0
or
sin φ − tan
R
−1 500 &times; 1
i.e.,
sin φ − tan
=0
100
Hence,
φ = tan−1 (5) = 78.69◦
7. A 3-phase, 400 volts supply is given to a balanced star connected load of
impedance (8 + j6) Ω in each branch. Find the line current, power factor
and total power.
Line voltage VRY is taken as reference. For a phase sequence of RYB,
Line voltage,
Therefore, phase voltage
VRY = 400∠0◦
400∠0◦ − 30◦
√
VRN =
3
= 230.94∠ − 30◦ V
[In Y-connected networks, phase voltage lags behind the line voltage by 30◦ ]
Since, the star-connected load is a balanced one,
VYN = 230.94∠ − 150◦ V
VBN = 230.94∠ − 270◦ V
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In Y-connected networks line current and phase current is same.
Therefore,
IR =
=
VRN
Zph
230.94∠ − 30◦
8 + j6
= 23.09∠ − 66.87◦ A
IY = 23.09∠ − 186.87◦ A
IB = 23.09∠ − 306.87◦ A
So,
and
Power factor
cos φ = cos
−1
tan
X
R
−1 6
= cos tan
= 0.8
8
Total power
P =
√
√
3VL IL cos φ
= 3 &times; 400 &times; 23.09 &times; 0.8
= 12797.77 W
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21390
B.E/B.Tech. DEGREE EXAMINATION, MAY/JUNE 2013
Electrical and Electronics Engineering
EE2151 – CIRCUIT THEORY
1. Using mesh analysis, determine the current through 1 Ω resistor in the circuit
shown in Fig. 1.
10 Ω
2Ω
5Ω
I1
50 V
I2
+
10 V
+
5V
1Ω
+
3Ω
I3
Fig. 1
By mesh analysis method, the mesh equations are written as


 

18 −5 −3
I1
50
 −5
8 −1   I2  =  −10 
−3 −1
4
I3
−5
The currents I2 and I2 are found by using Cramer’s rule as follows:
∆=
18 −5 −3
−5
8 −1
−3 −1
4
= 18(32 − 1) + 5(−20 − 3) − 3(5 + 24) = 356
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∆I2 =
18
50 −3
−5 −10 −1
−3 −5
4
= 18(−40 − 5) − 50(−20 − 3) − 3(25 − 30) = 355
∆I3 =
18 −5
50
−5
8 −10
−3 −1 −5
= 18(−40 − 10) + 5(+25 − 30) + 50(5 + 24) = 525
Therefore,
∆I2
355
=
= 0.9972
∆
356
∆I3
525
I3 =
=
= 1.4747
∆
356
I2 =
Hence, the current flowing through 1 Ω resistor = I2 − I3 = (0.9972 − 1.4747) =
−0.4775 A.
2. Three loads A, B and C are connected in parallel to a 240 V source. Load
A takes 9.6 kW, Load B takes 60 A and Load C has a resistance of 4.8 Ω.
Calculate (1) RA and RB (2) the total current (3) the total power, and (4)
equivalent resistance.
(1) Resistances RA and RB
IA =
P
V
9.6 &times; 103
= 40 A
240
V
RA =
IA
=
240
=6Ω
40
V
RB =
IB
=
=
240
=4Ω
60
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(2) Total current
IC =
=
V
RC
240
= 50 A
4.8
Therefore, the total current, IT = IA + IB + IC
= 40 + 60 + 50 = 150 A
(3) Total power
Power, P = V IT = 240 &times; 150 = 36000 W
(4) Equivalent resistance
1 1
1
1
= + +
Req
6 4 4.8
= 0.625 Ω
1
= 1.6 Ω
Therefore, the equivalent resistance, Req =
0.625
3. By applying nodal analysis for the circuit shown in Fig. 2, determine the power
output of the source and the power in each resistor of the circuit.
3Ω
20∠30&deg; V
–j4 Ω
VA
j5 Ω
2Ω
Fig. 2
Writing KCL at node A,
VA − 20∠30◦
VA
VA
+
+
=0
3
−j4 2 + j5
1
1
1
20∠30◦
VA
−
+
=
3 j4 2 + j5
3
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VA [0.33 + j0.25 + 0.069 − j0.17] = 6.67∠30◦
Hence,
VA =
6.67∠30◦
= 16.39∠18.66◦
0.399 + j0.08
Current through 2 Ω resistor =
=
VA
2 + j5
16.39∠18.66◦
5.385∠68.2◦
= 3.04∠ − 49.54◦ A
Therefore, the power consumed by 2 Ω resistor = 3.042 &times; 2 = 18.48 W
VA − 20∠30◦
3
16.39∠18.66◦ − 20∠30◦
=
3
Current through 3 Ω resistor =
= 1.7∠ − 111◦ A
Therefore, the power consumed by 3 Ω resistor = 1.72 &times; 3 = 8.67 W
Hence, the power output of the source = 18.48 + 8.67 = 27.15 W
4. Use the superposition theorem to find the current through 4 Ω resistor in the
circuit shown in Fig. 3.
4Ω
50&deg; A
‒j2 Ω
+
j10 Ω
2090&deg; V
‒
8Ω
‒j2 Ω
Fig. 3
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Case (i) 5∠0◦ A current source is acting alone
The 5∠0◦ A current source is made only active and by short-circuiting the
20∠90◦ V voltage source, the circuit is redrawn as shown in Fig. 4.
4Ω
‒j2 Ω
50&deg; V
j10 Ω
‒j2 Ω
8Ω
Fig. 4
The ∆-connected network (comprising 8 Ω, j10 Ω and−j2 Ω) is converted
to its equivalent star-network as follows:
(j10)(8)
= 5 + j5
(8 + j8)
(j10)(−j2)
Z2 =
= 1.25 − j1.25
(8 + j8)
(−j2)(8)
Z3 =
= −1 − j1
(8 + j8)
Z1 =
Using the star equivalent impedance values, the circuit (Fig. 4) is redrawn
as shown in Fig. 5.
4Ω
‒j2 Ω
j1
‒
j5
25
+
1.
5
.2
5
50&deg; A
‒1 ‒ j1
Fig. 5
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The current through 4 Ω resistor due to 5∠0◦ A current source using current division rule is found as,
I 0 = 5∠0◦ &times;
[(1.25 − j1.25) + (−j2)]
[(1.25 − j1.25) + (−j2) + (−1 − j1) + (4)]
= (2.6471 − j1.1765) = 2.8967∠ − 23.96◦ A
Case (ii) 20∠90◦ V voltage source is acting alone
Similarly, only 20∠90◦ V source alone is kept in the circuit and the 5∠0◦
A current source is open-circuited as shown in Fig. 6.
4Ω
‒j2 Ω
+
j10 Ω
2090&deg; V
I2
8Ω
I1
‒j2 Ω
‒
Fig. 6
By inspection, the mesh equations are written as,
8 + j8
j2
I1
0
=
j2 4 − j4
I2
20∠90◦
Using Cramer’s rule, I2 which is nothing but I 00 is found as,
8 + j8
0
◦
j2
20∠90
∆I2
00
=
I = I2 =
∆
8 + j8
j2
j2 4 − j4
(20∠90◦ )(8 + j8)
=
(8 + j8)(4 − j4) − (j2)2
I 00 =
−160 + j160
68
= (−2.3529 + j2.3529) = 3.3276∠135◦ A
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By superposition theorem, the current flowing through 4 Ω resistor is,
I = I 0 + I 00 = (2.8967∠ − 23.96◦ ) + (3.3276∠135◦ )
I = 6.12∠ − 35.22◦ A
5. Find the current through branch a-b of the network shown in Fig. 7, using
Thevenin’s theorem.
j5 Ω
5Ω
a
+
3Ω
100&deg; V
5Ω
j4 Ω
–
b
Fig. 7
To find Thevenin’s voltage, Vth
The resistance, 5 Ω connected across a-b is first removed and the circuit shown
in Fig. 8(a) is used to find Thevenin’s voltage Vth .
j5 Ω
5Ω
j5 Ω
5Ω
a
+
a
3Ω
3Ω
 Rth
Vth
100&deg; V
j4 Ω
–
j4 Ω
b
(a)
b
(b)
Fig. 8
By voltage division rule,
3 + j4
5 + 3 + j4
= (5 + j2.5) = 5.59∠26.56◦ V
Vth = Vab = 10∠0◦ &times;
To find Thevenin’s impedance, Zth
The 10∠0◦ V voltage source is short-circuited as shown in Fig. 8(b) and the
Thevenin’s impedance Zth is found by looking back the circuit through the
terminals a-b.
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Zth = (j5) + (5||3 + j4)
= (2.5 + j6.25) Ω
Zth
2.5 Ω j6.25 Ω
Vth = 5.5926.56&deg; V
a
5Ω
b
Fig. 9
Using the above found values of Vth and Zth , the Thevenin’s equivalent circuit
is drawn as shown in Fig. 9. Thus, from Fig. 9, the current through branch
a-b is,
Vth
5.59∠26.56◦
=
=
5 + Zth
5 + 2.5 + j6.25
Iab = (0.5573 − j0.1312) = 0.5726∠ − 13.25◦ A
6. A series circuit with R = 10 Ω, L = 0.1 H and C = 50 &micro;F has an applied
voltage 50∠0◦ V with a variable frequency. Find (1) the resonant frequency, (2)
the value of frequency at which maximum voltage occurs across the inductor
(3) the value of frequency at which maximum voltage across capacitor and (4)
the quality factor.
1
Resonant frequency,
fr = √
2π LC
1
= √
2π 0.1 &times; 50 &times; 10−6
= 71.18 Hz.
The frequency at which maximum voltage occurs across the inductor is,
s
−1
R2 C
fL = (fr )
1−
2L
s
−1
102 &times; 50 &times; 10−6
= (71.18)
1−
2 &times; 0.1
= 72.09 Hz.
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The frequency at which maximum voltage occurs across the capacitor is,
r
1
1
R2
fC =
−
2π r LC 2L
1
1
102
=
−
2π 0.1 &times; 50 &times; 10−6 2 &times; 0.1
= 71.09 Hz.
r
1 L
Quality factor,
Q=
R rC
0.1
1
=
= 4.47
10 50 &times; 10−6
7. Consider the single tuned circuit shown below and determine,
(i) the resonant frequency,
(ii) the output voltage at resonance, and
(iii) the maximum output voltage.
Assume RS &gt;&gt; ωr L1 and k = 0.9.
M
10 Ω
15 V
RS
R2
L1
1 μH
L2
100 μH
–
0.1 μF

10 Ω
vo
Fig. 10
Mutual inductance,
At resonance,
Resonant frequency,
p
L1 L2
p
= 0.9 (1 &times; 10−6 )(100 &times; 10−6 )
= 9 &micro;H
1
ωr = √
L2 C
1
=p
−6
(100 &times; 10 )(0.1 &times; 10−6 )
M =k
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Output voltage,
Vo =
M Vi
C[R1 R2 + ω 2 M 2 ]
9 &times; 10−6 &times; 15
0.1 &times; 10−6 [(10)(10) + (316227.766)2 (9 &times; 10−6 )2 )]
= 12.49 V
=
Maximum output voltage,
Vo(max) =
V
√i
2ωr C R1 R2
15
√
2 &times; 316227.776 &times; 0.1 &times; 10−6 10 &times; 10
= 23.72 V
=
8. The two wattmeter method produces wattmeter readings P1 = 1560 W and
P2 = 2100 W, when connected to a delta connected load. If the line voltage is
220 V, calculate:(1) the per-phase active power (2) the per-phase reactive power
(3) the power factor and (4) the phase impedance.
Per-phase active power
Active power,
Also, active power,
Therefore, per-phase active power
√
P = 3VL IL cos φ
P = W1 + W2
= 1560 + 2100 = 3660 W
3660
= 1220 W
=
3
Per-phase reactive power
√
3VL IL sin φ
√
Also, reactive power,
Q = 3(W2 − W1 )
√
= 3(2100 − 1560) = 935.3
935.3
Therefore, per-phase reactive power
=
= 311.77 var
3
Power factor
√ W2 − W1
−1
Power factor,
cos φ = cos tan
3
W + W2
1
√
2100 − 1560
−1
= cos tan
3
1560 + 2100
−1
= cos tan 0.2555
= 0.97
Total reactive power
Q=
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Power,
P =
√
3VL IL cos φ
P
IL = √
3VL cos φ
3660
=√
= 9.9
3(220 &times; 0.97)
or
IL
For a ∆-connected load, Iph = √ and Vph = VL .
3
Vph
Iph
√ VL
= 3
I
L
√ 220
= 3
= 38.49 Ω
0.9
= 38.49∠14.33◦ Ω
Therefore, impedance per phase Zph =
In polar form,
Zph
9. An unbalanced star-connected load has balanced voltage of 100 V and RBY
phase sequence. Calculate the line currents and the neutral current.
Take: ZA = 15 Ω, ZB = (10 + j5) Ω, ZC = (6 − j8) Ω.
In star-connected networks, line current and phase current is same. Hence, for
the given phase sequence of RBY system,
1
IR = IRN = (VRN )
ZA
100∠(0◦ − 30◦ )
1
√
=
15
3
= 3.849∠ − 30◦ A
[In star-connected networks, the phase voltage lags behind the line voltage by 30◦ ]
IY = IYN = (VYN )
1
ZB
100∠(−240◦ − 30◦ )
√
=
3
= 5.164∠63.43◦ A
1
10 + j5
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IB = IBN = (VBN )
=
1
ZC
100∠(−120◦ − 30◦ )
√
3
1
6 − j8
= 5.7735∠ − 96.87◦ A
The neutral current,
IN = −(IR + IY + IB )
= −(3.849∠ − 30◦ + 5.164∠63.43◦ + 5.7735∠ − 96.87◦ )
IN = 5.81∠148.47◦ A
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31390
B.E/B.Tech. DEGREE EXAMINATION, NOV./DEC. 2013
Electrical and Electronics Engineering
EE2151 – CIRCUIT THEORY
1. In the circuit of Fig. 1 find the current I by mesh method.
15 Ω
5Ω
I
10 V
+
–
+
–
1.25 Ω
20 V
Fig. 1
The direction of mesh currents I1 and I2 are assumed as shown in Fig. 2.
15 Ω
5Ω
I
10 V
+
–
I1
1.25 Ω
I2
+
–
20 V
Fig. 2
By inspection, the mesh equations are written as,
6.25 −1.25
I1
10
=
−1.25 16.25
I2
−20
Solving the above matrix equation, we get
I1 = 1.375 A
I2 = −1.125 A
Therefore, current I = I1 − I2 = 2.5 A
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2. Write the nodal equations for the network of Fig. 3. Hence find the potential
difference between nodes 2 and 4.
4A
3A

1Ω


2Ω
4Ω
5
Ω
5
Ω

4A
Fig. 3
Taking node 4 as reference, by inspection the node equations are written as,


1
1 1
+
−
0

 5 1

 
1
 V1

1

1 
1 1 1
  V2  =  −1 
 −1
+ +
−


1
1
2
4
2
 V3

0

1 1 
1
− +
0
−
2
5 2
Simplifying,


 

1.2
−1
0
V1
1
 −1 1.75 −0.5   V2  =  −1 
0 −0.5
0.7
V3
0
Solving the above matrix equation, we get
V1 = 0.5851 V
V2 = −0.2978 V
V3 = −0.2128 V
Hence, the potential difference between nodes 2 and 4, V24 = −0.2978 V
3. Using star-delta transformation, in the following wheatstone bridge circuit of
Fig. 4, find (i) the equivalent resistance between P and Q (ii) the total current
and (iii) the current through the 18 Ω resistor.
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2Ω
A
6
Ω
P
12
Ω
18 Ω
B
240 V
6
21
Ω
Q
C
Ω
D
Fig. 4
Equivalent resistance between P and Q
Using star-delta transformation technique, delta ABC is converted into star
[Fig. 5] and its values are obtained as,
6 &times; 12
=2Ω
6 + 12 + 18
12 &times; 18
=
=6Ω
6 + 12 + 18
18 &times; 6
=
=3Ω
6 + 12 + 18
RAN =
RBN
RCN
A
12
6Ω
Ω
A
2Ω

6Ω
B
18 Ω
C
B
N 3
Ω
C
Fig. 5
Replacing the ABC delta by its equivalent star, the circuit is redrawn as shown
in Fig. 6. Both 6 Ω resistances are in series. Also in series are 21 Ω and 3 Ω.
These two series combinations together form one parallel connection and its
equivalent value is,
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RDN = (6 + 6)||(21 + 3)
=
12 &times; 24
=8
12 + 24
2Ω
A
P
2Ω
N
6
240 V
3
Ω
Ω
B
C
6
Ω
Q
21
Ω
D
Fig. 6
Therefore, the equal resistance between P and Q, RPQ = 8 + 2 + 2 = 12 Ω
Total current
The total current, I =
VPQ
240
=
= 20 A
RPQ
12
Current through the 18 Ω resistor
The current through the 18 Ω resistor is found by using mesh analysis. With
reference to Fig. 7, the mesh equations by inspection are written as,


 

20 −12 −6
I1
240
 −12
36 −18   I2  =  0 
−6 −18
45
I3
0
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2Ω
P
A
6
Ω
12
I2
I1
18 Ω
B
I3
Ω
Ω
6
C
21
240 V
Ω
Q
D
Fig. 7
Solving the above matrix equation, we get
I1 = 20 A
I2 = 10 A
I3 = 6.6667 A
Hence, the current through 18 Ω resistor = I2 − I3 = 10 − 6.6667 = 3.3333 A
4. Find the current through the 10 Ω resistor in Fig. 8, using Thevenin’s theorem.
20
Ω
10 Ω
12
Ω
B
Ω
A
16
200 V
30
Ω
Fig. 8
The given circuit is redrawn first by opening the 10 Ω resistor between A and
B as shown in Fig. 9. Now, Thevenin’s voltage Vth is nothing but the voltage
between terminals A and B.
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20
Ω
30
Ω
B
Ω
16
12
Ω
A
200 V
Fig. 9
200
= 6.25 A
20 + 12
200
Current through 30 Ω and 16 Ω branch =
= 4.35 A
30 + 16
Current through 20 Ω and 12 Ω branch =
Therefore,
Voltage across 12 Ω resistor = 12 &times; 6.25 = 75 V
Voltage across 16 Ω resistor = 16 &times; 4.35 = 69.6 V
Hence, Thevenin’s voltage, Vth = 75 − 69.6 = 5.4 V
20 Ω
A
12 Ω
30 Ω
20 Ω
B
A
12 Ω
16 Ω
30 Ω
B
16 Ω
Fig. 10
Figures 10 and 11 are used to calculate Rth .
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A
A
12 Ω
20 Ω
7.5 Ω
16 Ω
30 Ω
10.43 Ω
B
B
Fig. 11
Rth = (12||20) + (16||30)
12 &times; 20
16 &times; 30
=
+
12 + 20
16 + 30
= 7.5 + 10.43
= 17.93 Ω
Using the values of Vth and Rth found above, the Thevenin’s equivalent circuit
is drawn as shown in Fig. 12.
Rth = 17.93 Ω
A
Vth = 5.4 V
+
10 Ω
B
Fig. 12
Thus, current through 10 Ω resistor =
Vth
5.4
=
= 0.1933 A
Rth + RL
17.93 + 10
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5. A RLC series circuit has R = 60 Ω, L = 160 mH and C = 160 &micro;F. Find the
resonant frequency, under resonant condition obtain the current, power and the
voltage drops across the various elements if the applied voltage is 300 V.
1
Resonant frequency,
fr = √
2π LC
=
1
p
2π (160 &times; 10−3 )(160 &times; 10−6 )
= 31.46 Hz.
At resonant condition, the sum of reactances is zero, therefore,
Current at resonance,
I=
300
V
=
=5A
R
60
P = I 2 R = 52 &times; 60 = 1500 W
VR = IR = 5 &times; 60 = 300 V
Power at resonance,
Voltage drop across resistor,
At resonant condition, the voltage drops across inductor and capacitor are equal,
therefore,
r
1 L
VL = VC = V Q = V
R C
!
r
1 160 &times; 10−3
= (300)
60 160 &times; 10−6
= 158.11 V
6. In the series RL circuit shown in Fig. 13, the switch is closed on position 1
at t = 0. At t = 1 millisecond, the switch is moved to position 2. Obtain the
equations for the current in both intervals and draw the transient curve.
50 Ω

0.2 H

100 V
50 V
Fig. 13
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At t = 0, the switch is connected to position 1 and the differential equation is
50i + 0.2
di
= 100
dt
Taking Laplace transformation on both sides, we get
50I(s) + 0.2[sI(s) − I(0)] =
100
s
(i)
At t = 0, i(t) = 0, and in s-domain, I(0) = 0. Therefore, Eq. (i) becomes
100
s
100
I(s)[50 + 0.2s] =
s
50I(s) + 0.2sI(s) =
100
s(0.2s + 50)
100
500
=
=
0.2s(s + 250)
s(s + 250)
I(s) =
By applying partial fraction technique,
=
A
B
+
s
(s + 250)
(ii)
Solving for the constants A and B,
A=
500
s + 250
=2
and
B=
s=0
500
s
= −2
s=−250
Substituting the values of A and B into Eq. (ii), we get
I(s) =
2
2
−
s (s + 250)
Taking inverse Laplace transformation on both sides, the current equation is
obtained as:
i(t) = 2 − 2e−250t A
(iii)
The switch is kept at this position for 0.001 s. Therefore, at t = 0.001 s,
the current through the circuit is determined by putting t = 0.001 in Eq. (iii),
i.e. at t = 0.001 s.
i(t) = 2 − 2e−250(0.001) = 0.4424 A
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This value becomes initial current for the circuit when the switch is put in
position 2. Then, the switch is moved to position 2 at t = 0.001 s. Due to this
action, 50 V source is connected in the circuit and 100 V source is disconnected.
This moment (t = 0.001 s) can be redefined as t0 = 0.
Applying KVL,
50i + 0.2
di
= 50
dt0
Taking Laplace transformation on both sides, we get
50I(s) + 0.2[sI(s) − I(0)] =
50
s
(iv)
At t0 = 0, I(0) = 0.4424 A. Therefore, Eq. (iv) becomes
50
s
50
I(s)[50 + 0.2s] =
+ 0.08848
s
50 + 0.08848s
I(s) =
s(0.2s + 50)
50I(s) + 0.2sI(s) − 0.08848 =
=
50 + 0.08848s
0.2s(s + 250)
=
250 + 0.4424s
s(s + 250)
By applying partial fraction technique,
=
A
B
+
s
(s + 250)
(v)
Solving for the constants A and B,
A=
250 + 0.4424s
s + 250
=1
s=0
and
B=
250 + 0.4424s
s
= −0.5576
s=−250
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i
2A
1A
t
0
t = 0.001
(or)
t′ = 0
Fig. 14
Substituting the values of A and B into Eq. (v), we get
I(s) =
1
0.5576
−
s (s + 250)
Taking inverse Laplace transformation on both sides, the current equation is
obtained as:
0
i(t) = 1 − 0.5576e−250t A
or
i(t) = 1 − 0.5576e−250(t−0.001) A
Therefore,
i(t) = 2 − 2e−250t A
= 1 − 0.5576e
−250(t−0.001)
for 0 &lt; t &lt; 0.001
A
for t &gt; 0.001
The transient current is shown in Fig. 14.
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7. In Fig. 15, find the current. Assume initial charge on the capacitor is zero.
t=0
v = 200 sin 500t
100 Ω
25 μF
i(t)
Fig. 15
Applying KVL to the circuit after closing the switch at t = 0, we get
Z
1
idt = 200 sin 500t
100i +
25 &times; 10−6
Taking Laplace transformation on both sides,
500
1
100I(s) + 40000 I(s) = 200 &times; 2
s
s + 5002
40000
100000
I(s) 100 +
= 2
s
s + 5002
s
100000
I(s) =
100s + 40000
s2 + 5002
100s
=
(s + 400)(s2 + 5002 )
By applying partial fraction technique,
I(s) =
100s
A
Bs + C
=
+ 2
2
2
(s + 400)(s + 500 )
s + 400 s + 5002
(vi)
Solving for the values of A, B and C, we get
and
A = −0.9756
B = 0.9756
C = 609.76
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Substituting the values of A, B and C into Eq. (vi), we obtain
−0.9756 0.9756s + 609.76
+
s + 400
s2 + 5002
−0.9756
0.9756s
609.76
=
+ 2
+ 2
2
s + 400
s + 500
s + 5002
−0.9756
0.9756s
500
609.76
=
+ 2
&times;
+
s + 400
s + 5002
500
s2 + 5002
Taking inverse Laplace transformation on both sides, we get
I(s) =
i(t) = −0.9756e−400t + 0.9756 cos 500t + 1.2195 sin 500t
(vii)
0.
97
56 2
1
.2
19
52
1
.5
61
7
A right-angled triangle is constructed using 0.9756 and 1.2195 as two sides
as shown in Fig. 16.
0.9756

1.2195
Fig. 16
Referring to Fig. 16,
−1
φ = tan
Also,
or
0.9756
1.2195
= 38.66◦
cos φ =
1.2195
1.5617
1.2195 = 1.5617 cos 38.66◦
Similarly,
0.9756 = 1.5617 sin 38.66◦
Therefore, Eq. (vii) becomes,
i(t) = −0.9756e−400t + 1.5617 sin 38.66◦ cos 500t + 1.5617 cos 38.66◦ sin 500t
= −0.9756e−400t + 1.5617(sin 38.66◦ cos 500t + cos 38.66◦ sin 500t)
i(t) = −0.9756e−400t + 1.5617 sin(500t + 38.66◦ ) A
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11358
B.E/B.Tech. DEGREE EXAMINATION, NOV./DEC. 2012
Electrical and Electronics Engineering
EE2151 – CIRCUIT THEORY
1. Determine the current IL in the circuit shown below.
4V
3Ω
3Ω
3Ω
IL
8V
6V
5Ω
1Ω
1Ω
Fig. 1
The mesh currents are assumed in the directions as indicated in Fig. 2.
4V
3Ω
I3
3Ω
3Ω
IL
I1
8V
5Ω
1Ω
6V
I2
1Ω
Fig. 2
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By mesh analysis method, the mesh equations are written as


  
9 5 −3
I1
8
 5 9




3
I2 = 6 
−3 3
9
I3
4
Using Cramer’s rule, I1 is found as, I1 =
∆I1 =
∆I1
.
∆
8 5 −3
6 9
3
4 3
9
= 8(81 − 9) − 5(54 − 12) − 3(18 − 36) = 420
∆=
9 5 −3
5 9
3
−3 3
9
= 9(81 − 9) − 5(45 + 9) − 3(15 + 27) = 252
∆I1
420
Therefore, I1 =
=
= 1.6667
∆
252
Similarly, I2 is found as, I2 =
∆I2 =
∆I2
.
∆
9 8 −3
5 6
3
−3 4
9
= 9(54 − 12) − 5(54 − 12) − 3(20 + 18) = −168
Therefore,
I2 =
−168
∆I2
=
= −0.6667
∆
252
Hence, the current IL = I1 + I2 = (1.6667 − 0.6667) = 1 A.
2. For the circuit shown in figure below, determine the total current IT , phase
angle and power factor.
100 μF
10 Ω
IT
50 V,
100 Hz
30 Ω
0.1 H
Fig. 3
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Capacitive reactance of 100 &micro;F,
XC =
1
2πf C
1
2π(100)(100 &times; 10−6 )
= 15.92 Ω
Inductive reactance of 0.1 H,
XL = 2πf L
= 2π(100)(0.1)
= 62.83 Ω
Equivalent impedance of the circuit
= (10 − j15.92) + (30)||(j62.83)
= (10 − j15.92) + (24.43 + j11.66)
= (34.43 − j4.26) Ω
50
Total current,
IT =
34.43 − j4.26
= (1.43 + j0.17) = 1.44∠7.05◦
So, phase angle,
φ = 7.05◦
and power factor,
cos φ = cos 7.05◦ = 0.99.
=
3. For the circuit shown in figure below, determine the value of V2 such that the
current through (3 + j4) Ω impedance is zero.
4Ω
V1 = 200&deg;
I1 j3 Ω
j4 Ω
3Ω
I2 ‒j5 Ω
5Ω
I3
V2
Fig. 4
Referring to Fig. 4, the current through (3 + j4) Ω is I2 . Mesh analysis is used
here to find out the expression for I2 which will be then equated to zero.


 

4 + j3
−j3
0
I1
20∠0◦
 −j3 3 + j2
−j5   I2  =  0 
0
−j5 5 − j5
I3
V2
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Using Cramer’s rule, I2 is found as,
∆I2
=0
(given)
∆
∆I2 = 0


4 + j3 20∠0◦
0
0
−j5  = 0
∆I2 =  −j3
0
V2
5 − j5
I2 =
or
So,
or
= (4 + j3)(−V2 )(−j5) − (20∠0◦ )(−j3)(5 − j5) = 0
= V2 (−15 + j20) − (−300 − j300) = 0
−300 − j300
V2 =
−15 + j20
V2 = (−2.4 + j16.8) = 16.97∠98.13◦
Thus, the value of V2 which makes the current through (3 + j4) Ω impedance
be zero is 16.97∠98.13◦ V
4. Using source transformation, replace the current source in the circuit shown
below by a voltage source and find the current delivered by the 50 V voltage
source.
3Ω
50 V
Ω
10
2
A
5Ω
+
+
10 V
Fig. 5
The 10 A current source is converted into a voltage source as shown in Fig. 5.
I1
I1 ‒ I2
I2
5Ω
50 V
2Ω
+
20 V
3Ω
+
+
10 V
Fig. 6
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Applying KVL to the two meshes with the assumed current directions, we get
−50 + 5I1 + 2I2 + 20 = 0
5I1 + 2I2 = 30
−20 − 2I2 + 3(I1 − I2 ) + 10 = 0
3I1 − 5I2 = 10
or
or
(i)
(ii)
Solving Eqs. (iii) and (iv), we obtain the current delivered by 50 V source,
I1 = 5.4838 A.
5. Calculate the equivalent resistance Rab when all the resistance values are equal
to 1 Ω for the circuit shown below.
a
c
Req 
e
b
d
Fig. 7
Since the ∆ - connected resistances are having of equal values, their star equivalent resistances are obtained as:
Rstar =
1Ω
a
1
1&times;1
=
1+1+1
3
c
a
c
1/3
1Ω
1Ω
1/3
Ω
Ω

1/3 Ω
e
e
Fig. 8
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Thus, the two ∆ − networks ace and bde are converted into equivalent star
networks and are replaced in the original circuit as shown in Fig. 8.
a
1/
3
c
Ω
3
1/
Ω
1/3 Ω
1Ω
1Ω
e
1/3 Ω
3
1/
Ω
1/
b
3
Ω
d
Fig. 9
Figures shown below illustrate the step-by-step reduction of given network.
Hence, the equivalent resistance is Rab = 0.5333 Ω.
1/3 Ω
1/3 Ω
1/3 Ω
a
c
1Ω
2/3 Ω
a
1Ω
1/3 Ω
1Ω
1/3 Ω
2/3 Ω
5/3 Ω
1/3 Ω
b
b
d
1/3 Ω
a
a
1Ω
10/21 Ω
a
1Ω
8/7 Ω
0.5333 Ω
1/3 Ω
b
b
b
Fig. 10
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6. Verify Reciprocity theorem for the circuit shown below.
3Ω
12 Ω
14 Ω
100 V
4Ω
4Ω
Fig. 11
Let the current through the load 4 Ω be Ix and is found using mesh current
analysis.
3Ω
12 Ω
IX
100 V
I1 14 Ω
I2
4Ω
IX
4Ω
Fig. 12
The mesh equations by inspection [Fig. 12] are written as,


 

17 −14
0
I1
100
 −14
30 −4   I2  =  0 
0 −4
8
Ix
0
x
.
Using Cramer’s rule Ix is found as, Ix = ∆I
∆


17 −14 100
30
0  = 5600
∆Ix =  −14
0 −4
0


17 −14
0
30 −4  = 2240
∆ =  −14
0 −4
8
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∆Ix
5600
=
= 2.5 A
(iii)
∆
2240
To verify reciprocity theorem, the excitation 100 V source and the response Ix
are interchanged as shown in Fig. 13.
Therefore,
Ix =
3Ω
12 Ω
100 V
IX
IX 14 Ω
I2
4Ω
I3
4Ω
Fig. 13
Again mesh analysis is used to find Ix .
Ix =
∆Ix
∆


0 −14
0
30 −4  = 5600
∆Ix =  0
100 −4
8


17 −14
0
30 −4  = 2240
∆ =  −14
0 −4
8
5600
∆Ix
=
= 2.5 A
(iv)
∆
2240
Comparing Eqns. (iii) and (iv), it is observed that in both results the ratio of
excitation to response is same. Hence, reciprocity theorem is verified.
Therefore,
Ix =
7. Find the current through various branches of the circuit shown below.
5Ω
10 Ω
50 V
10 Ω
5A
Fig. 14
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According to superposition theorem, the current through any element is computed by adding all the individual currents delivered by the sources independently acting one source at a time.
Case 1 50 V source is acting alone
The 50 V source is made active and the 5 A current source is open-circuited.
Let the currents be I10 , I20 and I30 as shown in Fig. 15.
10 Ω
I 1′
I 3′ 5 Ω
I 2′
10 Ω
50 V
Fig. 15
With reference to Fig. 15,
50
= 2.5
10 + 10
I30 = 0
I10 = I20 =
Case 2 5 A source is acting alone
Similarly, only 5 A current source is kept in the circuit and the 50 V voltage
source is short-circuited. Now let the currents be I100 , I200 and I300 as shown
in Fig. 16.
10 Ω
I1′′
I3′′ 5 Ω
I2′′
10 Ω
5A
Fig. 16
By current division rule,
10
= −2.5
10 + 10
10
I200 = 5 &times;
= 2.5
10 + 10
I300 = −5 A
I100 = −5 &times;
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Thus, by superposition theorem,
I1 = I10 + I100 = 2.5 − 2.5 = 0 A
I2 = I20 + I200 = 2.5 + 2.5 = 5 A
I3 = I30 + I300 = 0 − 5 = −5 A
8. A series RLC circuit consists of 50 Ω resistance, 0.2 H inductance and 10 &micro;F
capacitance with the applied voltage of 20 V. Determine the resonant frequency,
the Q factor, the lower and upper frequency limits and the bandwidth of the
circuit.
Resonant frequency,
fr =
1
√
2π LC
1
√
2π 0.2 &times; 10 &times; 10−6
= 112.54 Hz.
r
1 L
Q=
R rC
1
0.2
= 2.828
=
50 10 &times; 10−6
fr
BW =
Q
=
Quality factor,
Bandwidth,
=
Lower cut-off frequency,
Upper cut-off frequency,
112.54
2.828
= 39.79
R
f1 = fr −
4πL
50
= 112.54 −
4π &times; 0.1
= 72.75 Hz.
R
f2 = fr +
4πL
50
= 112.54 +
4π &times; 0.1
= 152.33 Hz.
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9. Obtain a conductively coupled equivalent circuit for the magnetically coupled
circuit shown below.
j6 Ω
j5 Ω

500&deg;
j10 Ω
3
I2
I1
V
–
5
–j4 Ω
Fig. 17
Applying KVL to mesh 1 in the given magnetically coupled circuit, we get
or
j5I1 − j6I2 + (3 − j4)(I1 − I2 ) = 50∠0◦
(3 + j1)I1 + (−3 − j2)I2 = 50∠0◦
(v)
Similarly, by applying KVL to mesh 2 in the given circuit, we get
or
j10I2 − j6I1 + 5I2 + (3 − j4)(I2 − I1 ) = 0
(−3 − j2)I1 + (8 + j6)I2 = 0
Equations (v) and (vi) are written in matrix form as,
(3 + j1) −(3 + j2)
I1
50∠0◦
=
0
−(3 + j2)
(8 + j6)
I2
(vi)
(vii)
The following points are noted carefully when constructing the conductively
coupled equivalent circuit.
If the two currents do not enter at dotted ends of coupled coils,
(i) Negative value of mutual reactance is connected in series with the individual elements of a particular mesh.
(ii) Positive value of mutual reactance is connected in series with the common
elements of two meshes.
Alternately, if the two currents enter at dotted ends of coupled coils,
(i) Positive value of mutual reactance is connected in series with the individual elements of a particular mesh.
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(ii) Negative value of mutual reactance is connected in series with the common
elements of two meshes.
By above reasoning, the conductively coupled circuit is drawn as shown in Fig.
18.
50∠0&deg; V
j5 Ω
‒j6 Ω
j10 Ω
‒j6 Ω
j6 Ω
+
I1
‒j4 Ω
I2
5Ω
–
3Ω
Fig. 18
By inspection the following mesh equations are obtained.
(3 + j5 − j4 + j6 − j6)
−(3 − j4 + j6)
I1
50∠0◦
=
0
−(3 − j4 + j6) (8 + j10 − j4 + j6 − j6)
I2
(viii)
Simplifying, we obtain
(3 + j1) −(3 + j2)
I1
50∠0◦
(ix)
=
0
−(3 + j2)
(8 + j6)
I2
Comparing Eqs. (vii) and (ix), it is observed that both are same.
10. Two coupled coils have self inductances of L1 = 100 mH and L2 = 400 mH.
The coupling coefficient is 0.8. Find M. If N1 is 1000 turns, what is the value
of N2 . If a current i1 = 2 sin (500t) A through the coil 1, find the flux φ1 and
the mutually induced voltage V2M .
p
Mutual inductance,
M = k L1 L2
p
= 0.8 (100 &times; 10−3 )(400 &times; 10−3 )
= 0.16
r
N1
L1
=
Turns ratio,
N2
L2
r
100 &times; 10−3
=
= 0.5
400 &times; 10−3
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N1
0.5
1000
N2 =
= 2000 turns
0.5
N1 φ1
L1 =
i1
Therefore,
N2 =
Number of turns in coil 2
Self inductance of coil 1,
Therefore, the flux
φ1 =
L1 i1
N1
(100 &times; 10−3 ) &times; (2 sin 500t)
1000
φ1 = 0.2 sin 500t mWb
=
di1
dt
d
= 0.16 (0.2 sin 500t)
dt
Mutually induced emf,
V2M = M
= 16 cos 500t V
11. Solve for i and V as functions of time in the circuit shown below, when the
switch is closed at t = 0.
10 Ω
10 Ω
+
10 V
S
v
t=0
i
10 mH
‒
Fig. 19
Before closing the switch at t = 0, the circuit should have attained steady state
condition. Therefore,
10
= 0.5 A
I(0) =
20
This is the initial condition for the circuit for t ≥ 0. Figure 18 shows the
equivalent circuit for t ≥ 0. With reference to Fig. 18, i1 (t) is the forced
response and i2 (t) is the free response due to the internal energy stored in the
inductance when the circuit was in steady state. Hence,
i1 (t) =
10
= 1 A.
10
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To calculate i2 (t), KVL is applied in the mesh which has the closed switch and
10 mH inductance.
di2
=0
dt
10I2 (s) + 0.01[sI2 (s) − I2 (0)] = 0
10I2 (s) + 0.01[sI2 (s) − 0.5] = 0
10I2 (s) + 0.01[sI2 (s)] = 0.005
I2 (s)[10 + 0.01s] = 0.005
10i2 + 0.01
I2 (s) =
0.005
0.005
=
(10 + 0.01s)
0.01(s + 1000)
I2 (s) =
0.5
(s + 1000)
Taking inverse Laplace transformation on both sides, we obtain
i2 (t) = 0.5e−1000t A
Referring to Fig. 18,
Voltage across inductor,
i = i1 (t) − i2 (t)
= 1 − 0.5e−1000t A
di2 (t)
V =L
dt
d
= (10 &times; 10−3 ) (0.5e−1000t )
dt
= −5e−1000t V
12. For the circuit shown below, calculate the line current, the power and the power
factor. The value of R, L and C in each phase are 10 Ω, 1 H and 100 &micro;F
respectively.
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R
I1
C
R
L
400 V, 50 Hz
C
L
R
B
Y
R
I3
C
L
I2
Fig. 20
RYB phase sequence is assumed.
Inductive reactance,
Capacitive reactance,
∴ Impedance of each phase,
Phase voltage,
XL = 2πf L
= 2π &times; 50 &times; 1
= 314.16 Ω
1
XC =
2πf C
1
=
2π &times; 50 &times; 100 &times; 10−6
= 31.83 Ω
1
1
1
+
+
Yph =
10 j314.16 −j31.83
= (0.1 + j0.0282) f
1
1
Zph =
=
Yph
0.1 + j0.0282
= 9.2633 − j2.6123 = 9.6246∠ − 15.75◦
400∠0◦
VRN = √
3
= 231∠ − 30◦ V
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[It is to be noted that in star-connected networks, the phase voltage lags behind the line voltage
by 30◦ ]
Phase current in R phase,
IRN =
VRN
231∠ − 30◦
=
Zph
9.6246∠ − 15.75◦
= 24∠ − 14.25◦ A
In star-connected networks, line current and phase current is same. Therefore,
the line current in each line are:
IR = 24∠ − 14.25◦ A
IY = 24∠ − 134.25◦ A
IB = 24∠ − 254.25◦ A
Xph
−1
φ = tan
Rph
2.6123
−1
= tan
= 15.75◦
9.2633
cos φ = cos 15.75◦ = 0.96
Phase angle,
Power factor,
Power,
P =
=
√
√
3VL IL cos φ
3 &times; 400 &times; 24 &times; 0.96 = 15962.58 W
13. A 3-phase, 3-wire, 120 V, ABC system feeds a ∆-connected load whose phase
impedance is 30∠45◦ Ω. Find the phase and line currents in this system and
draw the phasor diagram.
Phase currents are computed as follows:
IRY =
120∠0◦
VRY
=
= 4∠ − 45◦ A
ZRY
30∠45◦
Similarly,
IYB = 4∠ − 165◦ A
IBR = 4∠ − 285◦ A
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Applying KCL the line currents are found as follows:
IR = IRY − IBR
= (4∠ − 45◦ ) − (4∠ − 285◦ )
= 6.9282∠ − 75◦ A
IY = IYB − IRY
= (4∠ − 165◦ ) − (4∠ − 45◦ )
= 6.9282∠165◦ A
VBR
IBR
IY
IB
VRY
45&deg;
30&deg;
IRY
IYB
IR
VYB
Fig. 21 Phasor diagram for Question No. 13
IB = IBR − IYB
= (4∠ − 285◦ ) − (4∠ − 165◦ )
= 6.9282∠45◦ A
14. A three-phase, four-wire, 120 V, RYB system feeds an unbalanced Y-connected
load with ZA = 5∠0◦ Ω, ZA = 10∠30◦ Ω and ZA = 20∠60◦ Ω. Obtain the four
line currents.
For the given phase sequence, phase voltages are computed as follows:
VRY
120∠0◦
VRN = √ = √
= 69.28∠ − 30◦ V
3
3
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[It is to be noted that in star-connected networks, the phase voltage lags behind the line voltage
by 30◦ ] Similarly,
VYN = 69.28∠ − 150◦ V
VBN = 69.28∠ − 270◦ V
Using the above found phase voltages, the line currents (also phase currents, in
star-connected networks) are obtained as follows:
IR = IRN =
=
IY = IYN =
=
IB = IBN =
=
VRN
ZR
69.28∠ − 30◦
= 13.856∠ − 30◦ A
5∠0◦
VYN
ZY
69.28∠ − 150◦
= 6.928∠180◦ A
10∠30◦
VBN
ZB
69.28∠ − 270◦
= 3.464∠ − 330◦ A
20∠60◦
Neutral current,
IN = −(IR + IY + IB )
= −(13.856∠ − 30◦ + 6.928∠180◦ + 3.464∠ − 330◦ )
= 9.6∠147.23◦ A
15. Three impedances Z1 = (17.32 + j10), Z2 = (20 + j34.64) and Z3 = (0 − j10)
ohms are delta connected to a 400 V, three phase system. Determine the phase
currents, line currents and total power consumed by the load.
A phase sequence of RYB is assumed. Taking VRY = 400∠0◦ as reference, the
phase currents are obtained as,
IRY =
400∠0◦
17.32 + j10
= 17.32 − j10 = 20∠ − 30◦ A
400∠ − 120◦
20 + j34.64
= −10 − j0.000127 = 10∠ − 180◦ A
IYB =
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IBR =
400∠ − 240◦
0 − j10
= 34.64 − j20 = 40∠ − 150◦ A
Applying KCL the line currents are obtained as,
IR = IRY − IBR = (20∠ − 30◦ ) − (40∠ − 150◦ )
= 51.96 + j10 = 52.92∠10.89◦ A
IY = IYB − IRY = (10∠ − 180◦ ) − (20∠ − 30◦ )
= −27.32 + j10 = 29.09∠159.9◦ A
IB = IBR − IYB = (40∠ − 150◦ ) − (10∠ − 180◦ )
= −24.64 − j20 = 31.74∠ − 140.94◦ A
Total power consumed by the load,
2
2
2
P = IRY
RRY + IYB
RYB + IBR
RBR
= (202 &times; 17.32) + (102 &times; 20) + (402 &times; 0)
= 8928 W
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10314
B.E/B.Tech. DEGREE EXAMINATION, MAY/JUNE 2012
Electrical and Electronics Engineering
EE2151 – CIRCUIT THEORY
1. In the circuit shown in Fig. 1, VR and VL were measured and found to be 10 V
each.
VR
VL
100 Ω
0.1 H
i(t)
+
e(t)
‒
Fig. 1
Assuming i(t) as the reference waveform, find
(a) the frequency f and current i(t).
(b) ZT , the total impedance and e(t).
Current through the circuit,
Reactive inductance of the coil,
Also,
Therefore,
Total impedance,
VR
R
10
=
= 0.1 A
100
VL
XL =
I
10
=
= 100 Ω
0.1
XL = 2πf L
I=
XL
2πL
100
=
= 159.15 Hz.
2π &times; 0.1
ZT = R + jXL
f=
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= (100 + j100) = 141.42∠45◦ Ω
√
i(t) = Im sin ωt = (0.1 &times; 2) sin (2πf t)
= 0.1414 sin 1000t A
E = IZ = (0.1)(100 + j100)
= (10 + j10) = 14.14∠45◦
Given that i(t) is taken as reference and so e(t) leads the current by φ = 45◦ .
Therefore,
√
e(t) = Em sin (ωt + φ) = (14.14 &times; 2) sin (2πf t + 45◦ )
= 20 sin (1000t + 45◦ ) V
2. What is the voltage across A and B in the circuit shown in Fig. 2?
6Ω
12 V
6V
4Ω
12
V
A
4Ω
10 Ω
B
Fig. 2
The current through 6 Ω and 4 Ω circuit,
Therefore, voltage across 4 Ω resistor
The current through 4 Ω and 10 Ω circuit,
Therefore, voltage across 4 Ω resistor
6
= 0.6 A
10
= 0.6 &times; 4
= 2.4 V (A being negative)
12
I2 =
= 0.86 A
14
= 0.86 &times; 4
= 3.44 V (B being positive)
I1 =
Replacing the resistances connected between A and B with their respective
voltage drops, the given circuit is redrawn as shown in Fig. 3.
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6Ω
12 V
6V
2.4 V
12
V
A
3.44 V
10 Ω
B
Fig. 3
From Fig. 3 it is found that,
VAB = −2.4 + 12 + 3.44 = 13.04 V
3. Find the current in 4 Ω resistor in the circuit shown in Fig. 4.
3Ω
57 V
5Ω
4Ω
7Ω
6Ω
42 V
4V
25 V
70 V
Fig. 4
Mesh currents are assumed with their direction as shown in Fig. 5.
3Ω
57 V
5Ω
4Ω
42 V
I1
6Ω
I2
25 V
7Ω
I3
4V
70 V
Fig. 5
By mesh analysis

7
 −4
0
method, the mesh equations are written as

 

−4
0
I1
42 + 25 = 67
15 −6   I2  =  −25 − 57 − 70 = −152 
−6 13
I3
70 + 4 = 74
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Solving the above equations, we get
I1 = 5 A
I2 = −8 A
I3 = 2 A
Therefore, current through 4 Ω resistor, I1 − I2 = 5 − (−8) = 13 A
4. Convert the network shown below, into a π− connected equivalent circuit.
a
‒j2 Ω
c
j4 Ω
d
2Ω
a′
2Ω
b
‒j4 Ω
e
b′
f
Fig. 6
Using star-delta transformation, the star connection with c as star-point can be
converted into an equivalent delta network as follows:
(−j2)(2) + (2)(j4) + (j4)(−j2)
= 4 + j2
(2)
Zae =
(−j2)(2) + (2)(j4) + (j4)(−j2)
= 1 − j2
(j4)
Zed =
(−j2)(2) + (2)(j4) + (j4)(−j2)
= −2 + j4
(−j2)
a
(1 ‒ j2) Ω
a′
(4 + j2) Ω
d
(‒2 + j4) Ω
2Ω
b
‒j4 Ω
e
f
b′
Fig. 7
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The delta network thus found is replaced the star network in the given circuit
as shown in Fig. 7, where (−2+j4) and (−j4) are found parallel. The equivalent
value of impedance is,
Zeq1 =
(−2 + j4)(−j4)
= (−8 − j4)
(−2 + j4 − j4)
Using Zeq1 , Fig. 7 is redrawn as illustrated in Fig. 8.
(4 + j2) Ω
a
(1 ‒ j2) Ω
a′
2Ω
d
b
(‒8 ‒ j4) Ω
b′
f
Fig. 8
Again using star-delta transformation, the star connection with d as star-point
can be converted into an equivalent delta network as follows:
Zab =
(4 + j2)(−8 − j4) + (−8 − j4)(2) + (2)(4 + j2)
= 5 + j2
(−8 − j4)
Zaf =
(4 + j2)(−8 − j4) + (−8 − j4)(2) + (2)(4 + j2)
= −16 − j18
(2)
Zbf =
(4 + j2)(−8 − j4) + (−8 − j4)(2) + (2)(4 + j2)
= −10 − j4
(4 + j2)
a
(1 ‒ j2) Ω
a′
(5 + j2) Ω
(‒16 ‒ j18) Ω
f
b
(‒10 ‒ j4) Ω
b′
Fig. 9
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Using the delta equivalent values the simplified network is redrawn as shown
in Fig. 9, in which (1 − j2) and (−16 − j18) are found parallel. The equivalent
value of impedance is, therefore,
(1 − j2)(−16 − j18)
Zeq2 =
= (−32 − j36)
(1 − j2) + (−16 − j18)
(5 + j2) Ω
a
(‒32 ‒ j36)
Ω
b
(‒10 ‒ j4) Ω
b′
a′
Fig. 10
Figure 9 is further modified using the value of (−32 − j36) as found in Fig. 10
which is the π− equivalent circuit of the given network.
5. Calculate the current through the 2 Ω resistor in the circuit shown in Fig. 11,
using superposition theorem.
1Ω
10 V
+
1Ω
2Ω
1Ω
1A
Fig. 11
10 V source is acting alone
The given circuit is redrawn by open-circuiting 1 A current source.
1Ω
10 V
+
I1
1Ω
2Ω
I2
1Ω
Fig. 12
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By mesh analysis method, the mesh equations are written as
3 −2
I1
10
=
−2
4
I2
0
Solving, we get
I1 = 5 A
I2 = 2.5 A
Therefore, current through 2 Ω resistor due to 10 V source is I 0 = 5 − 2.5 = 2.5
A.
1 A source is acting alone
1Ω
I3
1Ω
1Ω
I4
2Ω
+
1V
Fig. 13
The given circuit is redrawn by short-circuiting 10 V source. By source transformation technique, the 1 A current source is converted into voltage source as
shown in Fig. 13. By mesh analysis method, the mesh equations are written as
0
3 −2
I3
=
−2
4
I4
1
Solving, we get
I3 = 0.25 A
I4 = 0.375 A
Therefore, current through 2 Ω resistor due to 1 A source is I 00 = 0.375 – 0.25
= 0.125 A.
By the principle of superposition, current through 2 Ω resistor,
I = I 0 + I 00 = 2.5 + 0.125 = 2.625 A.
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6. Calculate the current through the 2 Ω resistor in the circuit shown in Fig. 14,
using Thevenin’s theorem.
1Ω
a
2V
+
1Ω
1A
2Ω
b
Fig. 14
To find Thevenin’s voltage, Vth
1Ω
Va
a
2V
+
1Ω
1A
Vth
b
Fig. 15
The 2 Ω resistor connected between terminals a and b is removed first. The
voltage, Vth across a and b is obtained by using Nodal analysis as follows:
Va − 2 Va
+
=1
1
1
or
Va = Vth = 1.5 V
To find Thevenin’s resistance, Rth
1Ω
a
 Rth
1Ω
b
Fig. 16
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Figure 16 is used to find Rth . As seen from Fig. 16,
Rth =
1
= 0.5 Ω
1+1
Using the values of Vth , and Rth found above, the Thevenin’s equivalent circuit
is drawn as shown in Fig. 17.
Rth = 0.5 Ω
a
Vth =
1.5 V
+
2Ω
b
Fig. 17
As seen in Fig. 17, current through 2 Ω resistor is =
1.5
= 0.6 A
0.5 + 2
7. The signal voltage in the circuit shown in Fig. 18 is e(t) = 0.01 sin(2π &times;
455 &times; 103 t) V. What would be the value of C in order that the circuit would
resonate at this signal frequency? At this condition, find the values of I, VC , Q
and bandwidth of the circuit.
0.9 Ω
15 μH
+
e(t)
C
‒
Fig. 18
Inductive reactance,
XL = 2πf L
= 2π &times; 455000 &times; 15 &times; 10−6
= 42.88 Ω
At resonance, inductive reactance = capacitive reactance, i.e., XL = XC
Therefore, capacitive reactance
XC = 42.88 Ω
Also,
XC =
1
2πf C
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1
1
=
2πf XC
2π &times; 455 &times; 103 &times; 42.88
or
C=
At resonance,
C = 8.15 &times; 10−9 F
Upon comparing the signal voltage e(t) = 0.01 sin(2π&times;455&times;103 t) with standard
sinusoidal form of e(t) = Vmax sin ωt, we get
Therefore,
Vmax = 0.01
0.01
Vrms = √
2
= 0.00707
Hence, the current
Voltage across capacitor,
Quality factor,
Bandwidth,
I=
0.00707
Vrms
=
R
0.9
= 7.8567 &times; 10−3 A
VC = IXC
= (7.8567 &times; 10−3 )(42.88)
= 0.3369 V
r
1 L
Q=
R C
r
15 &times; 10−6
1
=
0.9 8.5 &times; 10−9
= 46.68
R
BW =
2πL
=
0.9
2π &times; 15 &times; 10−6
= 9549.29 Hz.
8. (RL + j20) Ω and (20 − j10) Ω are connected in parallel. Determine the value
of RL for resonance.
Y = Y1 + Y2
=
1
1
+
RL + j20 20 − j10
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1
RL − j20
1
20 + j10
=
&times;
+
&times;
RL + j20 RL − j20
20 − j10 20 + j10
RL
20
j10
j20
=
+
+
−
RL2 + 202 RL2 + 202
500 500
Separating real and imaginary terms,
Y =
RL
20
+
2
RL + 202 500
+j
10
20
− 2
500 RL + 202
At resonance, the imginary part of above equation is zero, therefore,
10
20
− 2
=0
500 RL + 202
or
RL2
10
20
=
+ 202
500
RL2 + 202 − 1000 = 0
or
Solving the above equation, we get
RL = 24.49 Ω
9. Consider the single tuned circuit shown below and determine,
(i) the resonant frequency, and
(ii) the output voltage at resonance.
Assume RS &gt;&gt; ωr L1 and k = 0.9.
M
10 Ω
15 V
RS
R2
L1
1 μH
L2
100 μH
–
0.1 μF

10 Ω
vo
Fig. 19
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Mutual inductance,
At resonance
p
L1 L2
p
= 0.9 (1 &times; 10−6 )(100 &times; 10−6 )
= 9 &micro;H
1
=
ωr C
1
=√
L2 C
1
=p
−6
(100 &times; 10 )(0.1 &times; 10−6 )
M Vi
=
C[R1 R2 + ω 2 M 2 ]
9 &times; 10−6 &times; 15
=
0.1 &times; 10−6 [(10)(10) + (316227.766)2 (9 &times; 10−6 )2 )]
= 12.49 V.
M =k
ωr L2
or
ωr
Resonant frequency,
ωr
Output voltage,
Vo
10. A 3-phase balanced delta-connected load of (4 + j8) Ω is connected across a 400
V, 3-phase supply. Calculate the phase currents and line currents. Assume the
RYB phase sequence. Also calculate the power drawn by the load.
The three phase currents are found as follows:
IRY
400∠0◦
VRY
=
= (20 − j40) A
=
VRY
4 + j8
IYB
VYB
400∠ − 120◦
=
=
= (−44.64 + j2.68) A
VYB
4 + j8
IBR
VBR
400∠ − 240◦
=
=
= (24.64 + j37.32) A
VBR
4 + j8
The three line currents are found by using KCL as follows:
IR = IRY − IBR
= (20 − j40) − (24.64 + j37.32)
= (−4.64 − j77.32) = 77.46∠ − 93.43◦ A
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IY = IYB − IRY
= (−44.64 + j2.68) − (20 − j40)
= (−64.64 + j42.68) = 77.46∠146.56◦ A
IB = IBR − IYB
= (24.64 + j37.32) − (−44.64 + j2.68)
= (69.28 + j34.64) = 77.46∠26.56◦ A
8
−1
Phase angle,
φ = tan
4
◦
= 63.43
So, power factor,
cos φ = cos 63.43◦
√
Power,
P = 3VL IL cos φ
√
= 3 &times; 400 &times; 77.46 &times; 63.43◦
So, power drawn by the load,
P = 24005 W
11. Three equal inductors connected in star, take 5 kW at 0.7 pf when connected
to a 400 V, 50 Hz, three-phase, three-wire supply. Calculate the line currents,
(i) if one of the inductors is disconnected, and
(ii) if one of the inductors is shortcircuited
A phase sequence of RYB is assumed. For a balanced three-phase load,
√
Power,
P = 3VL IL cos φ
So, line current,
IL = √
=√
P
3VL cos φ
5000
3(400)(0.7)
= 10.31 A
In star connected loads, line current(IL ) = phase current(Iph )
Therefore,
Impedance per phase,
Iph = 10.31 A
400
√
Vph
3
=
Zph =
Iph
10.31
= 22.4 Ω
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The elements of inductor are found as follows:
R = Z cos φ
= 22.4 &times; 0.7 = 15.68 Ω
X = Z sin φ
= 22.4 &times; 0.7141 = 16 Ω
Case 1 If one of the inductors is disconnected
IR
R
(15.68 + j16) Ω
4000&deg; V
(15
Y
IY
.68
+j
16
)Ω
400‒120&deg; V
B
IB
Fig. 20
If one of the inductors is disconnected, the load becomes a two-phase load
as illustrated in Fig. 20. Here, the inductor in line Y is removed. Thus,
VRB
400∠ − 240◦
=
Zph
(15.68 + j16) + (15.68 + j16)
= 2.39 − j8.6 = 8.93∠74.42◦
IR = 8.93∠74.42◦ A
IB = 8.93∠ − 105.58◦ A
IY = 0 A
IR = −IB =
Therefore,
and
Clearly,
Case 2 If one of the inductors is shortcircuited
The inductor of Y branch is shortcircuited as shown in Fig. 21, and the
line currents are calculated by mesh analysis method. By inspection, the
mesh equations are written as:
15.68 + j16
0
I1
400∠0◦
=
0
15.68 + j16
I2
400∠ − 120◦
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IR
R
I1
(15.68 + j16) Ω
4000&deg; V
(15
Y
IY
.68
+j
16
)Ω
I2
400‒120&deg; V
IB
B
Fig. 21
Using Cramer’s rule, the loop current I1 in the assumed direction is found
as follows:
∆I1
I1 =
∆
∆I1 =
400∠0◦
0
◦
400∠ − 120 15.68 + j16
= (400∠0◦ )(15.68 + j16)
= (6272 + j6400)
∆=
15.68 + j16
0
0
15.68 + j16
= (15.68 + j16)(15.68 + j16)
= −10.1376 + j501.76
Therefore,
I1 =
∆I1
6272 + j6400
=
∆
−10.1376 + j501.76
= (12.4974 − j12.7525) = 17.8553∠ − 45.58◦ A
Similarly, the loop current I2 is found as follows:
∆I2
∆
15.68 + j16
400∠0◦
∆I2 =
0
400∠ − 120◦
I2 =
= (15.68 + j16)(400∠ − 120◦ )
= (2406.56 − j8631.71)
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Therefore,
I2 =
∆I2
2406.56 − j8631.71
=
∆
−10.1376 + j501.76
= (−17.2927 − j4.4469) = 17.8553∠ − 165.58◦ A
As seen in Fig. 21,
IR = I1 = 17.8553∠ − 45.58◦ A
IY = I2 − I1 = 17.8553∠74.42◦ A
IB = −I2 = 17.8553∠14.42◦ A
12. If W1 and W2 are the readings of the two wattmeters which measures power in
W1
the three phase balanced system and if
= a, show that the power factor of
W2
a+1
the circuit is given by cos φ = √
.
2 a2 − a + 1
By employing two wattmeter method of measuring three phase power, we have
√ W2 − W1
tan φ = 3
W1 + W2
W1
√ W2 1 − W2
= 3
W1
W2 1 +
W2
√ (1 − a)
3
(1 + a)
(1 − a)2
φ = (3)
(1 + a)2
(1 − a)2
φ = 1 + (3)
(1 + a)2
(1 + a)2 + 3(1 − a)2
φ=
(1 + a)2
(1 + a2 + 2a) + 3(1 + a2 − 2a)
=
φ
(1 + a)2
(1 + a2 + 2a + 3 + 3a2 − 6a)
=
φ
(1 + a)2
(4a2 − 4a + 4)
=
(1 + a)2
=
tan2
1 + tan2
sec2
1
cos2
1
cos2
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4(a2 − a + 1)
(1 + a)2
(1 + a)2
cos2 φ =
4(a2 − a + 1)
=
Therefore,
1+a
cos φ = √
2 a2 − a + 1
13. Obtain the readings of two wattmeters connected to a three-phase, three-wire
120 V system feeding a balanced ∆-connected load with a load impedance of
12∠30◦ Ω. Assume either phase sequence. Find the phase power and compare
the total power to the sum of the wattmeter readings.
A phase sequence of RYB is assumed.
Phase current,
IRY =
=
VRY
ZRY
120∠0◦
12∠30◦
= 10∠ − 30◦ A
Similarly, the other phase currents
and
Power factor,
IYB = 10∠ − 150◦ A
IBR = 10∠ − 270◦ A
cos φ = cos 30◦ = 0.866
Line current (IL ) in R-line is calculated by using KCL as follows:
So, total power
Also, total power
IL = IRY − IBR
= (10∠ − 30◦ ) − (10∠ − 270◦ )
= 17.32∠ − 60◦
√
P = 3VL IL cos φ
√
= 3 &times; 120 &times; 17.32 &times; 0.866 = 3117.6 W
P = W1 + W2
W1 = VL IL cos (30◦ + φ)
= 120 &times; 17.32 &times; cos (30◦ + 30◦ )
W1 = 1039.2 W
(i)
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Per phase power
W2 = VL IL cos (30◦ − φ)
= 120 &times; 17.32 &times; cos (30◦ − 30◦ )
W2 = 2078.4 W
= Vph Iph cos φ
= 120 &times; 10 &times; 0.866
= 1039.2 W
(ii)
(iii)
Upon comparing Eqs. (i) and (iii), it is observed that when the power factor is
0.866, one of the wattmeters reads per phase power, i.e., one-third of the total
power, while the other wattmeter reads two-third of the total power.
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22115
B.E/B.Tech. DEGREE EXAMINATION, APRIL/MAY 2011
Electrical and Electronics Engineering
EE2151 – CIRCUIT THEORY
1. Determine the current delivered by the source in the circuit shown in figure
below.
2Ω
4Ω
A
30 V
+
2Ω
B
2Ω
2Ω
1Ω
C
2Ω
2Ω
D
1Ω
Fig. 1
The current delivered by the source is found by network reduction technique.
Equivalent resistance across CD, RCD =(2 + 1)||2 = 1.2 Ω
Equivalent resistance across AC, RAC =(2 + 2)||4 = 2 Ω
Using the above found equivalent values the reduced network is drawn as shown
in Fig. 2.
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2Ω
A
30 V
+
2Ω
2Ω
B
C
1.2 Ω
D
1Ω
Fig. 2
Equivalent resistance across AD, RAD = (2 + 1.2)||2 = 1.23 Ω
Equivalent resistance across AB, RAB = Req = (1.23 + 1)||2 = 1.0544 Ω
V
30
Hence, the current delivered by the 30 V source =
=
= 28.4522 A
Req
1.0544
2. For the network shown in Fig. 3, obtain the current ratio (I1 /I3 ) using mesh
analysis.
j2 Ω
‒j4 Ω
5Ω
+
I1
V1
5Ω
I2
I3
j2 Ω
–
Fig. 3
By mesh analysis method, the mesh equations


5 + j2
−5
0
 −5
5 − j2 −j2  
0
−j2 5 + j2
are written as

 
V1
I1
I2  =  0 
0
I3
Using Cramer’s rule, the ratio I1 /I3 is found as,
I1
∆I1 /∆
=
I3
∆I3 /∆
=
∆I1
∆I3
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V1
−5
0
0 5 − j2 −j2
0
−j2 5 + j2
5 + j2
−5
V1
−5
5 − j2 0
0
−j2
0
=
=
V1 [(5 − j2)(5 + j2) − (−j2)2 ]
33
=
V1 [(−5)(−j2) − 0]
j10
Hence, the current ratio,
I1
= −j3.3
I3
3. Use source transformation to find I0 in the circuit shown below.
4Ω
12 V
Io 2 Ω
+
6Ω
3Ω
6A
+
24 V
Fig. 4
Step 1 The two voltage sources are converted into current sources as shown in
Fig. 5.
Io 2 Ω
3A
4Ω
3Ω
6A
6Ω
4A
Fig. 5
Step 2 Current sources 3 A and 6 A are combined and made as a single current
source of 3 A (Fig. 6).
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Io 2 Ω
4Ω
6Ω
3Ω
3A
4A
Fig. 6
Step 3 Resistances 3 Ω and 6 Ω are found in parallel; hence, they are simplified
and the result is 2 Ω as shown in Fig. 7.
Io 2 Ω
4Ω
2Ω
3A
4A
Fig. 7
Step 4 The two current sources are converted into voltage sources as shown in
Fig. 8.
4Ω
12 V
Io 2 Ω
+
2Ω
+
8V
Fig. 8
Step 5 By Ohm’s law, the current
I0 =
−12 + 8
= −0.5 A
4+2+2
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4. Use the technique of ∆-Y connection to find the equivalent resistance between
terminals A-B of the circuit shown below.
A
C
1
4
Ω
Ω
3Ω
D
E
2
Ω
5
Ω
F
B
Fig. 9
The delta network, CDE comprising 1 Ω, 3 Ω and 4 Ω is converted to Y-network
as follows:
RCN =
(1 &times; 4)
= 0.5
(1 + 3 + 4)
RDN =
(1 &times; 3)
= 0.375
(1 + 3 + 4)
REN =
(3 &times; 4)
= 1.5
(1 + 3 + 4)
The star equivalent network thus found is fit into the original circuit as shown
in Fig. 10. From the Fig. 10, RAB is calculated as,
= 0.5 + (0.375 + 2)||(1.5 + 5)
2.375 &times; 6.5
= 0.5 +
2.375 + 6.5
RAB = 2.2394 Ω
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C
A
0.5 Ω
N
0
7
.3
5
1.
Ω
5
Ω
E
Ω
5
2
Ω
D
B
F
Fig. 10
5. Determine the voltage across 20 Ω resistance in the circuit shown below using
superposition theorem.
1.5 A
20 Ω
16 V
10 V
3A
80 Ω
Fig. 11
According to superposition theorem, the current through any element is computed by adding all the individual currents delivered by the sources independently acting one source at a time.
Case 1 16 V source is acting alone
Making 16 V source as only active by short-circuiting the 10 V source and
open-circuiting the other two current sources the circuit thus obtained is
shown in Fig. 12 (a).
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20 Ω
10 V
20 Ω
16 V
80 Ω
80 Ω
(a)
(b)
Fig. 12
By Ohm’s law, the current through 20 Ω resistance is found as,
I1 =
16
= 0.16 A
20 + 80
Case 2 10 V source is acting alone
Similarly, only 10 V source is excited in the circuit and other three sources
are made inactive as shown in Fig. 12 (b). By Ohm’s law, the current
through 20 Ω resistance is found as,
I2 =
10
= 0.1 A
20 + 80
1.5 A
20 Ω
20 Ω
3A
80 Ω
(a)
80 Ω
(b)
Fig. 13
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Case 3 3 A source is acting alone
In this case, the two voltage sources are replaced by short-circuits and
the 1.5 A current source is open-circuited. With 3 A source only acting
the current through 20 Ω resistance is found by current division rule as
follows:
80
I3 = 3 &times;
= 2.4 A
20 + 80
Case 4 1.5 A source is acting alone
Under this condition, the 1.5 A is short-circuited when the 10 V is replaced
by a short-circuit. So, the entire current flows through the short-circuit
only. Thus,
I4 = 0 A
Here, I3 is flowing in one direction whereas I1 and I2 are flowing in opposite
direction. Thus, by the principle of superposition, current through 20 Ω
resistance is,
I = −I1 − I2 + I3 + I4
= −0.16 − 0.1 + 2.4 + 0
= 2.14 A
Therefore, voltage across 20 Ω resistance is,
= 20 &times; 2.14 = 42.8 V
6. Find the voltage drop across 12 Ω resistance using Norton’s theorem for the
circuit shown below.
4V
4Ω
14 V
+
2Ω
12 Ω
4Ω
Fig. 14
To find the Norton’s current (IN ), 12 Ω resistance is replaced by a short-circuit
as shown in Fig. 15. The current, IN is then found by applying mesh analysis
method.
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[The short-circuit across 4 Ω resistor makes the resistor insignificant, because the entire current would flow
through the short-circuit only]
4V
4Ω
14 V
+
2Ω
I1
I2
4Ω
IN
Fig. 15
6 −2
−2
2
I1
I2
=
14
4
Using Cramer’s rule, I2 , which is nothing but the Norton’s current is found as,
IN = I2 =
=
∆I2
=
∆
6 14
−2 4
6 −2
−2
2
24 + 28
12 − 4
IN = 6.5 A
Thevenin’s resistance, Rth is found by looking at the network from terminals
A-B in Fig. 16.
4Ω
2Ω
Rth
4Ω
Fig. 16
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Rth = (4||2)||4
8
=
||6
6
=1Ω
By using the Norton’s equivalent circuit shown in Fig. 17, current through 12 Ω
is,
6.5 A
1Ω
12 Ω
Fig. 17
= 6.5 &times;
1
(1 + 12)
= 0.5 A
Therefore, voltage drop across 12 Ω resistance is,
= 12 &times; 0.5 = 6 V
7. Determine the quality factor of a coil for the series circuit consisting of R =
10 Ω, L = 0.1 H and C = 10 &micro; F. Derive the formula used.
r
1 L
Q=
R rC
1
0.1
=
10 10 &times; 10−6
= 10
[For derivation of the formula for Q-factor, please refer Section 10.1.6 in the book mentioned at footer ]
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8. In the coupled circuit shown below, find the voltage across 5 Ω resistor.
k = 0.8
j5 Ω

500&deg;
j10 Ω
3
I1
V
–
I2
5
–j4 Ω
Fig. 18
Mutual reactance,
p
(jωL1 )(jωL2 )
p
= 0.8 (j5)(j10)
= j5.6568 Ω
M =k
Referring to Fig. 18, the KVL equations are written around loop 1 as,
j5I1 − j5.6568I2 + 3(I1 − I2 ) − j4(I1 − I2 ) = 50∠0◦
j5I1 − j5.6568I2 + 3I1 − 3I2 − j4I1 + j4I2 = 50∠0◦
I1 (j5 + 3 − j4) + I2 (−j5.6568 − 3 + j4) = 50∠0◦
I1 (3 + j1) + I2 (−3 − j1.6568) = 50∠0◦
(i)
Similarly, KVL equations are written around loop 2 as follows:
j10I2 − j5.6568I1 + 5I2 − j4(I2 − I1 ) + 3(I2 − I1 ) = 0
j10I2 − j5.6568I1 + 5I2 − j4I2 + j4I1 + 3I2 − 3I1 ) = 0
I1 (−j5.6568 + j4 − 3) + I2 (j10 + 5 − j4 + 3) = 0
I1 (−3 − j1.6568) + I2 (8 + j6) = 0
(ii)
Equations (i) and (ii) are put in matrix form as,
3 + j1
−3 − j1.6568
I1
50 ∠0◦
=
−3 − j1.6568
8 + j6
I2
0
Using Cramer’s rule, the current I2 is found as,
I2 =
∆I2
∆
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3 + j1
50 ∠0◦
−3 − j1.6568
0
3 + j1
−3 − j1.6568
−3 − j1.6568
8 + j6
=
(3 + j1.6568)(50 ∠0◦ )
(3 + j1)(8 + j6) − (−3 − j1.6568)2
=
Hence the current,
I2 = 7.8114 − j3.6275
= 8.6126∠ − 25◦ A
Hence, voltage across 5 Ω resistor is = 5 &times; 8.6126∠ − 25◦ = 43.063∠ − 25◦ V
9. Using Laplace transform, obtain the expression for i1 and i2 in the circuit shown
below, when DC voltage is applied suddenly. Assume that the initial energy
stored in the circuit is zero.
t=0
8.4 H
10 H
i1
i2
42 Ω
336 V
48 Ω
Fig. 19
Applying KVL around the outer loop,
−336 + 8.4
di1
di2
+ 10
+ 48i2 = 0
dt
dt
Taking Laplace transformation on both sides, we get
8.4sI1 (s) + 10sI2 (s) + 48I2 (s) = 336
336
I1 (s)[8.4s] + I2 (s)[10s + 48] =
s
(iii)
Similarly, applying KVL around the second mesh,
10
di2
+ 48I2 − 42(i1 − i2 ) = 0
dt
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Taking Laplace transformation on both sides, we get
10sI2 (s) + 48I2 (s) − 42I1 (s) + 42I2 (s) = 0
I1 (s)[−42] + I2 (s)[10s + 48 + 42] = 0
(iv)
Writing Eqs. (iii) and (iv) in matrix form, we obtain
#
# &quot;
#&quot;
&quot;
336/s
I1 (s)
8.4s 10s + 48
=
0
I2 (s)
−42 10s + 90
Using Cramer’s rule, the current I1 (s) is found as,
I1 =
=
∆I1 (s)
∆
336
10s + 48
s
0
10s + 90
8.4s 10s + 48
−42 10s + 90
336
(10s + 90)
s
=
(8.4s)(10s + 90) + 42(10s + 48)
30240
3360 +
s
=
2
84s + 1176s + 2016
3360s + 30240
=
84s(s2 + 14s + 24)
=
I1 (s) =
40s + 360
+ 14s + 24)
s(s2
40s + 360
s(s + 2)(s + 12)
By applying partial fraction technique,
I1 (s) =
40s + 360
A
B
C
= +
+
s(s + 2)(s + 12)
s
s + 2 s + 12
(v)
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Solving for the constants,
A=
B=
and
C=
40s + 360
(s + 2)(s + 12)
= 15
s=0
40s + 360
s(s + 12)
s=−2
40s + 360
s(s + 2)
s=−12
= −14
= −1
Substituting the values of A, B and C into Eq. (v), we get
I1 (s) =
14
1
15
−
−
s
s + 2 s + 12
Taking inverse Laplace transformation on both sides, the current i1 (t) is obtained as:
i1 (t) = 15 − 14e−2t − 1e−12t A
Similarly, by using Cramer’s rule, the current I2 (s) is found as,
I2 =
=
∆I2 (s)
∆
8.4s
336
s
−42
0
8.4s 10s + 48
−42 10s + 90
336
42
s
=
(8.4s)(10s + 90) + 42(10s + 48)
14112
s
=
84s2 + 1176s + 2016
14112
=
2
84s(s + 14s + 24)
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=
I2 (s) =
s(s2
168
+ 14s + 24)
168
s(s + 2)(s + 12)
By applying partial fraction technique,
I2 (s) =
168
A
B
C
= +
+
s(s + 2)(s + 12)
s
s + 2 s + 12
(vi)
Solving for the constants,
A=
B=
and
C=
168
(s + 2)(s + 12)
168
s(s + 12)
168
s(s + 2)
=7
s=0
= −8.4
s=−2
= 1.4
s=−12
Substituting the values of A, B and C into Eq. (v), we get
I2 (s) =
7
8.4
1.4
−
+
s s + 2 s + 12
Taking inverse Laplace transformation on both sides, the current i2 (t) is obtained as:
i2 (t) = 7 − 8.4e−2t + 1.4e−12t A
10. In the circuit shown below, find the expression for the transient current. The
initial current is as shown in the figure.
3H
i(0) = 6 A
i(t)
5Ω
100 V
Fig. 20
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Applying KVL after closing on the switch to position-2,
5i + 3
di
= 100
dt
Taking laplace transformation on both sides, we get
5I(s) + 3[sI(s) − I(0)] =
100
s
At t = 0, i(t) = −6 A and in s − domain, I(0) = −6 A. Therefore, the above
equation becomes
100
s
100
− 18
I(s)[3s + 5] =
s
5I(s) + 3[sI(s) + 6] =
( 100
) − 18
I(s) = s
3s + 5
100 − 18s
=
s(3s + 5)
=
100 − 18s
33.33 − 6s
=
3s(s + 1.6667)
s(s + 1.6667)
By applying partial fraction technique,
I(s) =
A
B
33.33 − 6s
= +
s(s + 1.6667)
s
s + 1.6667
(vii)
Solving for constants A and B,
A=
33.33 − 6s
(s + 1.6667)
= 20 and B =
s=0
33.33 − 6s
s
= −26
s=−1.6667
Substituting the values of A and B into Eq. (vii), we get
I(s) =
20
26
−
s
s + 1.6667
Taking inverse Laplace transformation on both sides, the current equation is
obtained as:
i(t) = 20 − 26e−1.6667t A
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11. A balanced ∆-connected load has one phase current IBC = 2∠ − 90◦ A. Find
the other phase currents and the three line currents if the system is an ABC
system. If the line voltage is 100 V, what is the load impedance?
VBC
100∠ − 120◦
Impedance, ZBC =
=
IBC
2∠ − 90◦
= 50∠ − 30◦
In balanced load the impedances of all branches are of equal values. So,
ZAB = ZBC = ZCA = 50∠ − 30◦ Ω
The phase currents in all the three phases are calculated as follows:
100∠0◦
IAB =
= 2∠30◦ A
◦
50∠ − 30
IBC =
100∠ − 120◦
= 2∠ − 90◦ A
50∠ − 30◦
100∠ − 240◦
= 2∠ − 210◦ A
50∠ − 30◦
By applying KCL the line currents are found as follows:
ICA =
IA = IAB − ICA
= 2∠30◦ − 2∠ − 210◦
= 3.4641∠0◦ A
IB = IBC − IAB
= 2∠ − 90◦ − 2∠30◦
= 3.4641∠ − 120◦ A
IC = ICA − IBC
= 2∠ − 210◦ − 2∠ − 90◦
= 3.4641∠ − 240◦ A.
12. The power consumed in a three phase balanced star connected load is 2 kW at
a power factor of 0.8 lagging. The supply voltage is 400 V, 50 Hz. Calculate
the resistance and reactance of each phase.
√
Power,
P = 3 VL IL cos φ
So, the line current,
2000
IL = √
3 &times; 400 &times; 0.8
IL = 3.6084
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In star connected loads, line current = phase current. Therefore,
IL = Iph = 3.6084 A
400
Vph = VRN = √
3
Vph = 230.94 V
Vph
Zph =
Iph
Also, phase voltage,
Phase impedance,
230.94
3.6084
= 64
= 64∠ cos−1 0.8 = 64∠36.87◦
= 51.2 + j38.4
= 51.2 Ω
= 38.4 Ω
=
Zph
Zph
Zph
Rph
Xph
∴
In complex form,
or
Hence,
13. A three phase, 220 V, 50 Hz, 11.2 kW induction motor has a full load efficiency
of 88% and draws a line current of 38 A under full load, when connected to
three phase, 220 V supply. Find the reading on two wattmeters connected in
the circuit to measure the input to the motor. Determine also the power factor
at which the motor is working.
Power output = 11.2 kW
Efficiency = 88%
power output
11.2
Power input =
=
efficiency
0.88
= 12.73 kW
If two wattmeters are connected to measure total power, then,
Power,
So, the power factor,
or
Therefore,
W1 + W2 = 12.73 kW
√
P = 3 VL IL cos φ
(viii)
11200
cos φ = √
3 &times; 220 &times; 38
cos φ = 0.77
φ = cos−1 0.77 = 39.65
tan φ = tan 39.65 = 0.83
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0.83 =
or
W2 − W1 =
√
3
W2 − W1
W1 + W2
=
12.73 &times; 0.83
√
= 6.1
3
√
3
W2 − W1
12.73
(ix)
Solving Eqs. (viii) and (ix), we get
W1 = 3.315 kW
W2 = 9.415 kW
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66184
B.E/B.Tech. DEGREE EXAMINATION, NOV./DEC. 2011
Electrical and Electronics Engineering
EE2151 – CIRCUIT THEORY
1. In the circuit shown in Fig. 1, determine the total resistance across the supply
voltage.
18 Ω
82 Ω
100 Ω
60 Ω
76 Ω
E
40 Ω
Fig. 1
60 Ω and 40 Ω resistances are in parallel, i.e., (60||40) =
Now, 24 Ω and 76 Ω resistances are in series, i.e.,
82 Ω and 18 Ω resistances are in series, i.e.,
60 &times; 40
= 24 Ω
60 + 40
(24 + 76) = 100 Ω
(82 + 18) = 100 Ω
(i)
(ii)
Equivalent resistances calculated in Eqs. (i) and (ii) are found in parallel, i.e.,
(100||100) =
100 &times; 100
= 50 Ω
100 + 100
50 Ω and 100 Ω resistances are in series, hence,
(50 + 100) = 150 Ω
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2. A wheatstone bridge ABCD has the following details: AB = 1 kΩ, BC =
100 Ω, CD = 450 Ω and DA = 5 kΩ. A galvanometer of 500 Ω is connected
between B and D. A 4.5 V battery of negligible resistance is connected between
A and C with A at higher potential. Find the magnitude and direction of the
current through the galvanometer.
kΩ
10
0
Ω
1
500 Ω
5
kΩ
45
0
Ω
+–
4.5 V
Fig. 2
Fig. 2 shows the arrangement. To find the current through the galvanometer, mesh analysis is used and the direction of mesh currents are assumed as
indicated in Fig. 2.
By inspection, mesh equations are written as

 


0
6.5 −0.5
−5
I1

 −0.5
0
1.05 −0.45   I2  = 
0.0045
−5 −0.45
5.45
I3
Solving the above equations, we obtain
I1 = 4.0943 mA
I2 = 4.0569 mA
Therefore, current through galvanometer is
I1 − I2 = 0.0374 mA
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3. Find loop currents by mesh analysis for the circuit shown below.
5Ω
4Ω
+
2Ω
+
1000&ordm; V
j2 Ω
5090&deg; V
–j2 Ω
‒
‒
Fig. 3
By inspection, the mesh equations are written as

 

5 + j2 −j2
0
I1
100∠0◦
 −j2

4
j2   I2  = 
0
◦
0
j2 2 − j2
I3
50∠90

Using Cramer’s rule, the loop currents in the assumed direction I1 , I2 and I3
are found as follows:
∆I1
∆
5 + j2 −j2
0
4
j2
∆ = −j2
0
j2 2 − j2
I1 =
= [5 + j2][(4)(2 − j2) − (j2)2 ] + [j2][(−j2)(2 − j2)]
= 84 − j24
∆I1 =
100∠0◦ −j2
0
0
4
j2
50∠90◦ j2 2 − j2
= [100∠0◦ ][(4)(2 − j2) − (j2)2 ] + [j2][(−j2)(50∠90◦ )]
= 1200 − j600
∆I1
1200 − j600
=
∆
84 − j24
Therefore,
I1 =
or
= (15.0943 − j2.8302) A
I1 = 15.3573∠ − 10.62◦ A
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Similarly,
Therefore,
∆I2 =
5 + j2 100∠0◦
0
−j2
0
j2
0
50∠90◦ 2 − j2
= [5 + j2][(−j2)(50∠90◦ )] − [100∠0◦ ][(−j2)(2 − j2)]
= 900 + j600
∆I2
900 + j600
I2 =
=
∆
84 − j24
= (8.0189 + j9.4339) A
I2 = 12.3815∠49.64◦ A
5 + j2 −j2 100∠0◦
4
0
∆I3 = −j2
0
j2 50∠90◦
or
and
= [5 + j2][(4)(50∠90◦ )] + [j2][(−j2)(50∠90◦ )]
+ [100∠0◦ ][(−j2)(j2)]
= j1200
Therefore,
I3 =
∆I3
j1200
=
∆
84 − j24
= (−3.7736 + j13.21) A
I3 = 13.7384∠105.95◦ A
or
4. Using source transformation, replace the current source in the circuit shown
below by a voltage source and find the current delivered by the 50 V voltage
source.
5Ω
3Ω
2
10
+
A
Ω
50 V
+
10 V
Fig. 4
The 10 A current source is converted into a voltage source as shown in Fig. 5.
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I1
I1 ‒ I2
I2
5Ω
50 V
2Ω
+
20 V
3Ω
+
+
10 V
Fig. 5
Applying KVL to the two meshes with the assumed current directions, we get
or
or
−50 + 5I1 + 2I2 + 20 = 0
5I1 + 2I2 = 30
−20 − 2I2 + 3(I1 − I2 ) + 10 = 0
3I1 − 5I2 = 10
(iii)
(iv)
Solving Eqs. (iii) and (iv), we obtain the current delivered by 50 V source,
I1 = 5.4838 A.
5. Calculate the equivalent resistance Rab when all the resistance values are equal
to 1 Ω for the circuit shown below.
a
c
Req 
e
b
d
Fig. 6
Since the ∆ - connected resistances are having of equal values, their star equivalent resistances are obtained as:
Rstar =
1&times;1
1
=
1+1+1
3
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1Ω
a
c
a
c
1/3
1/3
Ω
Ω
1Ω
1Ω

1/3 Ω
e
e
Fig. 7
Thus, the two ∆ − networks ace and bde are converted into equivalent star
networks and are replaced in the original circuit as shown in Fig. 8.
a
3
1/
1/
3
c
Ω
Ω
1/3 Ω
1Ω
1Ω
e
1/3 Ω
3
1/
Ω
1/
3
b
Ω
d
Fig. 8
Figures 9 and 10 illustrate the step-by-step reduction of given network.
1/3 Ω
1/3 Ω
1/3 Ω
a
c
1Ω
2/3 Ω
1/3 Ω
a
1Ω
1Ω
1/3 Ω
b
2/3 Ω
5/3 Ω
1/3 Ω
d
(a)
b
(b)
Fig. 9
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1/3 Ω
a
a
1Ω
a
10/21 Ω
1Ω
8/7 Ω
0.5333 Ω
1/3 Ω
b
b
b
(a)
(b)
(c)
Fig. 10
Hence, from Fig. 10(c), the equivalent resistance is Rab = 0.5333 Ω.
6. For the circuit shown below, determine the voltage across (2 + j5) Ω impedance
by using the superposition theorem.
j4 Ω
– j3 Ω
2Ω
+
2030&deg; A
500&deg; V
j5 Ω
–
Fig. 11
50∠0◦ V source is acting alone
The given circuit is redrawn by open-circuiting 20∠30◦ A source as shown in
Fig. 11(a). By Ohm’ s law,
50∠0◦
= 1.1765 − j5.2941
I =
2 + j9
0
20∠30◦ A source is acting alone
The given circuit is redrawn by short-circuiting 50∠0◦ V source as shown in
Fig. 11(b). By current division rule,
j4
00
◦
I = 20∠30 &times;
= 6.3946 + j5.8655
2 + j9
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j4 Ω
j4 Ω
– j3 Ω
2Ω
+
– j3 Ω
2Ω
2030&deg; A
500&deg; V
j5 Ω
–
j5 Ω
(a)
(b)
Fig. 12
By superposition principle, the current through (2 + j5) Ω impedance is
= I 0 + I 00
= (1.1765 − j5.2941) + (6.3946 + j5.8655)
= (7.5711 + j0.5714) A
= 7.5926∠4.32◦ A
Therefore, voltage across (2 + j5) impedance = (7.5926∠4.32◦ )(2 + j5) =
40.8874∠72.52◦ V
7. A series RLC circuit has R = 2 Ω, L = 2 mH and C = 10 &micro;F. Calculate the
resonant frequency, the Q-factor, bandwidth and half power frequencies of the
circuit.
1
Resonant frequency,
fr = √
2π LC
1
= p
2π (2 &times; 10−3 )(10 &times; 10−6 )
= 1125.39 Hz.
r
1 L
Q-factor,
Q=
Rs C
=
Bandwidth,
1
2
(2 &times; 10−3 )
(10 &times; 10−6 )
= 7.07
fr
BW =
Q
=
1125.39
7.07
= 159.18 Hz.
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Lower half power frequency, f1 = fr −
R
4πL
= 1125.39 −
= 1045.81 Hz.
R
Upper half power frequency, f2 = fr +
4πL
= 1125.39 +
2
4π &times; 2 &times; 10−3
2
4π &times; 2 &times; 10−3
= 1204.97 Hz.
8. Find the value of L at which the circuit shown below resonates at a frequency
5Ω
L
10 Ω
–j12 Ω
Total admittance, Y = Y1 + Y2 Fig. 13
1
1
=
+
5 + jωL 10 − j12
1
5 − jωL
10 + j12
1
=
&times;
&times;
+
5 + jωL 5 − jωL
10 − j12 10 + j12
jωL
j12
5
10
−
+
=
+
25 + ω 2 L2 25 + ω 2 L2
100 + 144 244
Separating real and imaginary terms,
5
10
12
ωL
+j
Y =
+
−
25 + ω 2 L2 244
244 25 + ω 2 L2
At resonance, the imaginary part of above equation is zero, therefore,
12
ωL
−
=0
244 25 + ω 2 L2
or
ω 2 L2 − 20.33ωL + 25 = 0
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Solving the above quadratic equation, we get
ωL = 19.015 Ω or 1.315 Ω
Hence, L = 19.015 mH or 1.315 mH (∵ ω = 1000 rad/s.)
9. Calculate the effective inductance of the circuit shown below.
i
a
+
8H
4H
v
‒
b
10 H
5
H
6H
Fig. 14
Vab = V1 + V2 + V3
di
di
di
di
di
di
di
− 4 + 10 − 4 + 5 + 6 + 5
dt
dt
dt
dt
dt
dt
dt
di
Vab = 26
dt
di
di
Leff = 26
dt
dt
=8
or
Hence,
Leff = 26 H
10. Using Laplace transform, obtain the expression for i1 and i2 in the circuit shown
below, when DC voltage is applied suddenly. Assume that the initial energy
stored in the circuit is zero.
t=0
8.4 H
10 H
i1 (i1 ‒ i2)
336 V
42 Ω
i2
48 Ω
Fig. 15
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Applying KVL around the outer loop,
−336 + 8.4
di1
di2
+ 10
+ 48i2 = 0
dt
dt
Taking Laplace transformation on both sides, we get
8.4sI1 (s) + 10sI2 (s) + 48I2 (s) = 336
336
I1 (s)[8.4s] + I2 (s)[10s + 48] =
s
(v)
Similarly, applying KVL around the second mesh,
10
di2
+ 48I2 − 42(i1 − i2 ) = 0
dt
Taking Laplace transformation on both sides, we get
10sI2 (s) + 48I2 (s) − 42I1 (s) + 42I2 (s) = 0
I1 (s)[−42] + I2 (s)[10s + 48 + 42] = 0
(vi)
Writing Eqs. (iii) and (iv) in matrix form, we obtain
&quot;
#&quot;
# &quot;
#
8.4s 10s + 48
I1 (s)
336/s
=
−42 10s + 90
I2 (s)
0
Using Cramer’s rule, the current I1 (s) is found as,
I1 =
=
∆I1 (s)
∆
336
10s + 48
s
0
10s + 90
8.4s 10s + 48
−42 10s + 90
336
(10s + 90)
s
=
(8.4s)(10s + 90) + 42(10s + 48)
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30240
3360 +
s
=
2
84s + 1176s + 2016
=
=
I1 (s) =
3360s + 30240
84s(s2 + 14s + 24)
40s + 360
+ 14s + 24)
s(s2
40s + 360
s(s + 2)(s + 12)
By applying partial fraction technique,
I1 (s) =
40s + 360
A
B
C
= +
+
s(s + 2)(s + 12)
s
s + 2 s + 12
(vii)
Solving for the constants,
A=
B=
and
C=
40s + 360
(s + 2)(s + 12)
= 15
s=0
40s + 360
s(s + 12)
s=−2
40s + 360
s(s + 2)
s=−12
= −14
= −1
Substituting the values of A, B and C into Eq. (v), we get
I1 (s) =
15
14
1
−
−
s
s + 2 s + 12
Taking inverse Laplace transformation on both sides, the current i1 (t) is obtained as:
i1 (t) = 15 − 14e−2t − 1e−12t A
Similarly, by using Cramer’s rule, the current I2 (s) is found as,
I2 =
∆I2 (s)
∆
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=
8.4s
336
s
−42
0
8.4s 10s + 48
−42 10s + 90
336
42
s
=
(8.4s)(10s + 90) + 42(10s + 48)
14112
s
=
84s2 + 1176s + 2016
=
=
I2 (s) =
84s(s2
s(s2
14112
+ 14s + 24)
168
+ 14s + 24)
168
s(s + 2)(s + 12)
By applying partial fraction technique,
I2 (s) =
168
A
B
C
= +
+
s(s + 2)(s + 12)
s
s + 2 s + 12
(viii)
Solving for the constants,
A=
B=
and
C=
168
(s + 2)(s + 12)
168
s(s + 12)
168
s(s + 2)
=7
s=0
= −8.4
s=−2
= 1.4
s=−12
Substituting the values of A, B and C into Eq. (v), we get
I2 (s) =
7
8.4
1.4
−
+
s s + 2 s + 12
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Taking inverse Laplace transformation on both sides, the current i2 (t) is obtained as:
i2 (t) = 7 − 8.4e−2t + 1.4e−12t A
11. In the circuit shown below, find the expression for the transient current. The
initial current is as shown in the figure.
3H
i(0) = 6 A
i(t)
5Ω
100 V
Fig. 16
Applying KVL after closing on the switch to position-2,
5i + 3
di
= 100
dt
Taking laplace transformation on both sides, we get
5I(s) + 3[sI(s) − I(0)] =
100
s
At t = 0, i(t) = −6 A and in s − domain, I(0) = −6 A. Therefore, the above
equation becomes
100
s
100
I(s)[3s + 5] =
− 18
s
5I(s) + 3[sI(s) + 6] =
( 100
) − 18
s
3s + 5
100 − 18s
=
s(3s + 5)
I(s) =
=
100 − 18s
33.33 − 6s
=
3s(s + 1.6667)
s(s + 1.6667)
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By applying partial fraction technique,
I(s) =
33.33 − 6s
A
B
= +
s(s + 1.6667)
s
s + 1.6667
(ix)
Solving for constants A and B,
A=
33.33 − 6s
(s + 1.6667)
= 20 and B =
s=0
33.33 − 6s
s
= −26
s=−1.6667
Substituting the values of A and B into Eq. (ix), we get
I(s) =
26
20
−
s
s + 1.6667
Taking inverse Laplace transformation on both sides, the current equation is
obtained as:
i(t) = (20 − 26e−1.6667t ) A
12. In the circuit shown below, find the transient current after switch is closed at
time t = 0, given that an initial charge of 100 &micro;C is stored in the capacitor.
t=0
15 Ω
50 V
200 μF
i(t)
Fig. 17
Initial voltage on the capacitor,
V =
Q
100 &times; 10−6
=
= 0.5 V
C
200 &times; 10−6
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Applying KVL after closing the switch,
Ri + vc = V
dvc
+ vc = V
dt
dvc
+ vc = 50
15 &times; 200 &times; 10−6
dt
dvc
0.003
+ vc = 50
dt
RC
Taking Laplace transformation on both sides, we get
0.003[sVC (s) − VC (0)] + VC (s) =
50
s
At t = 0, VC (0) = 0.5 V. Therefore, the above equation becomes
50
s
50
+ 0.0015
VC (s)[1 + 0.003s] =
s
0.003[sVC (s) − 0.5] + VC (s) =
VC (s) =
( 50
) + 0.0015
s
(1 + 0.003s)
=
50 + 0.0015s
s(1 + 0.003s)
=
50 + 0.0015s
16666.67 + 0.5s
=
0.003s(s + 333.33)
s(s + 333.33)
By applying partial fraction technique,
I(s) =
16666.67 + 0.5s
A
B
= +
s(s + 333.33)
s
s + 333.33
(x)
Solving for constants A and B,
A=
16666.67 + 0.5s
(s + 333.33)
= 50 and B =
s=0
16666.67 + 0.5s
s
= −49.5
s=−333.33
Substituting the values of A and B into Eq. (vii), we get
VC (s) =
50
49.5
−
s
s + 333.33
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Taking inverse Laplace transformation on both sides, the voltage across the
capacitor is obtained as
vC (t) = 50 − 49.5e−333.33t V
The transient current is given by
i(t) = C
dvC
dt
d
(50 − 49.5e−333.33t )
dt
A
= (200 &times; 10−6 )
Hence,
i(t) = 3.3e−333.33t
13. A sinusoidal voltage of 10 sin 100t is connected in series with a switch and
R = 10 Ω and L = 0.1 H. If the switch is closed at t = 0, determine the
transient current i(t).
Applying KVL after closing on the switch to position-2,
10i + 0.1
di
= 10 sin 100t
dt
Taking laplace transformation on both sides, we get
100
10I(s) + 0.1[sI(s) − I(0)] = 10 2
s + 1002
At t = 0, i(t) = 0 A and in s − domain, I(0) = 0 A. Therefore, the above
equation becomes
100
10I(s) + 0.1[sI(s)] = 10 2
s + 1002
I(s)[0.1s + 10] =
s2
I(s) =
=
1000
+ 1002
1000
2
s + 1002
1
0.1s + 10
10000
(s + 100)(s2 + 1002 )
By applying partial fraction technique,
I(s) =
10000
A
Bs + C
=
+ 2
2
2
(s + 100)(s + 100 )
s + 100 s + 1002
(xi)
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The constants A, B and C are evaluated as follows:
A=
10000
+ 1002 )
(s2
= 0.5
s=−100
On cross-multiplying Eq. (ix), we get
10000 = A(s2 + 1002 ) + (Bs + C)(s + 100)
10000 = As2 + 10000A + Bs2 + 100Bs + Cs + 100C
10000 = s2 (A+B) + s(100B+C) + 10000A + 100C
(xii)
On equating the coefficients of s2 of Eq. (xii), we get
or
A+B=0
B = −A = −0.5
On equating the coefficients of s of Eq. (xii), we get
or
100B + C = 0
C = 50
Substituting the values of A, B and C into Eq. (xi), we get
0.5
0.5s + 50
− 2
s + 100 s + 1002
0.5
0.5s
50
=
− 2
+ 2
2
s + 100 s + 100
s + 1002
0.5
0.5s
50
100
=
−
+
s + 100 s2 + 1002
100
s2 + 1002
I(s) =
Taking inverse Laplace transformation on both sides, the above equation becomes
i(t) = (0.5e−100t − 0.5 cos 100t + 0.5 sin 100t) A
14. If W1 and W2 are the readings of the two wattmeters which measures power in
W1
= a, show that the power factor of
the three phase balanced system and if
W2
a+1
the circuit is given by cos φ = √
.
2 a2 − a + 1
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For two wattmeter method of measuring three phase power, we have
√
W2 − W1
tan φ = 3
W1 + W2
W1
√ W2 1 − W2
= 3
W1
W2 1 +
W2
√ (1 − a)
3
(1 + a)
(1 − a)2
φ = (3)
(1 + a)2
(1 − a)2
φ = 1 + (3)
(1 + a)2
(1 + a)2 + 3(1 − a)2
φ=
(1 + a)2
(1 + a2 + 2a) + 3(1 + a2 − 2a)
=
φ
(1 + a)2
(1 + a2 + 2a + 3 + 3a2 − 6a)
=
(1 + a)2
(4a2 − 4a + 4)
=
(1 + a)2
4(a2 − a + 1)
=
(1 + a)2
(1 + a)2
φ=
4(a2 − a + 1)
=
tan2
1 + tan2
sec2
1
cos2
cos2
Therefore,
1+a
cos φ = √
2 a2 − a + 1
15. Obtain the readings of two wattmeters connected to a three phase, 3 wire, 120 V
system feeding a balanced ∆ connected load with a load impedance of 12∠30◦ Ω.
Assume either phase sequence. Determine the phase power and compare the
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total power to the sum of wattmeter readings.
Phase current,
IRY =
=
VRY
ZRY
120∠0◦
12∠30◦
= 10∠ − 30◦ A
Similarly, the other phase currents
and
Power factor,
IYB = 10∠ − 150◦ A
IBR = 10∠ − 270◦ A
cos φ = cos 30◦ = 0.866
Line current (IL ) in R-line is calculated by using KCL as follows:
So, total power
Also, total power
Per phase power
IL = IRY − IBR
= (10∠ − 30◦ ) − (10∠ − 270◦ )
= 17.32∠ − 60◦
√
P = 3VL IL cos φ
√
= 3 &times; 120 &times; 17.32 &times; 0.866 = 3117.6 W
P = W1 + W2
W1 = VL IL cos (30◦ + φ)
= 120 &times; 17.32 &times; cos (30◦ + 30◦ )
W1 = 1039.2 W
(xiii)
W2 = VL IL cos (30◦ − φ)
= 120 &times; 17.32 &times; cos (30◦ − 30◦ )
W2 = 2078.4 W
= Vph Iph cos φ
= 120 &times; 10 &times; 0.866
= 1039.2 W
(xiv)
(xv)
Upon comparing Eqs. (xiii) and (xv), it is observed that when the power factor
is 0.866, one of the wattmeters reads per phase power, i.e., one-third of the total
power, while the other wattmeter reads two-third of the total power.
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64010
B.E/B.Tech. DEGREE EXAMINATION, NOV. /DEC. 2010
Electrical and Electronics Engineering
EE2151 – CIRCUIT THEORY
1. Determine the current IL in the circuit shown below.
4V
3Ω
3Ω
3Ω
IL
8V
6V
5Ω
1Ω
1Ω
Fig. 1
The mesh currents are assumed as indicated in Fig. 2.
4V
3Ω
I3
3Ω
3Ω
IL
I1
8V
5Ω
1Ω
6V
I2
1Ω
Fig. 2
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By mesh analysis method, the mesh equations are written as


  
9 5 −3
I1
8
 5 9




3
I2 = 6 
−3 3
9
I3
4
Firstly, I1 and I2 are found by using Cramer’s rule as follows:
∆=
9 5 −3
5 9
3
−3 3
9
= 9(81 − 9) − 5(45 + 9) − 3(15 + 27) = 252
∆I1 =
8 5 −3
6 9
3
4 3
9
= 8(81 − 9) − 5(54 − 12) − 3(18 − 36) = 420
∆I2 =
9 8 −3
5 6
3
−3 4
9
= 9(54 − 12) − 5(54 − 12) − 3(20 + 18) = −168
420
∆I1
=
= 1.6667
∆
252
−168
∆I2
=
= −0.6667
I2 =
∆
252
Therefore, I1 =
Similarly,
Hence, the current IL = I1 + I2 = (1.6667 − 0.6667) = 1 A
2. For the circuit shown in figure below, determine the total current IT , phase
angle and power factor.
100 μF
10 Ω
IT
50 V,
100 Hz
30 Ω
0.1 H
Fig. 3
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Capacitive reactance of 100 &micro;F, XC =
1
2πf C
1
2π(100)(100 &times; 10−6 )
= 15.92 Ω
Inductive reactance of 0.1 H, XL = 2πf L
= 2π(100)(0.1)
= 62.83 Ω
Equivalent impedance of the circuit = (10 − j15.92) + (30)||(j62.83)
= (10 − j15.92) + (24.43 + j11.66)
= (34.43 − j4.26) Ω
50
Total current, IT =
34.43 − j4.26
= (1.43 + j0.17) = 1.44∠7.05◦
So, phase angle, φ = 7.05◦
and power factor, cos φ = cos 7.05◦ = 0.99.
=
3. Find the current through the source and capacitance in the network shown
below, using mesh current analysis.
Using mesh analysis method, the mesh equations by inspection are written as,
96 Ω
+
100&deg;
‒
1Ω
104 Ω
I1
100 Ω
I2
‒j50 Ω
j200 Ω
Fig. 4
200 + j200 −104 − j200
−104 − j200
205 + j150
I1
I2
Using Cramer’s rule,
Current through source, I1 =
=
10∠0
0
∆I1
∆
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=
=
10∠0 −104 − j200
0
205 + j150
200 + j200 −104 − j200
−104 − j200
205 + j150
2050 + j1500
40184 + j29400
= (0.051 + j2.4202 &times; 10−6 ) A = 0.051∠0.0027◦ A
Similarly, current through capacitance, I2 =
200 + j200 10∠0
−104 − j200
0
200 + j200 −104 − j200
−104 − j200
205 + j150
=
=
∆I2
∆
1040 + j2000
40184 + j29400
= (0.04 + j0.02) A = 0.045∠26.34◦ A
4. Compute V1 and V2 in the circuit shown below, using node analysis.
1045&deg; V
+
V1
30&deg; A
4Ω
‒j3 Ω
V2
j6 Ω
12 Ω
Fig. 5
[As the 4 Ω resistance is connected at the supernode it doesn’t have any significance in the
calculations].
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By supernode analysis, the nodal equations are written as,
−
V1 V2 V2
+
+
= 3∠0◦
j3 j6 12
j0.3333V1 + (0.0833 − j0.1666)V2 = 3
V1 − V2 = 10∠45◦
(i)
(ii)
Putting Eqs. (i) and (ii) in matrix form, we get
j0.3333 0.0833 − j0.1666
V1
3
=
1
−1
V2
10∠45◦
By applying Cramer’s rule,
∆=
j0.3333 0.0833 − j0.1666
−1
1
= (−j0.3333) − (0.0833 − j0.1666)
= −0.0833 − j0.1667
∆V1 =
3
0.0833 − j0.1666
10∠45◦
−1
◦
= (−3) − (10∠45 )(0.0833 − j0.1666)
= −4.7671 + j0.589
∆V2 =
j0.3333
3
1
10∠45◦
= (−3) − (10∠45◦ )(0.0833 − j0.1666)
= −4.7671 + j0.589
∆V1
V1 =
∆
−4.7671 + j0.589
=
−0.0833 − j0.1667
= (8.6073 − j24.29) A = 25.78∠ − 70.49◦ V
∆V2
V2 =
∆
−5.3568 + j2.3568
=
−0.0833 − j0.1667
= (1.536 − j31.37) A = 31.4044∠ − 87.2◦ V
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5. Convert the network shown below, into a π− connected equivalent circuit.
a
‒j2 Ω
j4 Ω
c
d
2Ω
a′
2Ω
b
‒j4 Ω
e
b′
f
Fig. 6
Using star-delta transformation, the star connection with c as star-point can be
converted into an equivalent delta network as follows:
(−j2)(2) + (2)(j4) + (j4)(−j2)
= 4 + j2
(2)
Zae =
(−j2)(2) + (2)(j4) + (j4)(−j2)
= 1 − j2
(j4)
Zed =
(−j2)(2) + (2)(j4) + (j4)(−j2)
= −2 + j4
(−j2)
a
(1 ‒ j2) Ω
a′
(4 + j2) Ω
d
(‒2 + j4) Ω
2Ω
b
‒j4 Ω
e
f
b′
Fig. 7
The delta network thus found is replaced the star network in the given circuit
as shown in Fig. 7, where (−2+j4) and (−j4) are found parallel. The equivalent
value of impedance is,
Zeq1 =
(−2 + j4)(−j4)
= (−8 − j4)
(−2 + j4 − j4)
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Using Zeq1 , Fig. 7 is redrawn as illustrated in Fig. 8.
(4 + j2) Ω
a
(1 ‒ j2) Ω
a′
2Ω
d
b
(‒8 ‒ j4) Ω
b′
f
Fig. 8
Again using star-delta transformation, the star connection with d as star-point
can be converted into an equivalent delta network as follows:
Zab =
(4 + j2)(−8 − j4) + (−8 − j4)(2) + (2)(4 + j2)
= 5 + j2
(−8 − j4)
Zaf =
(4 + j2)(−8 − j4) + (−8 − j4)(2) + (2)(4 + j2)
= −16 − j18
(2)
Zbf =
(4 + j2)(−8 − j4) + (−8 − j4)(2) + (2)(4 + j2)
= −10 − j4
(4 + j2)
(5 + j2) Ω
a
(1 ‒ j2) Ω
a′
(‒16 ‒ j18) Ω
f
b
(‒10 ‒ j4) Ω
b′
Fig. 9
Using the delta equivalent values the simplified network is redrawn as shown
in Fig. 9, in which (1 − j2) and (−16 − j18) are found parallel. The equivalent
value of impedance is, therefore,
Zeq2 =
(1 − j2)(−16 − j18)
= (0.8 − j2)
(1 − j2) + (−16 − j18)
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(5 + j2) Ω
a
(0.8 ‒ j2) Ω
b
(‒10 ‒ j4) Ω
b′
a′
Fig. 10
Figure 9 is further modified using the value of (0.8 − j2) as found in Fig. 10
which is the π− equivalent circuit of the given network.
6. Calculate the current through the 2 Ω resistor in the circuit shown in Fig. 11,
using superposition theorem.
1Ω
10 V
+
1Ω
2Ω
1Ω
1A
Fig. 11
10 V source is acting alone
The given circuit is redrawn by open-circuiting 1 A current source.
1Ω
10 V
+
I1
1Ω
2Ω
I2
1Ω
Fig. 12
By mesh analysis method, the mesh equations are written as
3 −2
I1
10
=
−2
4
I2
0
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Solving, we get
I1 = 5 A
I2 = 2.5 A
Therefore, current through 2 Ω resistor due to 10 V source is I 0 = 5 − 2.5 = 2.5
A.
1 A source is acting alone
1Ω
I3
1Ω
1Ω
I4
2Ω
+
1V
Fig. 13
The given circuit is redrawn by short-circuiting 10 V source. By source transformation technique, the 1 A current source is converted into voltage source as
shown in Fig. 13. By mesh analysis method, the mesh equations are written as
3 −2
I3
0
=
−2
4
I4
1
Solving, we get
I3 = 0.25 A
I4 = 0.375 A
Therefore, current through 2 Ω resistor due to 1 A source is I 00 = 0.375 – 0.25
= 0.125 A.
By the principle of superposition, current through 2 Ω resistor,
I = I 0 + I 00 = 2.5 + 0.125 = 2.625 A.
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7. Calculate the current through the 2 Ω resistor in the circuit shown in Fig. 14,
using Thevenin’s theorem.
1Ω
a
2V
+
1Ω
1A
2Ω
b
Fig. 14
To find Thevenin’s voltage, Vth
1Ω
Va
a
2V
+
1Ω
1A
Vth
b
Fig. 15
The 2 Ω resistor connected between terminals a and b is removed first. The
voltage, Vth across a and b is obtained by using Nodal analysis as follows:
Va − 2 Va
+
=1
1
1
or
Va = Vth = 1.5 V
To find Thevenin’s resistance, Rth
1Ω
a
 Rth
1Ω
b
Fig. 16
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Figure 16 is used to find Rth . As seen in Fig. 16,
Rth =
1
= 0.5 Ω
1+1
Using the values of Vth , and Rth found above, the Thevenin’s equivalent circuit
is drawn as shown in Fig. 17.
1.5
= 0.6 A
As seen in Fig. 17, current through 2 Ω resistor is =
0.5 + 2
Rth = 0.5 Ω
a
Vth =
1.5 V
+
2Ω
b
Fig. 17
8. Determine the quality factor of a coil for the series circuit consisting of R =
10 Ω, L = 0.1 H and C = 10 &micro;F. Derive the formula used.
1
1
=p
LC
(0.1)(10 &times; 10−6 )
ωr L
Q=
R
1000 &times; 0.1
=
10
Q = 10.
ωr = √
Therefore,
For derivation of the formula for Q-factor of a coil, please refer to Section 10.1.6.
9. Consider the single tuned circuit shown below and determine,
(i) the resonant frequency, and
(ii) the output voltage at resonance.
Assume RS &gt;&gt; ωr L1 and k = 0.9.
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M
10 Ω
10 Ω
R2
L1
15 V
1 μH
L2
100 μH
–
0.1 μF
RS

vo
Fig. 18
Mutual inductance,
At resonance
Resonant frequency,
Output voltage,
p
L1 L2
p
= 0.9 (1 &times; 10−6 )(100 &times; 10−6 )
= 9 &micro;H
1
ωr = √
L2 C
1
=p
−6
(100 &times; 10 )(0.1 &times; 10−6 )
M Vi
Vo =
C[R1 R2 + ω 2 M 2 ]
9 &times; 10−6 &times; 15
=
0.1 &times; 10−6 [(10)(10) + (316227.766)2 (9 &times; 10−6 )2 )]
= 12.49 V.
M =k
10. Derive the necessary equations for the response of RL circuit to step input
and also draw the current response curve. Using your result, write down the
expression for current in the circuit shown below, when the switch is closed at
time t = 0.
t=0
60 V
30 Ω
i(t)
15 H
Fig. 19
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Applying KVL to the circuit after closing the switch, we get
Ri + L
di
=V
dt
Taking laplace transformation on both sides,
RI(s) + L[sI(s) − I(0)] =
V
s
As there is no initial current through the inductor, I(0) = 0, therefore,
V
s
V
I(s)[R + sL] =
s
RI(s) + L[sI(s)] =
V
s(sL + R)
V
=
R
sL s +
L
(V /L)
=
s(s + R/L)
Hence,
I(s) =
Applying partial fraction technique,
I(s) =
(V /L)
A
B
= +
s(s + R/L)
s
s + R/L
(iii)
Solving for the values of A and B, we get
A=
V
R
B=−
V
R
Substituting the values of A and B into Eq. (iii), we get
V
V
−
sR R s + R
L
!
V 1
1
=
−
R s s+ R
L
I(s) =
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Taking inverse laplace transformation on both sides, we get
R
V i(t) =
1 − e− L t
R
Substituting the given values into the above equation, we get
30
60 1 − e− 15 t
i(t) =
30
Hence, the expression for current, i(t) = 2(1 − e−2t ) A.
i
2A
t
0
Fig. 20
The current response curve is shown in Fig. 20.
11. A step voltage v(t) = 100 u(t) is applied to a series RLC circuit with L = 10
H, R = 2 Ω and C = 5 F. The initial current in the circuit is zero, but there
is an initial voltage of 50 V on the capacitor in a direction, which opposes the
applied source. Find the expression for the current in the circuit.
By applying KVL to the given circuit,
Z
di 1
i dt = 100 − 50
2i + 10 +
dt 5
as the polarity of initial voltage on the capacitor is such that it opposes the
applied voltage.
Z
di 1
2i + 10 +
i dt = 50
dt 5
Taking laplace transformation on both sides, we get
2I(s) + 10[sI(s) − I(0)] +
1 I(s)
50
=
5 s
s
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As there is no initial current through the inductor, I(0) = 0. Therefore, the
above equation becomes
1 I(s)
2I(s) + 10sI(s) +
=
5
s
1
I(s) 2 + 10s +
=
5s
10s + 50s2 + 1
I(s)
=
5s
50
s
50
s
50
s
50
5s
I(s) =
s 10s + 50s2 + 1
250
=
2
50(s + 0.2s + 0.02)
5
= 2
s + 0.2s + 0.02
5
5
=
=
2
(s + 0.1) + 0.01
(s + 0.1)2 + 0.12
5
0.1
=
0.1 (s + 0.1)2 + 0.12
Taking inverse laplace transformation on both sides,
i(t) = 50e−0.1t sin 0.1t A
12. Determine the line current, power factor and total power when a 3-phase, 400
V supply is given to a balanced load of impedance (8 + j6) Ω in each branch is
connected in star.
Let the phase sequence be RYB.
Line currents
In star-connected networks, since IL = Iph , the line currents are found just by
calculating the phase currents as following:
400∠0◦
Phase voltage,
Vph = √
= 230.94∠ − 30◦
3
[It is to be noted that in star-connected networks, the phase voltage lags behind the line voltage
by 30◦ ]
So, phase current in R − phase,
IRN = IR =
Vph
Zph
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230.94∠ − 30◦
(8 + j6)
= (9.07 − j21.24) A
= 23.09∠ − 66.87◦ A
IYN = IY = 23.09∠ − 186.87◦ A
IBN = IB = 23.09∠ − 306.87◦ A
=
Phase current in Y − phase,
Phase current in B − phase,
Power factor
−1
φ = tan
So,
Total power,
6
8
= 36.87◦
cos φ = 0.8
√
P = 3 VL IL cos φ
√
= 3 (400)(23.09)(0.8)
= 12797.78 W.
13. A three-phase four-wire 120 V ABC system feeds an unbalanced Y-connected
load with ZA = 5∠0 Ω, ZB = 10∠30◦ Ωand ZC = 20∠60◦ Ω. Obtain the four
line currents.
Let the phase sequence be ABC. The line currents are found as follows:
120∠0
= (24∠0) A
5∠0
120∠120◦
= (12∠90◦ ) A
IB = IBN =
10∠30◦
120∠240◦
IC = ICN =
= (6∠180◦ ) A
◦
20∠60
Current through neutral wire, IN = −(IA + IB + IC )
= −(24∠0 + 12∠90◦ + 6∠180◦ )
= (−18 − j12) A = 21.6333 ∠ − 146.31◦ A.
IA = IAN =
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14. Three impedances Z1 = (17.32 + j10), Z2 = (20 + j34.64) and Z3 = (0 − j10)
ohms are delta connected to a 400 V, three phase system. Determine the phase
currents, line currents and total power consumed by the load.
Let the phase sequence be ABC.
The phase currents are found as follows:
IRY =
IYB
400∠0
= (17.32 − j10) = 20∠ − 30 A
17.32 − j10
400∠120◦
=
= (5 + j8.66) = 10∠60◦ A
20 + j34.64
IBR =
400∠240◦
= (34.64 − j20) = 40∠ − 30◦ A
0 − j10
The three line currents are found by using KCL as follows:
IR = IRY − IBR
= (17.32 − j10) − (34.64 − j20)
= (−17.32 + j10) = 20∠150◦ A
IY = IYB − IRY
= (5 + j8.66) − (17.32 − j10)
= (−12.32 + j18.66) = 22.36∠123.43◦ A
IB = IBR − IYB
= (34.64 − j20) − (5 + j8.66)
= 29.64 − j28.66 = 41.23∠ − 44◦ A.
2
2
Total power consumed by the load, P = IRY
RR + IYB
RY + IB2 RBR
= 202 (17.32) + 102 (20) + 402 (0)
= 8928 W.
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G 0662
B.E/B.Tech. DEGREE EXAMINATION, NOV./DEC. 2010
Electrical and Electronics Engineering
EE1151 – CIRCUIT THEORY
1. A coil having a resistance of 10 ohm and an inductance of 10 mH is connected
to a 100 V, 50 Hz supply. Calculate
(a)
(b)
(c)
(d)
The impedance
Circuit current
Phase angle
Power factor
(a) Impedance
Inductive reactance, XL = 2πf L
= 2π &times; 50 &times; 10 &times; 10−3 = 3.14 Ω
q
Impedance, Z = R2 + XL2
√
= 102 + 3.142 = 10.48 Ω
(b) Circuit current
V
Z
100
= 9.54 A
=
10.48
Circuit current, I =
(c) Phase angle
−1
Phase angle, φ = tan
X
R
−1
= tan
3.14
10
= 17.43◦
(d) Power factor
Power factor = cos φ = cos 17.43◦ = 0.95
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2. Using mesh analysis, determine the currents IA and IB in the following circuit.
1Ω
14 V
2Ω
6Ω
4Ω
8Ω
IA
4Ω
IB
Fig. 1
Mesh currents I1 , I2 and I3 are assumed as shown in Fig. 2.
1Ω
14 V
2Ω
6Ω
I1
I2
4Ω
8Ω
I3
4Ω
Fig. 2
By mesh analysis method, the mesh equations are written as


 

7 −6
0
I1
14
 −6 16 −8   I2  =  0 
0 −8 16
I3
0
Cramer’s rule is applied to solve the above matrix equation.
∆=
7 −6
0
−6 16 −8
0 −8 16
= 7(256 − 64) + 6(−96 − 0) = 768
∆I1 =
Therefore,
14 −6
0
0 16 −8
0 −8 16
= 14(256 − 64) = 2688
2688
∆I1
=
= 3.5
I1 =
∆
768
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Similarly, I2 is found as,
∆I2
∆
7 14
0
∆I2 = −6 0 −8
0 0 16
I2 =
= −14(−96) = 1344
Therefore,
I2 =
∆I2
1344
=
= 1.75
∆
768
and
∆I3
∆
7 −6 14
∆I3 = −6 16 0
0 −8 0
I3 =
= 14(48) = 672
Therefore,
I3 =
∆I3
672
=
= 0.875
∆
768
Hence,
IA = I1 − I2 = 3.5 − 1.75 = 1.75 A
IB = I2 − I3 = 1.75 − 0.875 = 0.875 A
3. Verify the reciprocity theorem by finding the current I.
10 Ω
20 V
30 Ω
5Ω
5Ω
I
Fig. 3
The current I is found by using mesh current analysis.
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10 Ω
20 V
5Ω
I1 30 Ω
I2
5Ω
I
Fig. 4
The mesh equations by inspection [Fig. 4] are written as,
40 −30
I1
20
=
−30
40
I2
0
Using Cramer’s rule I is found as,
I = I2 =
∆I2 =
∆I2
∆
40 20
−30 0
= 600
40 −30
∆=
= 1600 − 900 = 700
−30
40
Therefore,
I = I2 =
600
∆I2
=
= 0.8571 A
∆
700
(i)
To verify reciprocity theorem, the excitation 20 V source and the response I are
interchanged as shown in Fig. 5.
10 Ω
I
5Ω
I1 30 Ω
I2
5Ω
20 V
Fig. 5
The mesh equations by inspection [Fig. 5] are written as,
40 −30
I1
0
=
−30
40
I2
20
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Again mesh analysis is used to find I.
∆I1
∆
0 −30
∆I1 =
= 600
20
40
∆I1
600
I = I1 =
=
= 0.8571 A
∆
700
I = I1 =
Therefore,
(ii)
Comparing Eqns. (i) and (ii), it is observed that in both results the ratio of
excitation to response is same. Hence, reciprocity theorem is verified.
4. Using star to delta transformation obtain the equivalent resistance between A
and B.
12 Ω
2Ω
3Ω
A
6Ω
Requ
18 Ω
B
Fig. 6
The star network comprising resistors 2 Ω, 3 Ω and 6 Ω [Fig. 7(a)] is converted
into its equivalent delta network as shown in Fig. 7(b).
3Ω
6Ω
A
12

6Ω
C
Ω
C
Ω
18
2Ω
A
B
B
(a)
(b)
Fig. 7
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RAB =
(2 &times; 6) + (6 &times; 3) + (3 &times; 2)
= 12
3
(2 &times; 6) + (6 &times; 3) + (3 &times; 2)
= 18
2
(2 &times; 6) + (6 &times; 3) + (3 &times; 2)
=6
=
6
RBC =
RCA
The modified network shown in Fig. 8(a) is further reduced to Fig. 8(b). From
Fig. 8(b), the equivalent resistance between A and B obtained as:
RAB = 12||(4 + 9)
12 &times; 13
=
12 + 13
= 6.24 Ω
12 Ω
4Ω
6Ω
C
12
18
Ω
A
A
18 Ω

C
12 Ω
9Ω
Ω
B
(b)
B
(a)
Fig. 8
5. A series RLC circuit consists of R = 100 Ω, L = 0.02 H and C = 0.02 &micro;F. If it
is excited by sinusoidal voltage voltage source of 50 V, determine
(a) the resonant frequency
(b) current at resonance
(c) impedance at resonance
(d) voltage across each element
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Resonant frequency,
1
√
2π LC
fr =
1
p
2π (0.02)(0.02 &times; 10−6 )
= 7957.75 Hz.
V
I=
R
50
= 0.5 A
=
100
=
Current at resonance,
Impedance at resonance,
Voltage across resistor,
Z = R = 100 Ω
VR = IR = 0.5 &times; 100 = 50 V
r
1 L
Q=
R s
C
Q-factor,
1
100
=
(0.02)
(0.02 &times; 10−6 )
= 10
At resonance, voltage across inductor = voltage across capacitor. Therefore,
VL = VC = QV = 10 &times; 50 = 500 V
6. Find the current I2 using mesh method in the following circuit.
j5 Ω
j2 Ω
j10 Ω
3
1000&deg; V
I1
I2
5
–j2 Ω
Fig. 9
Current I1 is leaving the dot and I2 is entering the dot, hence the polarities of
mutually induced voltage and self-induced voltage are opposite.
Applying KVL to the first mesh,
j5I1 − j2I2 + 3(I1 − I2 ) − j2(I1 − I2 ) = 100∠0◦
(3 + j3)I1 − 3I2 = 100∠0◦
(iii)
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Applying KVL to the second mesh,
j10I2 − j2I1 + 5I2 − j2(I2 − I1 ) + 3(I2 − I1 ) = 0
−3I1 + (8 + j8)I2 = 0
(iv)
Putting Eqs. (iii) and (iv) in matrix form,
3 + j3
−3
I1
100∠0◦
=
−3 8 + j8
I2
0
Applying Cramer’s rule, I2 is calculated as:
3 + j3 100∠0◦
−3
0
∆I2
I2 =
=
∆
3 + j3
−3
−3 8 + j8
100∠0◦
=
−9 + j48
I2 = 6.143∠ − 100.62◦ A
7. Obtain the expression for the voltage across C for the circuit shown below.
S
t=0
E
R
C
20 Ω
0.1 F
20 V
Fig. 10
Applying KVL after closing the switch,
Ri + vc = E
dvc
+ vc = E
dt
dvc
20 &times; 0.1
+ vc = 20
dt
dvc
0.2
+ vc = 20
dt
RC
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Taking Laplace transformation on both sides, we get
0.2[sVC (s) − VC (0)] + VC (s) =
20
s
At t = 0, VC (0) = 0 V. Therefore, the above equation becomes
20
s
20
VC (s)[1 + 0.2s] =
s
0.2[sVC (s) − 0] + VC (s) =
VC (s) =
=
20
s(1 + 0.2s)
20
100
=
0.2s(s + 5)
s(s + 5)
By applying partial fraction technique,
I(s) =
A
B
100
= +
s(s + 5)
s
s+5
(v)
Solving for constants A and B,
A=
100
(s + 5)
= 20 and B =
s=0
100
s
= −20
s=−5
Substituting the values of A and B into Eq. (v), we get
VC (s) =
20
20
−
s
s+5
Taking inverse Laplace transformation on both sides, the voltage across the
capacitor is obtained as
vC (t) = 20 − 20e−5t V
8. A three phase balanced system supplies 110 V to a delta connected load whose
phase impedance are equal to (3.54 + j3.54) Ω. Determine the line and phase
currents. Use the phase sequence RYB.
The phase currents in all the three phases are calculated as follows:
IRY =
110∠0◦
= 21.97∠ − 45◦ A
3.54 + j3.54
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IYB =
110∠ − 120◦
= 21.97∠ − 165◦ A
3.54 + j3.54
IBR =
110∠ − 240◦
= 21.97∠ − 285◦ A
3.54 + j3.54
By applying KCL the line currents are found as follows:
IR = IRY − IBR
= (21.97∠ − 45◦ ) − (21.97∠ − 285◦ )
= 38.05∠ − 75◦ A
IY = IYB − IRY
= (21.97∠ − 165◦ ) − (21.97∠ − 45◦ )
= 38.05∠165◦ A
IB = IBR − IYB
= (21.97∠ − 285◦ ) − (21.97∠ − 165◦ )
= 38.05∠45◦ A
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E3081
B.E/B.Tech. DEGREE EXAMINATION, JUNE 2010
Electrical and Electronics Engineering
EE2151 – CIRCUIT THEORY
1. In the circuit shown in Fig. 1, find the different mesh currents, power delivered
by each source and the current through RL .
1Ω
2Ω
RL
2Ω
1Ω
10 V
1Ω
12 V
IL
2Ω
1Ω
1Ω
Fig. 1
The mesh currents are assumed with respective direction as indicated in Fig. 2.
1Ω
10 V
I1
2Ω
1Ω
I2
1Ω
RL
2Ω
I3
12 V
IL
2Ω
1Ω
1Ω
Fig. 2
By mesh analysis method, the mesh equations are written as


 

4 −1
0
I1
10
 −1
6 −2   I2  =  0 
0 −2
4
I3
12
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Using Cramer’s rule, I1 , I2 and I3 are found as,
I1 = 2.9474 A
I2 = 1.7895 A
I3 = 3.8947 A
Power delivered by 10 V source = 10 &times; I1 = 29.474 W.
Power delivered by 12 V source = 12 &times; I3 = 46.7364 W.
Current through RL = I2 − I3 = −2.1052 A.
2. In the circuit shown in Fig. 3, find the different node voltages and the currents
I1 , I2 and I3 .
12 A
4Ω
2Ω
4Ω
8Ω
15 A
10 Ω
4Ω
I2
I3
8Ω
I1
32 V
Fig. 3
Using source transformation technique, the 32 V voltage source is converted
into current source as shown in Fig. 4.
12 A
V1
V2
V3
2Ω
4Ω
15 A
8Ω
I1
10 Ω
8Ω
I2
4Ω
4Ω
8A
I3
Fig. 4
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By node analysis method, the node equations by inspection are written as


1 1 1
1
+ +
−
0
 8 8 4

 

4

 V1
15
−
12


1
1 1
1
1

  V2  = 

0
−
+ +
−


4
4 2 10
2

 V3
12 + 8

1 1 1 
1
+ +
0
−
2
4 4 2
Solving the above equations, we get
V1 = 18.1053 V
V2 = 24.2105 V
V3 = 32.1053 V
Hence, the currents I1 , I2 and I3 are found as follows:
I1 = V1 /8 = 2.2632 A
I2 = V2 /10 = 2.4211 A
I3 = V3 /4 = 8.0263 A.
3. In the circuit shown in Fig. 5, find
(i) the equivalent resistance between P and Q
(ii) the total current from the 240 V source
(iii) the current through the 18 Ω .
2Ω
A
6
Ω
P
12
Ω
18 Ω
B
240 V
6
21
Ω
Q
C
Ω
D
Fig. 5
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Equivalent resistance between P and Q
Using star-delta transformation technique, delta ABC is converted into star
[Fig. 6] and its values are obtained as,
6 &times; 12
=2Ω
6 + 12 + 18
12 &times; 18
=6Ω
=
6 + 12 + 18
18 &times; 6
=
=3Ω
6 + 12 + 18
RAN =
RBN
RCN
A
12
6Ω
Ω
A
2Ω

N 3
Ω
6Ω
B
C
18 Ω
B
C
Fig. 6
Replacing the ABC delta by its equivalent star, the circuit is redrawn as shown
in Fig. 7.
2Ω
A
P
2Ω
N
6
240 V
3
Ω
Ω
B
C
6
Ω
Q
21
Ω
D
Fig. 7
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Both 6 Ω resistances are in series. Also in series are 21 Ω and 3 Ω. These two
series combinations together form one parallel connection and its equivalent
value is, hence,
RDN = (6 + 6)||(21 + 3)
12 &times; 24
=
=8
12 + 24
Therefore, the equal resistance between P and Q, RPQ = 8 + 2 + 2 = 12 Ω
Total current
The total current, I =
VPQ
240
=
= 20 A
RPQ
12
Current through the 18 Ω resistor
The current through the 18 Ω resistor is found by using
reference to Fig. 8, the mesh equations by inspection are
 


240
20 −12 −6
I1


 −12


0
36 −18
I2 =
0
I3
−6 −18
45
2Ω
A
12
6
Ω
I2
I1
Ω
18 Ω
B
6
C
I3
Ω
Q
Ω
240 V

21
P
mesh analysis. With
written as,

D
Fig. 8
Solving the above matrix equation, we get
I1 = 20 A
I2 = 10 A
I3 = 6.6667 A
Hence, the current through 18 Ω resistor = I2 − I3 = 10 − 6.6667 = 3.3333 A
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4. In the following circuit shown in Fig. 9, find the current through 5 Ω by
Thevenin’s theorem.
2Ω
1Ω
3Ω
10 V
5Ω
Fig. 9
To find Thevenin’s voltage, Vth
2Ω
1Ω
Vth
3Ω
10 V
a
b
Fig. 10
Fig. 10 shows the circuit used to determine Thevenin’s voltage. By Ohm’s law,
current through 3 Ω resistance is calculated as
10
= 2 A.
3+2
Therefore, Vth = 2 &times; 3 = 6 V.
To find Thevenin’s resistance, Rth
2Ω
1Ω
3Ω
a
 Rth
b
Fig. 11
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Thevenin’s resistance, Rth is calculated by short-circuiting the voltage source as
shown in Fig. 11 and looking back the circuit through the terminals a-b. Thus,
from Fig. 11,
Rth = (2||3) + 1 = 2.2 Ω
Thevenin’s equivalent circuit
Rth = 2.2 Ω
a
Vth = 6 V
+
5Ω
b
Fig. 12
From the Thevenin’s equivalent circuit drawn as shown in Fig. 12, current
through 5 Ω resistance is obtained as,
Vth
6
= 0.8333 A.
=
Rth + RL
2.2 + 5
5. Find the value of RL at which maximum power is transferred to RL and hence
the maximum power transferred to RL .
2Ω
12 V,
0.5 Ω
3Ω
1Ω
RL
Fig. 13
Value of RL at which maximum power is transferred
According to maximum power transfer theorem, maximum power is transferred
to RL , when Rth = RL . Thevenin’s resistance, Rth is calculated by making the
12 V as 0 V, however, keeping the internal resistance 0.5 Ω in the circuit as
shown in Fig. 14.
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2Ω
3Ω
 Rth
1Ω
0.5 Ω
a
b
Fig. 14
By looking back the circuit through the terminals a-b,
Rth = (2.5||1) + 3 = 3.7143 Ω
Thus, maximum power is transferred when RL = 3.7143 Ω.
Maximum power that is transferred to RL
2Ω
12 V,
0.5 Ω
3Ω
a
Vth
1Ω
b
Fig. 15
12
= 3.4286 A
0.5 + 2 + 1
Vth = 3.4286 &times; 1 = 3.4286 V.
Current through 1 Ω resistor
=
Therefore,
Using the values of Rth , Rth and RL , Thevenin’s equivalent circuit is drawn as
shown in Fig. 16.
Maximum power that is transferred to RL
3.7143 Ω
a
3.4286 V
+
3.7143 Ω
b
Fig. 16
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So, current flowing through RL during maximum power transfer
3.4286
3.7143 + 3.7143
= 0.4615 A.
IL =
Thus, maximum power transferred, Pmax = IL2 RL = 0.7911 W.
6. A series RLC circuit with R = 10 Ω, L = 10 mH and C = 1 &micro;F has an applied
voltage of 200 V at resonance frequency. Calculate the resonant frequency, the
current in the circuit and the voltage across the elements at resonance. Find
also the quality factor and bandwidth for the circuit.
Resonant frequency,
fr =
1
1
√
= p
2π LC
2π (10 &times; 10−3 )(1 &times; 10−6 )
= 1591.55 Hz.
Current at resonance,
Quality factor,
Bandwidth,
Ir =
V
200
=
R
10
= 20 A
r
1 L
Q=
R C
r
1 10 &times; 10−3
=
10 1 &times; 10−6
= 10
10
R
=
β=
2πL
2π &times; 10 &times; 10−3
= 159.15 Hz.
Voltage across resistor,
VR = Ir R = 20 &times; 10
= 200 V
Voltage across inductor,
VL = Ir XL = (20)(2π &times; 1591.55 &times; 10 &times; 10−3 )
= 2000 V
Voltage across capacitor,
VC = Ir XC = 20
1
2π &times; 1591.55 &times; 1 &times; 10−6
= 2000 V
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7. In the circuit shown in Fig. 17, find the values of I1 and I2 and also the real
power supplied by each source.
j4 Ω
4Ω
I1
1200&deg; V
j8 Ω
4Ω
j4 Ω
I2
12090&deg; V
Fig. 17
The two currents I1 and I2 enter at the dotted ends of the respective coils the
mutual inductance j4 Ω is taken as positive. Accordingly, the conductively
coupled equivalent circuit of the given circuit is drawn as shown in Fig. 18.
4
1200&deg; V
(j8 – j4) 
I1
(j4 – j4) 
j4
I2
4
12090&deg; V
Fig. 18
Using mesh analysis method, the mesh equations by inspection are written as,
4 + j8
j4
I1
120∠0◦
=
120∠90◦
j4
4 + j4
I2
Using Cramer’s rule, I1 is found as,
I1 =
=
∆I1
=
∆
120∠0
j4
◦
120∠90 4 + j4
4 + j8
j4
j4
4 + j4
(480 + j480) − (−480 + j0)
(−16 + j48) − (−16)
960 + j480
j48
= (10 − j20) A
=
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Similarly,
I2 =
=
∆I2
=
∆
4 + j8 120∠0◦
j4
120∠90◦
4 + j8
j4
j4
4 + j4
−960
−960
=
= (j20) A
(−16 + j48) − (−16)
j48
The complex power in an AC circuit is given by, S = P + jQ = V I ∗ . Thus,
the complex power supplied by 120∠0◦ V source
S1 = P1 + jQ1 = (120∠0◦ )(10 + j20) = 1200 + j2400
and the complex power supplied by 120∠90◦ V source
S2 = P2 + jQ2 = (120∠90◦ )(−j20) = 2400
Therefore, the power delivered by the two sources are:
P1 = 1200 W and P2 = 2400 W
8. In the circuit shown in Fig. 19, the switch S is closed at time t = 0 in position1 and changed over to position-2 after 1 millisecond. Find the time at which
the current is zero and reversing its direction. Assume that the change over of
switch from position 1 to 2 takes place in zero time.
 S

50 V
0.2 H
50 V
50 Ω
Fig. 19
At t = 0, the switch is at position-1 and the differential equation is
50i + 0.2
di
= 50
dt
Taking laplace transformation on both sides, we get
50I(s) + 0.2[sI(s) − I(0)] =
50
s
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At t = 0, i (t) = 0, and in s-domain, I (0) = 0. Therefore,
50
s
50
I(s)[50 + 0.2s] =
s
50I(s) + 0.2sI(s) =
50
s(0.2s + 50)
250
=
s(s + 250)
I(s) =
By applying partial fraction technique,
I(s) =
250
A
B
= +
s(s + 250)
s
s + 250
(i)
Solving for constants A and B,
A=
250
s + 250
= 1 and B =
s=0
250
s
= −1
s=−250
Substituting the values of A and B into Eq. (i), we get
I(s) =
1
1
−
s s + 250
Taking inverse laplace transformation on both sides
i(t) = 1 − e−250t A
(ii)
The switch is kept at this position for t = 1 ms. Therefore, at t = 1 ms, the
current through the circuit is found by putting t = 0.001 in Eq. (ii), we get
i(t) = 1 − e−250&times;0.001 = 0.2212 A.
This value becomes initial current for the circuit when the switch is moved to
position-2. The switch is moved to position-2 at t=0.001 s, and due to this
action, -50 V source is connected to the circuit and 50 V is disconnected. This
switching moment (t = 0.001 s) can be redefined as t0 = 0. Applying KVL after
closing on the switch to position-2,
50i + 0.2
di
= −50
dt
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Taking laplace transformation on both sides, we get
50I(s) + 0.2[sI(s) − I(0)] =
−50
s
At t0 = 0, i(t0 ) = 0.2212, and in s − domain, I(0) = 0.2212. Therefore,
−50
s
−50
I(s)[50 + 0.2s] =
+ 0.04424
s
50I(s) + 0.2[sI(s) − 0.2212] =
) + 0.04424
(− 50
s
0.2s + 50
−250 + 0.2212s
=
s(s + 250)
I(s) =
By applying partial fraction technique,
I(s) =
A
B
−250 + 0.2212s
= +
s(s + 250)
s
s + 250
(iii)
Solving for constants A and B,
A=
−250 + 0.2212s
s + 250
= −1 and B =
s=0
−250 + 0.2212s
s
= 1.2212
s=−250
Substituting the values of A and B into Eq. (i), we get
1
1.2212
I(s) = − +
s s + 250
Taking inverse laplace transformation on both sides, the current equation is
obtained as:
0
i(t) = −1 + 1.2212e−250t A
(iv)
The time when the current becomes zero is obtained by equating Eq. (iv) to
zero, i.e.
0
i(t) = −1 + 1.2212e−250t = 0
0
So,
or
1.2212e−250t = 1
t0 = 0.7993 &times; 10−3 s
t = 1.7993 ms.
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9. In the circuit shown in Fig. 20, find the expression for current if the switch is
closed at t = 0 and the value of current at t = 1 millisecond. Assume initial
charge on the capacitor is zero.
t=0
200 sin 500t volt
100 Ω
25 μF
i(t)
Fig. 20
Applying KVL to the circuit after closing the switch, we get
Z
1
100i +
i dt = 200 sin 500t
25 &times; 10−6
Taking laplace transformation on both sides,
500
1
100I(s) + 40000 I(s) = 200 2
s
s + 5002
40000
100000
I(s) 100 +
= 2
s
s + 5002
s
100000
I(s) =
100s + 40000
s2 + 5002
s
1000
=
s + 400
s2 + 5002
=
1000s
(s + 400)(s2 + 5002 )
Applying partial fraction technique,
I(s) =
1000s
A
Bs + C
=
+ 2
2
2
(s + 400)(s + 500 )
s + 400 s + 5002
(v)
Solving for the values of A, B and C, we get
and
A = −0.9756
B = 0.9756
C = 609.75
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Substituting the values of A, B and C into Eq. (v), we get
I(s) =
1000s
−0.9756
0.9756s + 609.75
=
+
2
2
(s + 400)(s + 500 )
(s + 400)
(s2 + 5002 )
−0.9756
0.9756s
609.75
+ 2
+ 2
2
(s + 400) (s + 500 ) (s + 5002 )
−0.9756
0.9756s
609.75
500
=
+ 2
+
2
2
(s + 400) (s + 500 )
500
(s + 5002 )
=
Taking inverse laplace transformation on both sides, we get
i(t) = −0.9756 e−400t + 0.9756 cos 500t + 1.2195 sin 500t A.
The value of current at t = 1 millisecond is obtained by putting t = 0.001 in
the above equation, and we get
i(0.001) = 0.7869 A.
10. Two wattmeters are connected to measure the power in a 3-phase 3-wire balanced load. Determine the total power and power factor if the two wattmeters
read (1) 1000 W each, both positive and (2) 1000 W each of opposite sign.
When both wattmeters read equal and positive
tan φ =
√
3
W2 − W1
W1 + W2
=
√
1000 − 1000
1000 + 1000
1000 + 1000
1000 − 1000
3
=0
φ = tan−1 0 = 0
Power factor, cos φ = 1
When both wattmeters read equal and opposite
tan φ =
√
3
W2 − W1
W1 + W2
=
√
3
=∞
φ = tan−1 ∞ = 90◦
Power factor, cos φ = 0
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11. Determine the line currents for the unbalanced delta connected load shown in
Fig. 21. Phase sequence is RYB.
IR
j 40
Ω
Ω
30
j 12
Y
Ω
3
balanced 200 V
supply
15
Ω
R
IY
8 Ω –j14 Ω
B
IB
Fig. 21
The three phase currents are found as follows:
IRY =
200∠0◦
= (2.4 − j3.2) A
30 + j40
IYB =
200∠120◦
= (−12.4 − j0.055) A
8 − j14
IBR =
200∠240◦
= (−9.6977 − j3.7888) A
15 + j12
The three line currents are found by using KCL as follows:
IR = IRY − IBR
= (2.4 − j3.2) − (−9.6977 − j3.7888)
= (12.0977 + j0.5888) = 12.112∠2.79◦ A
IY = IYB − IRY
= (−12.4 − j0.055) − (2.4 − j3.2)
= (−14.8 + j3.145) = 15.13∠168◦ A
IB = IBR − IYB
= (−9.6977 − j3.7888) − (−12.4 − j0.055)
= (2.7023 − j3.7338) = 4.6091∠ − 54.1◦ A.
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12. An unbalanced star-connected load is supplied from a 3-phase, 440 V symmetrical system. Determine the line currents and the power input to the circuit
shown in Fig. 22. Assume RYB sequence. Take phase voltage VRN as reference
in the supply voltage.
IR
R
3
balanced 440 V
supply
10 Ω
IY
Y
15
20
Ω
Ω
IB
B
Fig. 22
Given that, phase voltage VRN in the supply side is taken as reference and the
phase sequence is RYB. Since in Y-connected system, the line voltage leads
phase voltage by 30◦ , the three line voltages become
VRY = 440∠30◦
VYB = 440∠ − 90◦
VBR = 440∠ − 210◦
The mesh currents are assumed in the direction as shown in Fig. 23.
IR
R
3
balanced 440 V
supply
Y
B
10 Ω
I1
IY
15
IB
20
Ω
Ω
I2
Fig. 23
The mesh equations by inspection are written as
25 −15
I1
VRY = 440∠30◦
=
−15
35
I2
VYB = 440∠ − 90◦
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Using Cramer’s rule, I1 is found as,
I1 =
=
∆I1
=
∆
440∠30◦ −15
440∠ − 90◦
35
25 −15
−15
35
(13336.79 + j1100)
650
= (20.52 + j1.69) A = 20.59∠4.72◦ A
Similarly,
I2 =
=
∆I2
=
∆
25
440∠30◦
−15 440∠ − 90◦
25 −15
−15
35
(5715.77 − j7700)
650
= (8.79 − j11.85) A = 14.75∠ − 53.41◦ A.
Therefore,
IR = I1 = 20.59∠4.72◦ A
IY = I2 − I1 = (14.75∠ − 53.41◦ ) − (20.59∠4.72◦ )
= 17.91∠ − 130.9◦ A
IB = −I2 = (−1)(14.75∠ − 53.41◦ )
= 14.75∠126.59◦ A
Power input to the circuit
= |IR |2 RR + |IY |2 RY + |IB |2 RB
= (20.592 &times; 10) + (17.912 &times; 15) + (14.752 &times; 20)
= 13402 W
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U4004
B.E/B.Tech. DEGREE EXAMINATION, NOV./DEC. 2009
Electrical and Electronics Engineering
EE2151 – CIRCUIT THEORY
1. Determine the current in the 4 Ω branch in the circuit shown in Fig. 1. Use
mesh analysis method.
2Ω
2Ω
12 Ω
12 V
10 V
1Ω
3Ω
4Ω
24 V
Fig. 1
Mesh currents are assumed as shown in Fig. 2.
2Ω
12 V
2Ω
I1 12 Ω
1Ω
I2
I3
10 V
3Ω
4Ω
24 V
Fig. 2
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The mesh equations by inspection, are written as


 

15 −12 −1
I1
12
 −12
17 −3   I2  =  −10 
−1 −3
8
I3
24
Current in 4 Ω branch,
I3 =
∆I3
∆
(i)

15 −12
12

17 −10 
∆I3 = −12
−1 −3
24
= 15(408 − 30) + 12(−288 − 10) + 12(36 + 17) = 2730


15 −12 −1
17 −3 
∆ =  −12
−1 −3
8
Therefore from Eq. (i),
= 15(136 − 9) + 12(−96 − 3) − 1(36 + 17) = 664
2730
= 4.1114 A
I3 =
664
2. For the network shown in Fig. 3, find VS which makes I0 = 7.5 mA. Use node
voltage method.

4Ω

Io
8Ω
VS
+
7Ω
6Ω
6Ω
12 Ω

Fig. 3
Given that I0 = 7.5 mA, so, V2 = 6Io = 6 &times; 7.5 &times; 10−3 = 0.045V.
Applying KCL at node 2 first,
V2 − V1 V2 V2
+
+
=0
4
6
6
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0.01125 −
or
Solving we get,
V1
+ 0.0075 + 0.0075 = 0
4
V1 = 0.105 V.
Applying KCL at node 1,
V1 − VS V1 V1
+
+
− 0.01125 = 0
20 7
4
1
1 1
VS
V1
+ +
−
= 0.01125
20 7 4
20
Substituting V1 = 0.105 V into the above equation we get,
VS = 0.705 V.
3. Using the mesh current method, obtain the voltage Vx in the network of Fig. 4.
2Ω
530&deg; V
–
+
– j2 Ω
+
I1
100&deg; V
I2
j5 Ω
10 Ω
–
5Ω
2Ω
– j2 Ω
I3
10 Ω
+ Vx
–
Fig. 4
By mesh analysis method, the mesh equations


7 + j3
j5
5
 j5
12 + j3 −2 + j2  
5
−2 + j2 17 − j2
are written as
 

I1
10∠0◦
I2  =  5∠30◦ 
I3
0
Using Cramer’s rule, I3 is found as,
I3 =
∆I3
=
∆
7 + j3
j5
5
7 + j3
j5
5
j5
10∠0
12 + j3 5∠30◦
−2 + j2
0
j5
5
12 + j3 −2 + j2
−2 + j2 17 − j2
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∆I3 = (7 + j3)[0 − (−2 + j2)(5∠30◦ )] − j5[0 − 5(5∠30◦ )]
+(10∠0)[(j5)(−2 + j2) − 5(12 + j3)]
= −655.9 − j126.39
∆ = (7 + j3)[(12 + j3)(17 − j2) − (−2 + j2)(−2 + j2)]
−j5[(j5)(17 − j2) − (−10 + j10)] + 5[(j5)(−2 + j2) − 5(12 + j3)]
= 1390 + j650
∆I3
−655.9 − j126.39
I3 =
=
∆
1390 + j650
= (−0.422 + j0.1065) A = 0.4353∠165.85◦ A
Hence, the voltage across 10 Ω resistance,
Vx = 10I3 = 10 &times; 0.4353∠165.85◦ = 4.353∠165.85◦ V
4. Obtain the current in each resistor in Fig. 5 using network reduction method.
IA
6Ω
2Ω
IF
25 V
IC
3Ω
IT
IB
4Ω
+
ID
4Ω
IE
4Ω
4Ω
Fig. 5
2Ω
2Ω
2Ω
IC
IT
25 V
IF
IT
4Ω
+
4Ω
25 V
+
3Ω
2Ω
(a)
(b)
Fig. 6
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Fig. 6 illustrates successive reduction of given network. From Fig. 6(b),
The total current, IT =
25
=5A
5
Again with reference to Fig. 5, and using current division technique, the current
through the resistances are obtained as:
3
= 1.6667 A
IA = 5 &times;
6+3
6
IB = 5 &times;
= 3.3333 A
6+3
2+4
= 2.5 A
IC = 5 &times;
2 + 4 + 4 + (4||4)
4 + (4||4)
ID = 5 &times;
= 2.5 A
2 + 4 + 4 + (4||4)
4
IE = 2.5 &times;
= 1.25 A
4+4
4
= 1.25 A
IF = 2.5 &times;
4+4
5. In the network shown in Fig. 7, determine the current I.
100 Ω
A
I
C
D
E
160 V
60 Ω
40 Ω
80 Ω
88 Ω
B
Fig. 7
The elements of delta network CDE are converted into star equivalent values
as given below:
60 &times; 100
RCN =
= 30 Ω
100 + 40 + 60
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40 &times; 100
= 20 Ω
100 + 40 + 60
60 &times; 40
= 12 Ω
=
100 + 40 + 60
RDN =
REN
The delta network is replaced with the equivalent star as shown in Fig. 8.
A
I
C
30
D
Ω
N
20
Ω
12 Ω
160 V
80 Ω
E
88 Ω
B
Fig. 8
With reference to Fig. 8,
RAB = 30 + (12 + 88)||(20 + 80)
= 30 + 50 = 80 Ω
Therefore,
Current I =
160
VAB
=2A
=
RAB
80
6. Using the principle of superposition, calculate the current I in the network of
Fig. 9.
4Ω
5Ω
2Ω
I
1000&deg; V
j2 Ω
‒j2 Ω
10090&deg; V
Fig. 9
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100∠0◦ V source is acting alone
The given circuit is redrawn by short-circuiting 100∠90◦ V source.
4Ω
5Ω
2Ω
I
I1
1000&deg; V
I2 ‒j2 Ω
j2 Ω
I3
Fig. 10
By mesh analysis method, the mesh equations are written as


 

5 + j2 −j2
0
I1
100∠0◦
 −j2

4
j2   I2  = 
0
0
j2 2 − j2
I3
0
Using Cramer’s rule, I2 is found as,
5 + j2 100∠0◦
0
−j2
0
j2
0
0
2 − j2
∆I2
=
I2 =
∆
5 + j2 −j2
0
−j2
4
j2
0
j2 2 − j2
(−1)(100∠0◦ )(−j2)(2 − j2)
=
(5 + j2)[(4)(2 − j2) − (j2)2 ] + j2[(−j2)(2 − j2)]
=
400 + j400
= 6.4752∠60.95◦ A
84 − j24
100∠90◦ V source is acting alone
Next the given circuit is redrawn by short-circuiting the 100∠0◦ V source.
4Ω
5Ω
2Ω
I
I4
j2 Ω
I5 ‒j2 Ω
I6
10090&deg; V
Fig. 11
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By mesh analysis method, the mesh equations are written as


 

5 + j2 −j2
0
I4
0
 −j2

4
−j2   I5  = 
0
◦
0
−j2 2 − j2
I6
100∠90
Using Cramer’s rule, I5 is found as,
5 + j2
0
0
−j2
0
−j2
◦
0
100∠90
2
− j2
∆I5
=
I5 =
∆
5 + j2 −j2
0
−j2
4
−j2
0
−j2 2 − j2
(5 + j2)(j2)(100∠90◦ )
=
(5 + j2)[(4)(2 − j2) − (j2)2 ] + j2[(−j2)(2 − j2)]
=
−1000 − j400
= 12.3285∠ − 142.25◦ A
84 − j24
By the principle of superposition, I = I2 + I5 = 18.4572∠45.7◦ A.
7. For a two-branch parallel circuit RL = 15, RC = 30, XC = 30, E = 10 and f =
60 Hz. Find the condition of resonance, calculate (a) the two values of L and
(b) the two values of total current.
(a) For resonance
RL2
XC
XL
= 2
2
+ XL
RC + XC2
RL2
XL
30
1
= 2
=
2
2
+ XL
30 + 30
60
XL2 − 60XL + 225 = 0
XL =
=
Hence,
60 &plusmn;
60 &plusmn;
√
√
602 − 900
2
3600 − 900
= 56 Ω or 4 Ω
2
L = 0.149 H or 0.0106 H.
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(b) Total current
Total current = EY amperes.
(i) When XL = 56 Ω
Y = Y1 + Y2
1
1
+
15 + j56 30 − j30
1
15 − j56
1
30 + j30
=
&times;
+
&times;
15 + j56 15 − j56
30 − j30 30 + j30
15 − j56
30 + j30
=
+
152 + 562
302 + 302
15
56
30
30
=
−j
+
+j
3361
3361
1800
1800
15
30
30
56
Y =
+
+j
−
3361 1800
1800 3361
15
30
Therefore, total current = EY = 120
+
+ j0 = 2.5356 A
3361 1800
(ii) When XL = 4 Ω
=
Y = Y1 + Y2
1
1
+
15 + j4 30 − j30
1
15 − j4
30 + j30
1
=
&times;
&times;
+
15 + j4 15 − j4
30 − j30 30 + j30
15 − j4
30 + j30
=
+
152 + 42
302 + 302
15
4
30
30
=
−j
+
+j
241
241
1800
1800
15
30
30
4
Y =
+
+j
−
241 1800
1800 241
15
30
Therefore, total current = EY = 120
+
+ j0 = 9.4689 A.
241 1800
=
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8. Calculate the voltage V for the coupled circuit shown in Fig. 12. Repeat with
the polarity of one coil reversed.
k = 0.8
j5 Ω
j10 Ω

3
–
5 V
I2
I1
500&deg; V

–j4 Ω
‒
Fig. 12
Case (i) When the two coils have different polarity
Equivalent circuit of the given coupled circuit for this case is shown in Fig.
13. Since the two currents I1 and I2 do not enter at the dotted ends of
their respective coils, M is taken as negative. Accordingly, the circuit in
Fig. 13 contains negative jωM values for self elements and positive jωM
value for coupling element.
j5 Ω
‒j5.66 Ω
I1
–

j5.66 Ω

500&deg; V
j10 Ω
‒j5.66 Ω
I2
3
5 V
‒
–j4 Ω
Fig. 13
By inspection, mesh equations are written as:
3 + j1
−3 − j1.66
I1
50∠0◦
=
−3 − j1.66
8 + j6
I2
0
By Cramer’s rule I2 is given by,
I2 =
∆I2
∆
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3 + j1
50∠0◦
−3 − j1.66
0
=
3 + j1
−3 − j1.66
−3 − j1.66
8 + j6
=
(50∠0◦ )(3 + j1.66)
(3 + j1)(8 + j6) − (−3 − j1.66)2
= 7.8252 − j3.6166 = 8.6205∠ − 24.8◦
Therefore, voltage across 5 Ω resistor, V = 5I2 = 5 &times; 8.6205∠ − 24.8◦
= 43.1025∠ − 24.8◦ V
Case (ii) When the two coils have same polarity
Equivalent circuit of the given coupled circuit for this case is shown in Fig.
14. Since the two currents I1 and I2 do enter at the dotted ends of their
respective coils, M is taken as positive. Accordingly, the circuit in Fig. 14
contains positive jωM values for self elements and negative jωM value for
coupling element.
By inspection, mesh equations are written as:
3 + j1
−3 − j1.66
I1
50∠0◦
=
−3 − j1.66
8 + j6
I2
0
j5 Ω
j5.66 Ω
I1
–
j10 Ω

‒j5.66 Ω

500&deg; V
j5.66 Ω
I2
3
5 V
‒
–j4 Ω
Fig. 14
By Cramer’s rule I2 is given by,
I2 =
∆I2
∆
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3 + j1
50∠0◦
−3 + j9.66
0
=
3 + j1
−3 + j9.66
−3 + j9.66
8 + j6
=
(50∠0◦ )(3 − j9.66)
(3 + j1)(8 + j6) − (−3 + j9.66)2
= −1.4388 − j3.5399 = 3.8212∠ − 112.12◦
Therefore, voltage across 5 Ω resistor, V = 5I2 = 5 &times; 3.8212∠ − 112.12◦
= 19.106∠ − 112.12◦ V
9. In the series circuit shown in Fig. 15, the switch is closed on position-1 at t = 0.
At t = 1 milli second, the switch is moved to position-2. Obtain the equations
for the current in both intervals and draw the transient current curve.
50 Ω

0.2 H

100 V
50 V
Fig. 15
At t = 0, the switch is connected to position 1 and the differential equation is
di
50i + 0.2 = 100
dt
Taking Laplace transformation on both sides, we get
100
50I(s) + 0.2[sI(s) − I(0)] =
(ii)
s
At t = 0, i(t) = 0, and in s-domain, I(0) = 0. Therefore, Eq. (ii) becomes
100
50I(s) + 0.2sI(s) =
s
100
I(s)[50 + 0.2s] =
s
100
I(s) =
s(0.2s + 50)
100
500
=
=
0.2s(s + 250)
s(s + 250)
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By applying partial fraction technique,
=
A
B
+
s
(s + 250)
(iii)
Solving for the constants A and B,
A=
500
s + 250
=2
and
B=
s=0
500
s
= −2
s=−250
Substituting the values of A and B into Eq. (iii), we get
I(s) =
2
2
−
s (s + 250)
Taking inverse Laplace transformation on both sides, the current equation is
obtained as:
i(t) = 2 − 2e−250t A
(iv)
The switch is kept at this position for 0.001 s. Therefore, at t = 0.001 s,
the current through the circuit is determined by putting t = 0.001 in Eq. (iv),
i.e. at t = 0.001 s.
i(t) = 2 − 2e−250(0.001) = 0.4424 A
This value becomes initial current for the circuit when the switch is put in
position 2. Then, the switch is moved to position 2 at t = 0.001 s. Due to this
action, 50 V source is connected in the circuit and 100 V source is disconnected.
This moment (t = 0.001 s) can be redefined as t0 = 0.
Applying KVL,
di
= 50
dt0
Taking Laplace transformation on both sides, we get
50i + 0.2
50I(s) + 0.2[sI(s) − I(0)] =
50
s
At t0 = 0, I(0) = 0.4424 A. Therefore, Eq. (v) becomes
50
s
50
I(s)[50 + 0.2s] =
+ 0.08848
s
50 + 0.08848s
I(s) =
s(0.2s + 50)
50I(s) + 0.2sI(s) − 0.08848 =
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(v)
=
50 + 0.08848s
0.2s(s + 250)
=
250 + 0.4424s
s(s + 250)
By applying partial fraction technique,
=
B
A
+
s
(s + 250)
(vi)
Solving for the constants A and B,
A=
250 + 0.4424s
s + 250
=1
and
B=
s=0
250 + 0.4424s
s
= −0.5576
s=−250
Substituting the values of A and B into Eq. (vi), we get
I(s) =
0.5576
1
−
s (s + 250)
i
2A
1A
t
0
t = 0.001
(or)
t′ = 0
Fig. 16
Taking inverse Laplace transformation on both sides, the current equation is
obtained as:
0
i(t) = 1 − 0.5576e−250t A
or
i(t) = 1 − 0.5576e−250(t−0.001) A
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Therefore,
i(t) = 2 − 2e−250t A
for 0 &lt; t &lt; 0.001
= 1 − 0.5576e−250(t−0.001) A
for t &gt; 0.001
The transient current is shown in Fig. 16.
10. A series RC circuit with R = 100 Ω and C = 25 &micro;F is supplied with a source
of 200 sin 500t V. Find the current in the circuit. Assume initial charge on the
capacitor is zero.
Applying KVL to the circuit after closing the switch, we get
Z
1
100i +
i dt = 200 sin 500t
25 &times; 10−6
Taking laplace transformation on both sides,
1
500
100I(s) + 40000 I(s) = 200 2
s
s + 5002
100000
40000
= 2
I(s) 100 +
s
s + 5002
s
100000
I(s) =
100s + 40000
s2 + 5002
s
1000
=
s + 400
s2 + 5002
=
1000s
(s + 400)(s2 + 5002 )
Applying partial fraction technique,
I(s) =
A
Bs + C
1000s
=
+ 2
2
2
(s + 400)(s + 500 )
s + 400 s + 5002
(vii)
Solving for the values of A, B and C, we get
and
A = −0.9756
B = 0.9756
C = 609.75
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Substituting the values of A, B and C into Eq. (iv), we get
I(s) =
1000s
−0.9756
0.9756s + 609.75
=
+
2
2
(s + 400)(s + 500 )
(s + 400)
(s2 + 5002 )
−0.9756
0.9756s
609.75
+ 2
+
(s + 400) (s + 5002 ) (s2 + 5002 )
0.9756s
609.75
500
−0.9756
+ 2
+
=
2
2
(s + 400) (s + 500 )
500
(s + 5002 )
=
Taking inverse laplace transformation on both sides, we get
i(t) = −0.9756 e−400t + 0.9756 cos 500t + 1.2195 sin 500t
0.
97
56 2
1
.2
19
52
1
.5
61
7
Let us construct a right angled triangle using 0.9756 and 1.2195 as sides as
shown in Fig. 17. Referring to it,
0.9756

1.2195
Fig. 17
0.9756
= 38.66◦
1.2195
1.2195
cos φ =
1.5617
φ = tan−1
Also,
or
1.2195 = 1.5617 cos φ = 1.5617 cos 38.66◦
Similarly,
0.9756 = 1.5617 sin φ = 1.5617 sin 38.66◦
Therefore,
i(t) = −0.9756 e−400t + 1.5617 sin 38.66◦ cos 500t + 1.5617 cos 38.66◦ sin 500t
= −0.9756 e−400t + 1.5617(sin 38.66◦ cos 500t + cos 38.66◦ sin 500t)
= −0.9756 e−400t + 1.5617(sin 38.66◦ cos 500t + cos 38.66◦ sin 500t)
i(t) = −0.9756 e−400t + 1.5617 sin(500t + 38.66◦ ) A.
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11. The power input to a 2000 V, 50 Hz, 3 - phase motor is measured by two
wattmeters which 300 kW and 100 kW respectively. Calculate the input power,
power factor and the line current.
Input power
= W1 + W2 = 300 + 100 = 400 kW
√ W2 − W1
√ 300 − 100
tan φ = 3
= 3
W1 + W2
400
= 0.866
φ = tan−1 0.866 = 40.89
Power factor, cos φ = 0.76
√
P = 3 VL IL cos φ
√
= 3 (2000) (IL ) (cos 0.76)
Line current, IL = 0.1519 A.
12. Determine the line currents and the total power for the unbalanced ∆−connected
load shown in Fig. 18. A 3 phase supply with an effective line voltage of 240 V
is given to the circuit.
A
25
&deg;Ω

0
&deg;Ω
0
20
VAB
B
VBC
1530&deg; Ω
C
Fig. 18
Line currents
Phase currents are determined first as follows:
IAB =
VAB
240∠0◦
=
RAB
25∠90◦
= 9.6∠ − 90◦ A
IBC =
VBC
240∠ − 120◦
=
RBC
15∠30◦
= 16∠ − 150◦ A
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ICA =
VCA
240∠ − 240◦
=
RCA
20∠0◦
= 12∠ − 240◦ A
Applying KCL, the line currents are obtained from the above found phase currents as follows:
IA = IAB − ICA = (9.6∠ − 90◦ ) − (12∠ − 240◦ )
= 20.87∠ − 73.29◦ A
IB = IBC − IAB = (16∠ − 150◦ ) − (9.6∠ − 90◦ )
= 13.95∠173.41◦ A
IC = ICA − IBC = (12∠ − 240◦ ) − (16∠ − 150◦ )
= 20∠66.87◦ A
Total power
Resistive components are obtained by finding the rectangular form of three
impedances as given below.
ZAB = 25∠90◦ = 0 + j25
ZBC = 15∠30◦ = 13 + j7.5
ZCA = 20∠0◦ = 20 + j0
Therefore, the total power input
2
2
2
= IAB
RAB + IBC
RBC + ICA
RCA
2
2
= (9.6 &times; 0) + (16 &times; 13) + (122 &times; 20)
= 6208 W
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J3904
B.E/B.Tech. DEGREE EXAMINATION, MAY/JUNE 2009
Electrical and Electronics Engineering
EE2151 – CIRCUIT THEORY
1. For the circuit of Fig. 1, find the current in each branch by Nodal method.
5Ω
V1
5Ω
V2
8Ω
10 Ω
10 Ω
50 V
10 V
5V
Fig. 1
Applying KCL at node 1, we get
V1 − 50 V1 − 10 V1 − V2
+
+
=0
5
8
5
0.2V1 − 10 + 0.125V1 − 1.25 + 0.2V1 − 0.2V2 = 0
0.525V1 − 0.2V2 = 11.25
(i)
Applying KCL at node 2, we get
V2 − V1 V2 V2 − 5
+
+
=0
5
10
10
0.2V2 − 0.2V1 + 0.1V2 + 0.1V2 − 0.5 = 0
−0.2V1 + 0.4V2 = 0.5
(ii)
Solving Eqns. (i) and (ii) we get, V1 = 27.0588 V and V2 = 14.7794 V
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Therefore,
50 − V1
5
50 − 27.0588
= 4.5882 A
=
5
V1 − 10
Current through 8 Ω resistor =
8
27.0588 − 10
=
= 2.1324 A
8
V1 − V2
Current through 5 Ω (connected between nodes 1 &amp; 2) =
5
Current supplied by 50 V source =
=
27.0588 − 14.7794
= 2.4558 A
5
14.7794
V2
=
= 1.4779 A
10
10
V2 − 5
Current through 10 Ω (connected with 5 V source) =
10
Current through 10 Ω resistor =
=
14.7794 − 5
= 0.9779 A
10
2. Find the current in each branch of the circuit and the total power consumed by
the circuit of Fig. 2. Assume E = 50 sin (ωt + 45◦ ).
2 + j1
E
2 ‒ j5
3 + j6
Fig. 2
Given that,
So,
Therefore,
E = Emax sin (ωt + 45◦ )
Emax = 50
50
Erms = √ = 35.36 V
2
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Equivalent impedance, Zeq = (2 − j5)||(3 + j6) + (2 + j1)
= 8.8077 − j0.9615
= 8.86∠ − 6.23◦
(iii)
Current through (2 + j1) Ω impedance,
Itotal =
35.36∠45◦
8.8077 − j0.9615
= 2.4991 + j3.1116 = 3.9991∠51.23◦
3 + j6
5 + j1
Current through (2 − j5) Ω impedance
= 3.9991∠51.23◦ &times;
Current through (3 + j6) Ω impedance
= (3.0333 − j0.0473) A
2 − j5
= 3.9991∠51.23◦ &times;
5 + j1
= (3.7120 − j1.9969) A
From Eq. (iii), φ = −6.23◦
Therefore, cos φ = cos 6.23◦ = 0.99
Hence, power consumed by the circuit = Erms Itotal cos φ = 35.36 &times; 3.9991 &times;
0.99 = 140 W
3. Find the Thevenin’s equivalent circuit at (a,b) in Fig. 3.
4Ω
10 Ω
10 Ω
6Ω
a
5Ω
b
Fig. 3
Thevenin’s voltage, Vth :
Mesh analysis is used to find the current flowing through 5 Ω resistor. With
reference to Fig. 4, by inspection the mesh equations are written as
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15 −10
−10
24
I1
I2
=
V
0
4Ω
I2
10 Ω
V
I1
6Ω
10 Ω
a
5Ω
b
Fig. 4
Using Cramer’s rule, I1 and I2 are obtained as follows:
V −10
0
24
∆I1
6V
=
I1 =
=
= 0.0923V
∆
65
15 −10
−10
24
Similarly,
15 V
−10 0
∆I2
V
=
I2 =
=
= 0.0385V
∆
26
15 −10
−10
24
4Ω
10 Ω
‒
+
V
6Ω
10 Ω
5Ω
10I2
+
+
0V
‒
a
5I1
‒
b
Fig. 5
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Referring to Fig. 5, Vab (or Vth ) is found as,
10I2 + 5I1 = Vth
(10 &times; 0.0385V ) + (5 &times; 0.0923V ) = Vth
Vth = 0.8465V Volts
Thevenin’s resistance, Rth :
To find Thevenin’s resistance, first the voltage source is short-circuited and the
circuit is redrawn as shown in Fig. 6.
4Ω
10 Ω
10 Ω
6Ω
a
5Ω
b
Fig. 6
Using star-delta transformation the circuit is modified as shown in Fig. 7.
4Ω
6Ω
40 Ω
20 Ω
a
20 Ω
b
Fig. 7
Figure 8 shows step-by-step reduction of the network and from Fig. 8(b) the
equivalent resistance across a and b, is found as Rth = (3.6364||20) + 6 =
9.0769 Ω .
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6Ω
3.6364 Ω
20 Ω
6Ω
3.6364 Ω
a
20 Ω
a
20 Ω
b
(a)
(b)
b
Fig. 8
Vth = 0.8465V Volts
Using Vth and Rth found above, the Thevenin’s equivalent circuit at (a,b) is
drawn as shown in Fig. 9.
Rth = 9.0769 Ω
a
+
b
Fig. 9
4. For the circuit shown in Fig. 10, find the current flowing through the 10 Ω
resistor.
10 Ω
I
2Ω
2Ω
12 Ω
40 V
2Ω
2Ω
Fig. 10
The mesh currents are assumed as shown in Fig. 11.
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10 Ω
2Ω
2Ω
I2
12 Ω
I1
40 V
I3
2Ω
2Ω
Fig. 11
By inspection, the mesh equations are written as,


 

14 −2 −2
I1
40
 −2
16 −12   I2  =  0 
−2 −12
16
I2
0
Solving the above equations we get,
I = I1 = 3.3333 A.
5. A sinusoidal voltage of 50 Hz is connected in series with switch and R =
10 Ω and L = 0.1 H. Calculate the transient current i(t).
As instantaneous value of sinusoidal voltage is written in the form, v = Vm sin ωt,
then for a frequency of f = 50 Hz, v = Vm sin 2πf t = Vm sin 314.6 t.
Applying KVL around the only loop,
di
= Vm sin 314.6 t
dt
314.16
10I(s) + 0.1sI(s) = Vm
s2 + 314.162
314.16
I(s)[10 + 0.1s] = Vm
s2 + 314.162
314.16 Vm
1
I(s) =
s2 + 314.162
0.1s + 10
314.16 Vm
=
0.1(s + 100)(s2 + 314.162 )
10i + 0.1
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By partial fraction technique,
3141.6 Vm
A
Bs + C
=
+ 2
2
2
(s + 100)(s + 314.6 )
s + 100 s + 314.162
(iv)
Solving for the constants, A, B and C we obtain,
A = 0.0289 Vm
B = −0.0289 Vm
C = 2.89 Vm
Substituting the values of A, B and C obtained above into Eq. (iv) we get,
0.0289 Vm −0.0289 Vm s + 2.89 Vm
+
s + 100
s2 + 314.162
0.0289 Vm
0.0289 Vm s
2.89 Vm 314.16
=
− 2
+
s + 100
s + 314.162
314.16 s2 + 314.162
I(s) =
Taking inverse laplace transformation to the above equation we get,
i(t) = (0.0289 Vm e−100t − 0.0289 Vm cos 314.16t + 0.0092 Vm sin 314.16t) A.
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