Michał Tadeusiewicz
Electric Circuits
Tutorial Problems
International Faculty of Engineering
Łódź, 2010
3
DC analysis
Question 1
Find the total resistance RAB of the one port AB shown in Fig. 1.
A
R4
R1
VS
R2
RAB
i7
i6
R7
R6
R5
R3
B
Fig. 1
Fig. 2
Compute the currents i6 and i7 in the circuit of Fig. 2
Data:
R1 = R3 = 10 Ω, R2 = R4 = R5 = 20 Ω, R6 = 15 Ω,
R7 = 30 Ω,
VS = 60 V.
Solution
To find the total resistance of the one-port AB we replace the Δ-connection consisting of the
resistors R2, R3, R5 by Ү-connection (see Fig. 3)
A
R1
R4
R 23
R 25
R 35
B
Fig. 3
The resistances R23, R25, R35 are as follows:
R2 R3
= 4Ω
R2 + R3 + R5
R2 R5
R25 =
= 8Ω
R2 + R3 + R5
R23 =
4
R35 =
Hence, the total resistance is
R AB = R35 +
R3 R5
= 4Ω
R2 + R3 + R5
(R1 + R23 )(R4 + R25 ) = 13.33Ω
R1 + R23 + R4 + R25
Now we consider the circuit of Fig. 2 repeated below.
iAB
i6
VS
R6
RAB
i7
R7
Fig. 4
We compute the resistance faced by the voltage source
R = R AB +
R6 R7
R6 + R7
and the current flowing through the source
i AB =
R AB
VS
60
=
= 2.57 A
R6 R7
23.33
+
R6 + R7
Since the resistors R6 and R7 are connected in parallel, then
i6 = i AB
R7
= 1.72A
R6 + R7
Current i7 can be determined using KCL:
i7 = −i AB + i6 = −0.85A
Question 2
For the one-port AB shown in Fig. 3 determine Thevenin’s circuit.
(i) Compute the current flowing through the resistor R = 5Ω connected to terminals A and B.
(ii) Compute the voltage between terminals A and B when the parallel combination of R0 = 6Ω
and I0 = 1A is connected to terminals AB.
5
R3
A
E2
E1
R
R1
R0
I0
R2
B
Fig. 5
Data:
R1 = R2 = 12Ω, R3 = 6Ω, E1 = 6V, E2 = 3V
Solution
To determine Thevenin’s circuit for the one-port AB we need voltage v0C between terminals A, B
(see Fig.6) and the input resistance Req of the one-port after setting to zero the source voltages (see
Fig.7).
R3
R3
i
A
A
E2
E1
C
R1
v0C
R2
R2
R1
B
B
Fig. 6
Fig. 7
Note that the same current i traverses all elements in the circuit of Fig. 6. To find this current we
apply KVL and Ohm’s law:
(R1 + R2 + R3 )i − E1 + E2 = 0
i=
E1 − E 2
= 0.1A
R1 + R2 + R3
Now we can determine v0C using KVL in the closed node sequence ACBA and Ohm’s law:
− E 2 − R2 i + v 0 C = 0
v 0C = E 2 + R2 i = 4.2V
The total resistance of the one-port shown in Fig.7 is given by
Req =
(R1 + R3 )R2
R1 + R3 + R2
= 7.2Ω
6
The Thevenin circuit is depicted in Fig. 8
A
A
v0C=4.2V
v0C=4.2V
Req=7.2Ω
Req= 7.2 Ω
iR
R
B
B
Fig. 8
Fig. 9
(i) When the one-port AB is terminated by the resistor R = 5Ω, the current iR flowing through this
resistor can be computed using Thevenin’s circuit (see Fig. 9). Hence, we find
iR =
v0C
= 0.34A
Req + R
(ii) Now we connect to the one-port AB the parallel connection of R0 = 6Ω and I0 = 1A and replace
the one-port by Thevenin’s circuit. As a result we obtain the circuit shown in Fig. 10
i’
A
vOC=4.2V
Req=7.2 Ω
R0=6Ω
vAB
I0=1A
B
Fig. 10
Let us replace the parallel connection of R0 and I0 by an equivalent series connection of the same
resistor R0 and a voltage source E0 = R0I0 = 6V (see Fig. 11).
A
vOC
Req
i’
R0
vAB
C
E0
B
Fig. 11
7
Next we compute current i’ using KVL and Ohm’s law:
(R
eq
i' =
+ R0 )i '−vOC − E 0 = 0
vOC + E 0 4.2 + 6
=
= 0.77 A .
Req + R0 7.2 + 6
We apply KVL, to the closed node sequence ACBA, and Ohm’s law:
− R0 i '+ E 0 + v AB = 0
and find voltage vAB
v AB = R0 i '− E 0 = 6 ⋅ 0.77 − 6 = 1.38V
Question 3
Describe the circuit shown in Fig. 12 using the node method. Compute the currents i2, i3 and the
power consumed by the resistor R3.
I S2
i2
i3
R2
R1
R3
I S1
R4
Fig. 12
Data:
R1 = 20 Ω, R2 = 30 Ω, R3 = 10 Ω, R4 = 10 Ω,
I S1 = 1 A,
I S 2 = 4 A.
Solution
To write the node equations we introduce the node voltages e1, e2, e3 as shown in Fig. 13
8
I S2
i2
2
R2
3
i3
I S1
R1
R3
e2
1
e3
R4
e1
Fig. 13
The set of the node equations is as follows:
e1 e1 − e2
+
= − I S2 ,
R4
R1
e2 − e1 e2 − e3
+
= I S1 ,
R1
R2
e3 − e2 e3
+
= I S2 .
R2
R3
Substituting the resistances we obtain
e1 e1 − e2
+
= −4 ,
10
20
e2 − e1 e2 − e3
+
= 1,
20
30
e3 − e2 e3
+
= 4.
30
10
We solve the set of these equations using the substituting method. As a result we find
e1 = −22.86 V ,
e2 = 11.42 V,
e3 = 32.85 V.
Next we express currents i2 and i3 in terms of the node voltages
Power consumed by the resistor R3 is
i2 =
e 2 − e3
= −0.71 A,
R2
i3 =
e3
= 3.29 A .
R3
9
P3 = R i = 107.9 W.
2
3 3
Question 4
In the circuit of Fig. 14 replace the Y-connected circuit consisting of resistors R2, R3, R4 by an
equivalent Δ- connected circuit.
For the one-port AB determine Thevenin’s and Norton’s circuits.
IS
R2
R4
A
R3
R1
R5
B
Fig. 14
Data:
R1 = 2 Ω, R2 = R3 = 1 Ω, R4 = 4 Ω, R5 = 3 Ω, I S = 1 A .
Solution
The circuit after transformation of the
Fig. 15.
- connected circuit into Δ- connected circuit is shown in
IS
R24
A
R1
R34
R23
R5
VOC
B
Fig. 15
The resistances R24, R23, R34 are as follows:
R24 = R2 + R4 +
R2 R4
= 9Ω ,
R3
R23 = R2 + R3 +
R2 R3
= 2.25Ω ,
R4
R34 = R3 + R4 +
R3 R4
= 9Ω .
R2
10
To determine Thevenin’s circuit for the one-port AB we need the total resistance Req of the oneport after the current source IS is set to zero (see Fig. 16) and the voltage VOC, between open
circuited terminals A,B, as indicated in Fig. 15.
R24
A
R34
R23
R1
R5
B
Fig. 16
The circuit of Fig. 16 can be reduced as depicted in Fig. 17, where
R123 =
R1 R23
= 1.06Ω ,
R1 + R23
R345 =
R34 R5
= 2.25Ω.
R34 + R5
R24
A
R123
R345
B
Fig. 17
Hence, we have
Req =
(R24 + R123 )R345
R24 + R123 + R345
= 1.84Ω .
In order to find VOC we rearrange the circuit of Fig. 15 as shown in Fig. 18.
R24
ES
i
A
R345
R123
VOC
B
Fig. 18
11
In this circuit the series connection of R24 and ES replaces the parallel connection of R24 and IS
appeared in Fig. 15. Hence, it holds
E S = I S R24 = 9 V .
We compute the current i
i=
ES
= 0.73 A
R24 + R123 + R345
and the voltage VOC
VOC = R345 i = 1.65 V .
Thus, the Thevenin circuit is as shown in Fig. 19
A
EOC=1.65 V
Req=1.84 Ω
B
Fig. 19
Norton’s circuit is depicted in Fig. 20, where
I SC =
EOC
= 0.90 A .
Req
A
I SC
Req = 1.84Ω
B
Fig. 20
Question 5
In the circuit shown in Fig. 21 current i2 is given, i2 = 1.5 A .
(i)
Compute the source voltage VS and the power consumed by the resistor R4.
(ii)
Compute new current i2 , flowing through R2, when the resistor R4 is short circuited.
'
12
R1
i1
R3
i3
i2
VS
v2
R4
R2
Fig. 21
Data:
R1 = 10 Ω, R2 = 20 Ω, R3 = 15 Ω, R4 = 5 Ω.
Solution
(i) At first we compute the voltage v2 across the R2:
v 2 = R2 i2 = 30 V .
Since R3 and R4 are connected in series and the voltage across this connection is v2 then
i3 =
v2
= 1.5 A .
R3 + R4
Using KCL at the top node we have
i1 = i2 + i3 = 3 A .
To find VS we apply KVL in the loop consisting of VS, R1 and R2
VS = R1i1 + R2 i2 = 60 V.
The power consumed by the resistor R4 is
P4 = R4 i32 = 11.25 W.
(ii) The circuit with the short circuited resistor R4 is shown in Fig. 22.
i1'
R1
R3
i2'
VS
R2
Fig. 22
i3'
13
In this circuit we find:
i1' =
VS
60
=
= 3.23 A ,
20 ⋅ 15
R2 R3
10 +
R1 +
35
R2 + R3
i2' = i1'
R3
15
= 3.23 = 1.38 A .
R2 + R3
35
Question 6
In the circuit shown in Fig. 23 compute the current iR flowing through the resistor R in three cases:
(i) R = 2.5 Ω
(ii) R = 7.5 Ω
(iii) R = 20 Ω
Apply Thevenin’s theorem.
R2
A
iR
E3
R1
I0
R
R3
Fig. 23
B
Data:
R1 = 10 Ω,
R2 = 15 Ω, R3 = 25 Ω, E3 = 5 V, I0 = 1 A.
Solution
We replace the one-port AB, consisting of R1, R2, R3, I0, and E3, by the equivalent Thevenin circuit.
Thus, we need to compute voltage vOC (see Fig. 24) and the resistance Req (see Fig. 25)
iAB = 0
R2
A
R2
I0
A
E3
R1
vOC
R1
R3
R3
B
B
Fig. 24
Fig. 25
Let us consider the circuit of Fig. 24 and replace the parallel connection of I0 and R1 by an
equivalent series connection of the voltage source E1 = R1I0 = 10 V and the same resistor R1 (see
Fig. 26).
14
iAB= 0
i
A
E1
R2
E3
vOC
R3
R1
B
Fig. 26
Since the terminals A,B are left open-circuited, it holds
i=
E1 − E3
5
=
= 0.1 A .
R1 + R2 + R3 50
Using KVL we obtain
v 0C = E3 + R3i = 5 + 2.5 = 7.5 V .
In order to find Req we consider the circuit depicted in Fig. 25 and compute the imput resistance of
the one-port AB
Req =
(R1 + R2 )R3
R1 + R2 + R3
=
25 ⋅ 25
= 12.5 Ω .
50
Using the equivalent Thevenin circuit we replace the circuit of Fig. 23 by the circuit shown in Fig.
27.
iR
VOC=7.5 V
R
Req=12.5 Ω
Fig. 27
In this circuit it holds
iR =
Hence, we obtain:
VOC
.
Req + R
15
(i)
i R' =
7.5
= 0.5 A
12.5 + 2.5
(ii)
i R'' =
7 .5
= 0.375 A
12.5 + 7.5
(iii)
i R''' =
7 .5
= 0.230 A.
12.5 + 20
Question 7
i4
R4
I S1
R1
R2
i1
R3
I S2
i3
i2
Fig. 28
(i)
(ii)
In the circuit shown in Fig. 28 determine the branch currents i1, i2, i3, i4 using the node
method.
Write the node equations if the current source I S1 is replaced by series combination of a
voltage source E =12V, directed to the right, and a resistor R = 12Ω.
Data:
R1 = 10 Ω, R2 = R3 =20 Ω, R4 = 40 Ω, I S1 = 1A, I S 2 = 3A.
Solution
To write the node equations we choose the bottom node as the datum node and introduce the node
voltages e1 and e2 as shown in Fig. 29
1
i1
i2
R1
i4
R4
2
i3
I S1
R2
e1
e2
Fig. 29
R3
I S2
16
(i) The node equations have the form
e1 e1 e1 − e2
+
+
= − I S1 ,
R1 R2
R4
e2 − e1 e2
+
= I S1 + I S 2 .
R4
R3
We substitute the resistance and source values and perform simple rearrangements. As a result we
obtain the following set of equations in two unknown variables e1 and e2:
-e2 = -407e1,
-e1 + 3e2 = 160 .
This set of equations is solved using the substituting method. As a result we obtain:
e1 = 2V
e2 = 54V.
Using the node voltages we compute the branch currents as follows:
i1 =
e1
= 0.2 A ,
R1
i2 =
e1
= 0.1 A ,
R2
e2
= 2.7 A ,
R3
i3 =
i4 =
e2 − e1
= 1.3 A .
R4
(ii) The circuit with a new branch consisting of the resistor R and the voltage source E is shown in
Fig. 30.
1
i1
i2
2
R
R2
R1
R4
i
e1
E
e2
i3
R3
Fig. 30
Current i flowing through this branch is given by
i=
e1 − e2 + E
.
R
Hence, the node equations are
⎛ 1
1 ⎞ e −e e −e + E
e1 ⎜⎜ + ⎟⎟ + 1 2 + 1 2
=0 ,
R4
R
⎝ R1 R2 ⎠
I S2
17
e2 − e1 e2 − e1 − E e2
+
= I S2 .
+
R4
R
R3
Setting the resistance and source values we obtain:
31e1 – 13e2 = –120 ,
–13e1 + 19e2 = 480 .
Question 8
A
R4
E
I
R3
R1
R2
B
Fig. 31
(iv) For the one-port AB, shown in Fig. 31, determine Thevenin’s and Norton’s circuits.
(v) Find the current flowing through the resistor R = 3 Ω connected to the terminals A and B.
Compute the power consumed by this resistor and voltage across the resistor R3.
Data:
R1 = R2 = 5 Ω, R3 = 10 Ω, R4 = 2 Ω, E = 10V, I = 1A .
Solution
(i) Thevenin’s and Norton’s circuits are shown in Fig. 32.
A
VOC
A
Req
ISC
Req
B
B
(b)
(a)
Fig. 32
To find Req we consider the circuit after the voltage source is short-circuited and the current source
is open-circuited (see Fig. 33)
18
A
R4
R3
R1
R2
B
Fig. 33
Total impute resistance Req of this one-port is
Req = R4 +
R3 (R1 + R2 )
= 7Ω .
R1 + R2 + R3
V0C is the voltage between the open-circuited terminals A, B as shown is Fig. 34
i4=0
E
i3
A
R4
I
R3
R1
VOC
R2
B
Fig. 34
To find V0C we transform the one-port AB as depicted in Figs. 35 – 37.
i4=0
i3
E
= 2A
R1
R1
I=1A
A
R4
R3
VOC
R2
B
Fig. 35
19
i4=0
A
i3
E
Iˆ = + I = 3A
R1
R1
R4
R3
VOC
R2
B
Fig. 36
i4=0
Eˆ = IˆR1 = 15V
R4
i3
V3
R3
R1
A
VOC
R2
B
Fig. 37
In the circuit of Fig. 37 current i3 traverses the resistors R3, R2, R1 and the voltage source Ê . Hence,
it holds
Eˆ
i3 =
= 0.75A .
R1 + R2 + R3
Since the voltage across the resistor R4 is zero, the voltage VOC is the same as V3 = R3i3. Thus, we
have
VOC = R3i3 = 7.5 V.
(iii) To determine current flowing through the resistor R connected to terminals A and B we replace
the one-port AB by Thevenin’s circuit (see Fig. 38).
A
VOC
i
R
Req
B
Fig. 38
20
Using KVL and Ohm’s law we obtain
i=
V0 C
= 0.75 A .
Req + R
Hence, the power consumed by R is
P = Ri 2 = 1.687 W .
To find voltage v̂3 across resistor R3 we consider the one-port AB terminated by the resistor R (see
Fig. 39)
i3
E
A
R4
i
I
R3
R1
R
v̂3
R2
B
Fig. 39
KVL and Ohm’s law lead to the relationship
vˆ3 = i (R4 + R ) = 3.75 V .
Question 9
In the circuit shown in Fig. 40 the power consumed by the resistor R5 is P5 = 10 W.
(i) Compute the source current IS.
(ii) Replace the resistor R5 by resistor R5' = 30Ω and compute the power consumed by this resistor.
Data:
R1 = R2 = R3 = 20 Ω, R4 = R5 = 10 Ω.
R2
A
i2
i1
IS
i4
i3
R1
B
R4
C
R3
D
Fig. 40
R5
21
Solution
(i) Since the same current traverses resistors R4 and R5 then the power consumed by resistor R5 is
given by the equation
P5 = R5 i 42 .
Hence, we find
P5
i4 =
= 1A .
R5
Having i4 we compute the voltage between the nodes C and D
vCD = (R4 + R5 )i4 = 20V
and next current i3
i3 =
vCD
= 1A .
R3
Applying KCL at node C we obtain
i 2 = i3 + i 4 = 2 A .
Hence, we have
v AC = R2 i2 = 40 V .
KVL applied to the loop ACDBA leads to the equation
v AB = v AC + vCD = 60 V .
Consequently, it holds
i1 =
v AB
= 3A .
R1
Using KCL at node A we hawe
I S = i1 + i2 = 5 A .
(ii) We now consider the circuit shown in Fig. 41a and reduce it as depicted in Figs. 41b and c.
a)
A
20Ω
10Ω
C
i4'
IS=5 A
20Ω
B
20Ω
30Ω
D
Fig. 41a
Thus, the total resistor faced by the current source is RT = 12.5 Ω. Hence, using successively circuits
of Figs 41c, 41b, and 41a we find
22
v
= 12.5 ⋅ 5 = 62.5 V ,
'
AB
'
v AB
⋅ 13.33 = 25 V ,
20 + 13.33
v' CD
i4' =
= 0.625 A .
10 + 30
'
vCD
=
Since the same current traverses resistors R4 and R5' we obtain
( )
P5' = R5' i 4'
b)
A
B
= 11.72 W .
c)
20Ω
20Ω
IS
2
C
A
13.33Ω
'
vCD
IS
12.5Ω
'
vAB
B
D
Fig. 41b
Fig. 41c
Question 10
(i) Replace the one-port AB shown in Fig. 42 by the equivalent Thevenin circuit.
(ii) Compute the current flowing through the resistor R = 2 Ω connected to terminals A and B.
(iii) Determine the resistance of a resistor connected to terminals A and B knowing that the power
consumed by this resistor is 0.35W.
Data:
R1 = 2Ω, R2 = 3Ω, R3 = 2Ω, R4 = 4Ω, E1 = 1V, E2 = 3V.
R4
A
E1
E2
R3
R1
R2
B
Fig. 42
23
Solution
(i) The equivalent Thevenin circuit consists of a voltage source VOC and a resistor Req connected in
series. To find Req we consider the circuit shown in Fig. 43 obtained from the original circuit by
setting E1 = 0 and E2 = 0.
A
R4
R3
R2
R1
B
Fig. 43
Resistance Req is as flows
⎞
⎛ R1 R3
⎜⎜
+ R4 ⎟⎟ R2
R + R3
⎠ = 1.875 Ω .
Req = ⎝ 1
R1 R3
+ R4 + R2
R1 + R3
VOC is the voltage between terminals A and B that are left open-circuited (see Fig. 44).
It can be computed using the node method.
1
2
R4
A
E1
E2
e1
R3
VOC
e2
R1
R2
B
Fig. 44
The node equations are as flows
e1 e1 − E1 e1 − e2
+
+
= 0,
R3
R1
R4
e2 − e1 e2 − E 2
+
= 0.
R4
R2
24
We set the values of the resistances and voltage sources
e1 e1 − 1 e1 − e2
= 0,
+
+
2
2
4
e2 − e1 e2 − 3
+
= 0.
4
3
To solve this set of equations we use the substituting method. After eliminating e1 we obtain
32
e2 = 22 ⇒ e2 = 2.06 V .
3
Note that
VOC = e2 = 2.06 V .
(ii) To find the current flowing through the resistor R we exploit the equivalent Thevenin circuit as
shown in Fig. 45.
A
VOC=2.06 V
i
R=2 Ω
Req=1.875 Ω
B
Fig. 45
Using KVL we obtain
i=
VOC
2.06
=
= 0.53A .
Req + R 1.875 + 2
(iii) We now consider the circuit of Fig. 46 where Rx is an unknown variable.
A
ix
VOC
Rx
Req
B
Fig. 46
25
The power consumed by the resistor Rx is
Px = 0.35 = R x i x2 ,
where
ix =
VOC
2.06
=
.
Req + R x 1.875 + R x
Hence, we can write the equation
⎛ 2.06
0.35 = R x ⎜⎜
⎝ 1.875 + R x
which, after some rearrangements becomes
2
⎞
⎟⎟ ,
⎠
R x2 − 8.374 R x + 3.514 = 0 .
Solving this equations we obtain
R x1 = 7.931Ω , R x2 = 0.443 Ω .
27
Transient analysis
Question 1
The switch in the circuit shown in Fig. 1 is close until the steady state prevails and then it is
opened. Assuming that the opening occurs at t = 0 find the current i L (t ) and the voltage v L (t ).
R1
i1
iL
i2
R2
E
t=0
Data: E = 100V, R1= 1Ω, R2 = 8.1Ω,
R3
vL
L
Fig. 1
R3 = 9Ω, L = 0.1H.
Solution
When the switch is closed, the circuit is in the steady state, all currents are constant and the voltage
di
across the inductor, given by v L = L L , is equal to zero. Thus, the inductor can be replaced by a
dt
short circuit and the circuit becomes as shown in Fig. 2.
i1 (0− )
R1
( )
i2 0−
E
R2
iL (0 − )
R3
Fig. 2
In this circuit it holds
( )
iL 0 − =
E
R R
R1 + 2 3
R2 + R3
R2
= 9A .
R2 + R3
(1)
( )
Since the inductor current i L (t ) satisfies the continuity property i L (0) = i L 0 − = 9A . When the
switch is open the circuit has the form shown in Fig. 3.
28
iL (t)
R1
R3
E
vL(t)
L
Fig. 3
The current i L (t ) is specified by the equation
i L (t ) = i L (∞ ) + (i L (0) − i L (∞ )) e
−
t
τ
,
(2)
where
τ=
The steady state current i L (∞ ) is
L
= 0.01s .
R1 + R3
i L (∞ ) =
(3)
E
= 10 A .
R1 + R3
(4)
Substituting i L (0) = 9 A , i L (∞ ) = 10 A and τ = 0.01s into the equation (2) yields
i L (t ) = (10 − e −100 t )A .
(5)
di L
= 0.1 ⋅ 100 e −100 t = 10 e −100 t .
dt
(6)
Next we find v L (t )
v L (t ) = L
Question 2
The switch in the circuit shown in Fig. 4 is close until the steady state prevails and then it is opened.
Assuming that the opening occurs at t = 0 find the current iC (t ).
Data:
t=0
E = 120 V,
R1 = 10 Ω, R2 = 100 Ω,
R3 = 20 Ω, C = 10 μF.
E
R2
iC
R1
vC
Fig. 4
C
R3
29
Solution
When the switch is closed, the circuit is in the steady state, the current iC = C
dvC
= 0 . Thus, the
dt
capacitor can be replaced by an open circuit (see Fig. 5).
i (0− )
E
R2
R3
( )
vC 0 −
R1
Fig. 5
In the resistive circuit shown in Fig. 5 we find
( ) ( )
v C 0 − = i 0 − R3 =
E
R3 = 80V .
R1 + R3
(7)
Since the voltage vC (t ) across the capacitor satisfies the continuity property, vC (0 ) = vC (0 − ) = 80V.
When the switch is open the circuit has the form shown in Fig. 6.
R2
R3
C
vC(t)
iC(t)
Fig. 6
In this circuit it holds
vC (t ) = vC (∞ ) + (vC (0 ) − vC (∞ )) e
where
−
t
τ
(8)
,
τ = (R2 + R3 )C = 120 ⋅ 10 −5 = 1.2 ⋅ 10 −3 s .
(9)
The steady state voltage vC (∞ ) = 0 . Substituting vC (0 ) = 80 V, vC (∞ ) = 0, and τ = 1.2 ⋅ 10 −3 into
the equation (8) yields
vC (t ) = 80e
−
t
1.2⋅10 − 3
.
(10)
Hence, we find
t
t
dvC (t )
(
− 1) −1.2⋅10− 3 ⎛⎜ 2 −1.2⋅10 −3
−5
iC (t ) = C
= 10 ⋅ 80
e
= − e
⎜ 3
dt
1.2 ⋅ 10 −3
⎝
⎞
⎟A .
⎟
⎠
(11)
30
Question 3
The switch in the circuit shown in Fig. 7 is close until the steady state prevails and then it is opened.
Assuming that the opening occurs at t = 0 find vC (t ), i L (t ) and the voltage v(t) across the switch.
t=0
R1
iC(t)
E
C
iL(t)
v(t)
R2
vC(t)
L
vL(t)
Fig. 7
Data:
E = 100 V, R1 = R2 = 100 Ω, C = 14 μF, L = 200mH.
Solution
When the switch is close, the circuit is in the steady state, hence, iC = 0 and vL = 0. Consequently,
the circuit has the form shown Fig. 8.
R1
iL (0− )
E
R2
( )
vC 0
−
Fig. 8
In the circuit depicted in Fig. 8 we write
( ) RE = 1A = i (0),
v (0 ) = 0 = v (0 ).
iL 0 − =
L
1
−
C
C
When the switch is open the circuit consists of two parts which can be analysed separately. They are
shown in Figs. 9 and 10.
R1
iL(t)
iC(t)
E
C
Fig. 9
vC(t)
R2
L
Fig. 10
Voltage vC (t ) in the circuit depicted in Fig. 9 is given by the equation
vL(t)
31
vC (t ) = vC (∞ ) + (vC (0 ) − vC (∞ ))e
−
t
τ1
(12)
,
where
τ1 = R1C = 0.4 ⋅ 10−3 s ,
Since vC (0) = 0 , we have
vC (∞ ) = E = 100V .
t
−
⎛
⎞
0.4⋅10 − 3 ⎟
⎜
vC (t ) = 100 1 − e
V.
⎜
⎟
⎝
⎠
The circuit shown in Fig. 10 is described by the equation
i L (t ) = i L (∞ ) + (i L (0) − i L (∞ )) e
−
t
τ2
(13)
,
(14)
where
τ2 =
L
= 2 ⋅10 −3 s ,
R2
iL (∞ ) = 0.
Since iL (0) = 1A then
−
t
iL (t ) = e
A.
To find the voltage v(t), across the switch we apply KVL in the circuit shown in Fig. 7
2⋅10 − 3
v(t ) = vC (t ) − v L (t )
(15)
(16)
where
−
diL (t )
1 ⎞ − 2⋅10−3
−3
⎛
= 0.2⎜ −
e
= −100e 2⋅10 V .
−3 ⎟
dt
⎝ 2 ⋅10 ⎠
Substituting (13) and (17) into (16) yields
t
v L (t ) = L
t
⎛ ⎛
−
0.4⋅10 − 3
⎜
⎜
v(t ) = 100 1 − e
⎜ ⎜
⎝ ⎝
t
t
−
⎞
⎟ + 100e 2⋅10−3
⎟
⎠
(17)
⎞
⎟ V.
⎟
⎠
Question 4
The switch in the circuit depicted in Fig. 11 is on the terminal 1 until the steady state prevails.
At t = 0 it is flipped from terminal 1 to 2. Find the voltage v12(t) (between the terminals 1 and 2) and
the energy consumed by the resistor R4 from t = 0 to t → ∞ .
R1
iL(t)
1
L
E
2
R2
Fig. 11
iC(t)
t=0
R3
R4
C
vC(t)
Data: R1 = 10Ω
R2 = R3 = 20Ω
R4 = 10Ω
L = 0.1H
C = 10μF
E = 2V .
32
Solution
When the switch is on the terminal 1 and the circuit is in the steady state the inductor L can be
replaced by a short circuit and the capacitor C by an open circuit (see Fig. 12).
iL (0− )
R1
E
R2
R3
( )
vC 0−
R4
Fig. 12
In this circuit we find
iL (0 − ) =
E
2
=
= 0.12A = iL (0),
R2 R4
20 ⋅10
10 +
R1 +
30
R2 + R4
vC (0 − ) = iL (0 − )
R2 R4
20 ⋅10
= 0.12
= 0.8V = vC (0).
R2 + R4
30
When the switch is on the terminal 2 the circuit is separated into two parts (see Fig.13).
iL(t)
1
L
R1
E
2
v2(t)
R2
iC(t)
v12(t)
R3
R4
C
Fig. 13
To find iL(t) we rearrange the left part of the circuit as shown in Fig. 14.
iL(t)
L
E
R123 = R1 +
Fig. 14
R2 R3
= 20Ω
R2 + R3
vC(t)
33
In this circuit
i L (t ) = iL (∞ ) + (iL (0 ) − iL (∞ ))e
−
t
τ1
(18)
,
where
τ1 =
L
0 .1
=
= 0.005s ,
R123 20
iL (∞ ) =
E
2
=
= 0.1A .
R123 20
Since iL (0) = 0.12 , then
t
−
⎞
⎛
0.005 ⎟
⎜
iL (t ) = ⎜ 0.1 + 0.02e
⎟ A.
⎠
⎝
(19)
To find vC(t) we consider the right part of the circuit, consisting of the resistor R4 and the capacitor
C connected in parallel. We write
vC (t ) = vC (∞ ) + (vC (0) − vC (∞ ))e
where
τ 2 = R4 C = 10 ⋅10 −5 = 10 −4 s,
−
t
τ2
,
(20)
vC (∞ ) = 0 .
Hence, we have
t
− −4
⎛
vC (t ) = ⎜ 0.8 e 10
⎜
⎝
In order to find v12(t) we apply KVL
⎞
⎟V.
⎟
⎠
(21)
v12 (t ) = v2 (t ) − vC (t ),
(22)
where
v2 (t ) = iL (t )
t
−
⎞
⎛
R2 R3
= ⎜⎜1 + 0.2e 0.005 ⎟⎟ V.
R2 + R3 ⎝
⎠
(23)
Substituting (21) and (23) into (22) gives
t
t
− −4
⎛
−
v12 (t ) = ⎜1 + 0.2 e 0.005 − 0.8e 10
⎜
⎝
⎞
⎟V .
⎟
⎠
The energy consumed by the resistor R4 is specified by the equation
∞
− −4
⎛ 10 −4 ⎞ −10− 4
vC2 (t )
⎟⎟ e
dt =0.1∫ 0.64e 10 dt = 0.064⎜⎜ −
R
2
⎝
⎠
4
0
0
∞
W4 = ∫
2t
2t
∞
0
= 3.2 ⋅10 -6 J.
34
Question 5
A
B
10 V
3Ω
i1 (t)
t=0
i2 (t)
20 V
2Ω
6Ω
iL (t)
vL (t)
2H
3Ω
Fig. 15
For t < 0 the switch is on terminal A and the circuit is in steady state. At t = 0 the switch is flipped
to terminal B. Determine iL (t ) , i1 (t ) and vL (t ) for t ≥ 0 .
Solution
For t < 0 the current iL (t ) is constant, hence, it holds
vL (t ) = L
diL (t )
=0 .
dt
(24)
Thus, at t = 0− the inductor can be replaced by short circuit as shown in Fig. 16.
3Ω
i1 (0-)
10 V
i2 (0-)
2Ω
6Ω
iL (0-)
Fig. 16
( )
Since i2 0 − = 0 , we have
( ) ( )
iL 0− = i1 0 − =
10
= 2A .
2+3
(25)
On the basis of the continuity property
( )
iL (0 ) = iL 0− = 2 A .
When the switch is on terminal B the circuit becomes as shown in Fig. 17.
(26)
35
i1 (t)
3Ω
20V
iL (t)
i2 (t)
vL(t)
2H
6Ω
3Ω
Fig. 17
We take into account the one-port lying inside the dotted rectangle and replace it by Thevenin’s
circuit. To find V0C and Req we create two circuits depicted in Figs. 18 and 19.
3Ω
3Ω
20V
V0C
6Ω
6Ω
3Ω
Req
3Ω
Fig. 18
Fig. 19
Analyzing the circuits we obtain
V0C =
20
6 = 10 V ,
12
Req = 3 Ω .
(27)
Thus, the circuit shown in Fig. 17 is reduced to the one depicted in Fig. 20.
The time constant is given by
2
τ= s .
3
(28)
iL (t)
10V
2H
vL(t)
3Ω
Fig. 20
As t → ∞ the current i (∞ ) becomes constant, consequently the voltage across the inductor is
equal to zero and the inductor can be replaced by short circuit (see Fig. 21).
36
10V
iL (∞)
3Ω
Fig. 21
Current iL (∞ ) is
iL (∞ ) =
10
A .
3
(29)
Now we use the general formula
iL (t ) = iL (∞ ) + (iL (0) − iL (∞ ))e
−
t
τ
(30)
and substitute (26) and (29)
t
t
10 ⎛
10 ⎞ −
10 4 −
iL (t ) = + ⎜ 2 − ⎟ e τ = − e τ ,
3 ⎝
3⎠
3 3
(31)
2
where τ = s . The plot of iL against t is shown in Fig. 22.
3
iL(t)
10
3
2
0
t
Fig. 22
To find the current i1 (t ) , for t > 0 we replace the inductor in Fig. 17 by the current source iL (t ) .
As a result we obtain the resistive circuit shown in Fig. 23.
37
3Ω
i1 (t)
i2 (t)
20V
6Ω
iL(t)
3Ω
Fig. 23
We replace the parallel connection of the 6-ohm resistor and the current source iL (t ) by the series
connection of the resistor and a voltage source. This voltage source is 6iL (t ) (see Fig. 24).
3Ω
i1 (t)
20V
6Ω
3Ω
6iL(t)
Fig. 24
Using KVL and Ohm’s law we obtain
i1 (t ) =
20 + 6iL (t )
,
3+3+6
(32)
and substitute (31) into equation (32)
20 + 20 − 8 e
i1 (t ) =
12
−
t
τ
⎛ 10 2 − t
=⎜ − e τ
⎜3 3
⎝
⎞
⎟A
⎟
⎠
for
t>0 .
(33)
( )
At t = 0− the current i1 0− , flowing in the circuit shown in Fig. 16, equals
( )
i1 0− =
10
=2A .
2+3
To determine the voltage vL we use the formula
vL (t ) = L
Applying (31) and substituting τ =
2
we have
3
diL (t )
.
dt
(34)
38
⎛ 4 1 −t
vL (t ) = 2⎜
e τ
⎜3τ
⎝
t
⎞
−
⎟ = 4e τ .
⎟
⎠
(35)
For t < 0 the circuit is in steady state and vL = 0 . The plot of vL against t shows that at t = 0 the
voltage vL jumps from 0 to 4 (see Fig. 25).
vL (t)
4
0
t
Fig. 25
Alternatively, the voltage v1 (t ) , for t > 0 can be found applying KVL in the circuit depicted in
Fig. 17
vL (t ) + (3 + 3)i1 (t ) − 20 = 0
(36)
and substituting (33). As a result we obtain
⎛ 10 2 − t
vL (t ) = 20 − 6⎜ − e τ
⎜3 3
⎝
t
⎞
−
⎟ = 4e τ ,
⎟
⎠
t >0.
(37)
Question 6
R3 = 10 Ω
K
iC (t)
R1 = 30 Ω
A
vC (t)
C = 1 μF
E = 40 V
R2 = 10 Ω
B
t=0
L
Fig. 26
The switch was in position A for a long time prior to t = 0 . Find the voltage vC (t ) for t ≥ 0 .
39
Solution
Since the switch was in position A for a long time the circuit is in steady state at t = 0 − . Hence, the
voltage vC is constant and the current iC , given by
iC = C
dvC
dt
(38)
is equal to zero. As a result the capacitor can be removed and the circuit at t = 0 − is as shown in
Fig. 27.
K
i3 (0-)
R3
R1
vC (0-)
-
i1 (0 )
R2
L
i2 (0-)=0
E
Fig. 27
In this circuit
( ) ( )
i1 0 − = i3 0 − =
E
= 1A
R1 + R3
(39)
and
( )
( )
vC 0− = − R3i3 0− = −10 V = vC (0) .
(40)
For t > 0 the circuit is as shown in Fig. 28.
R3
K
iC (t)
R1
vC (t)
C
E
R2
L
Fig. 28
As t → ∞ , the voltage vC is constant and iC (∞ ) = 0 . Hence, the capacitor can be removed and the
circuit becomes as depicted in Fig. 29.
40
R3
K
i3 (∞)=0
R1
vC (∞)
i1 (∞)
R2
L
i2 (∞)
E
Fig. 29
In this circuit we write
i1 (∞ ) = i2 (∞ ) =
E
= 1A ,
R1 + R2
(41)
vC (∞ ) = R2 i2 (∞ ) = 10 V .
(42)
To find vC (t ) we substitute (40) and (42) into the equation
vC (t ) = vC (∞ ) + (vC (0) − vC (∞ )) e
−
t
τ
,
(43)
where
τ = ReqC .
(44)
To find Req we consider the circuit shown in Fig. 28. In this circuit we set E = 0 , remove the
capacitor C and find the resistance of the one-port KL
Req = R3 +
R1 R2
= 17.5 Ω .
R1 + R3
(45)
Thus, it holds τ = 17.5 ⋅ 10 −6 s and
vC (t ) = 10 + (− 10 − 10)e
Plot of vC (t ) is shown in Fig. 30.
−
t
τ
t
⎛
−
τ
⎜
= 10 − 20e
⎜
⎝
⎞
⎟V .
⎟
⎠
(46)
41
vC (t)
10
0
t
-10
Fig. 30
43
AC analysis
Question 1
i3(t)
i2(t)
vs(t)
L
vL(t)
C
vC(t)
v2(t)
R1
R2
i1(t)
Fig. 1
(i) In the circuit, depicted in Fig. 1, find the sinusoidal voltage vS(t) knowing that
v L (t ) = (10 cos(500t + 30°)) V.
(ii) Compute the reactive powers entering the inductor L and the capacitor C as well as the average
power consumed by the resistor R2.
(iii) Sketch a phasor diagram.
Data: R1 = R2 = 10Ω, L = 0.01H, C = 0.2mF.
Solution
(i) We redraw the circuit using the phasor format as shown in Fig. 2.
Vs
V2
I3
VL
ZL
VC
ZC
R2
R1
I2
I1
Fig. 2
The phasor of the voltage vL(t) is
VL = 10ej30˚ V.
To determine the voltage source we proceed as follows. At first we compute the current I3
I3 =
where
VL
,
ZL
ZL = jωL = j5Ω = 5ej90˚ Ω.
Hence, it holds
I3 =
10e j30°
= 2e −60° = 2(cos 60° − j sin 60°) = (1 − j1.732) A .
j90°
5e
Then we compute the voltage V2
V2 = VL + VC = I3(ZL + ZC),
where
ZC = − j
1
= − j10 Ω = 10e − j90° Ω .
ωC
According to the foregoing
V2 = 2e-j60˚ (j5-j10) = 10e-j150˚ = 10 (cos 150˚ - j sin 150˚) = (-8.66 – j5)V.
Using Ohm’s law we find the current I2
I2 =
V2
= e − j150° = (−0.866 − j0.5) A .
R2
KCL applied to the bottom node leads to the equation
–I2 + I1 – I3 = 0
or
I1 = I2 + I3 = (0.134 – j2.232) A .
The results obtained above enable us to determine the source voltage
VS = V2 + R1I1 = (-7.32 – j27.32) V = 28.28ej255˚ V.
Now we go from the phasor VS to the time varying voltage vS(t)
vS(t) = Re(VSejωt) = 28.28cos(500t + 255˚) = 28.28cos(500t - 105˚) V.
(ii)
The reactive powers entering the inductor L and the capacitor C are as follows:
PX L =
1 2
I 3 X L = 10 VAR ,
2
44
PX C
45
1
2
= I 3 X C = −20 VAR ,
2
whereas the average power consumed by the resistor R2 is
1
2
Pav2 = R2 I 2 = 5 W.
2
(iii) A phasor diagram is sketched in Fig. 3.
VL
I2
V2=VL+VC
VC
I3
I1
VS
I1R1
Fig. 3
Question 2
L2
i2
vs(t)
R1
L1
R2
iL
i1
1
C1
iC
1
Fig. 4
(i) Compute the resonant angular frequency ω0 of the parallel combination L1 and C1.
(ii) Determine the branch currents i2 (t ), i1 (t ), i L1 (t ), iC1 (t ) if
v S (t ) = 60 cos(ω 0 t + 60°) V.
(iii) Sketch a phasor diagram.
Data: R1 = 100 Ω, R2 = 20 Ω, L1 = 0.02 H, L2 = 0.01 H, C1 = 0.5 μF.
46
Solution
(i) The resonant angular frequency is given by
1
ω0 =
L1C1
=
1
−2
2 ⋅ 10 ⋅ 0.5 ⋅ 10
= 10 4
−6
rad
⋅
s
(ii) We redraw the circuit in the phasor format (see Fig. 5).
I=0
I2
Z L2
VS
V1
IC
I L1
I1
R1
1
Z L1
ZC
1
R2
Fig. 5
Having vs(t) we find the voltage source phasor
V s = 60 j60° V.
Since at resonance I = 0, the same current I1 = I2 traverses R1 , Z L2 , R2 and VS. Hence, we have
I 2 = I1 =
Vs
60e j60°
=
= 0.384e j20.2° A
R2 + Z L2 + R1 120 + j100
and in the time domain
i2 (t ) = i1 (t ) = Re(I 2 e jωt ) = 0.384 cos(10 4 t + 20.2°) A .
To determine the current I L1 we find the voltage V1
V1 = R1 I 1 = 38.4e j20.2° V
and apply Ohm’s law
I L1 =
V1
38.4e j20.2°
=
= 0.192e − j69.8° A .
j90°
jω o L1
200e
Hence, we obtain
(
)
(
)
i L1 (t ) = Re I L1 e jωt = 0.192 cos 10 4 t − 69.8° A .
Since at resonance i L1 (t ) = −ic (t ) then
(
)
(
)
ic (t ) = 0.192 cos 10 4 t − 69.8° + 180° = 192 cos 10 4 t + 110.2° A .
(i)
47
A phasor diagram is sketched in Fig. 6
I2jωoL2
VS
I2(R1+R2)
V1=I2R1
I C1
I2=I1
I L1
Fig. 6
Question 3
In the circuit shown in Fig. 7 find the sinusoidal voltage vc(t) using the phasor concept. Compute the
average and reactive power entering the branch consisting of R2 and L. Sketch a phasor diagram.
vs(t)
vc(t)
R2
C
ic(t)
R1
L
i1(t)
i2(t)
Fig. 7
Data:
v S (t ) = 20 cos(500t + 30° ) V, R1 = R2 = 20Ω, L = 0.02 H, C = 4 ⋅ 10 −5 F .
48
Solution
We redraw the circuit of Fig. 3 in the phasor format as depicted in Fig. 8.
A
Vs
Vc
Zc
Z2
R1
Ic
I1
I2
B
Fig. 8
The impedances Zc and Z2 are:
Zc = − j
1
1
= −j
= − j50 Ω ,
ωC
500 ⋅ 4 ⋅ 10 −5
Z 2 = R2 + jωL = 20 + j500 ⋅ 0.02 = (20 + j10 ) Ω .
The total impedance faced by the voltage source VS is
Z = R1 +
ZcZ2
= 45 Ω .
Zc + Z2
Since
V S = 20e j 30° V,
then
I1 =
VS
= 0.44e j30° A .
Z
Voltage Vc appears between nodes A and B, hence, it holds
Vc =
ZcZ2
I 1 = 11.1e j30° V
Zc + Z2
and in time domain
v c (t ) = Re(Vc e jωt ) = 11.1 cos(500t + 30°) V .
Next we find
Vc 11.1e j30°
I2 =
=
= 0.49e j3.5° A .
Z 2 20 + j10
49
Observe that the voltage across the branch R2, L is the same as VC. Hence we have
1 2
I 2 R 2 = 2 .4 W ,
2
1 2
PX = I 2 ωL = 1.2 VAR .
2
Pav =
A phasor diagram is sketched in Fig. 9.
VS
R1I1
VC
IC
jωLI2
I1
R2I2
I2
Fig. 9
Question 4
The circuit depicted in Fig.10 is supplied with the sinusoidal voltage source
v S (t ) = 12 cos (ω O t + 45°) V, ω O =
1
L1C1
.
Find the sinusoidal current i2(t) and the reactive power of the capacitor C2. Sketch a phasor diagram.
ic (t )
2
R1
vs(t)
L1
C2
R2
i2(t)
Fig. 10
Data:
R1 = R2 = 50Ω, L1=0.01H, C1 = 1μF, C2 = 4μF.
C1
50
Solution
We consider the circuit shown in Fig. 11.
I1
R1
I C2
VS
L1
VL1
VC2
C2
R2
C1
VC1
I2
Fig. 11
The angular frequency ω0 is
1
ωO =
L1C1
= 10 4
rad
.
s
Since ω0 is the resonant frequency of the series connection of R1, L1, C1 then the impedance of this
connection is
Z R1L1C1 = R1 = 50Ω .
Hence, the total impedance faced by the voltage source is given by
Z = R2 +
R1 Z c
,
R1 + Z c
where
Zc = − j
1
= −25Ω .
ωO C 2
Thus, we have
Z = 50 +
50(− j25)
= 60 − j20 = 63.2e − j18.4° Ω
50 − j25
and
I2 =
Vs
12e j45°
=
= 019e j63.4° A ,
− j18.4°
Z 63.2e
A
(
)
i2 (t ) = Re I 2 e jω0t = 0.19 cos(10 4 t + 63.4°)A .
51
Next we compute voltage VC2 using KVL:
VC2 = VS − R2 I 2 = 4.23V .
To find the reactive power of the capacitor C2 we use the relationship
1
PX = − B VC2
2
2
,
where
B = ω0 C 2 = 0.04S .
Hence, we obtain
PX = −0.36VAR .
A phasor diagram is sketched in Fig. 12.
VS
VL1
I C2
R2 I 2
I2
I1
R1 I 1 = VC2
VC1
Fig. 12
52
Question 5
In the circuit depicted in Fig. 13 determine the sinusoidal source voltage vS(t) if
i2 (t ) = 1.5 cos (1000t + 60°) V .
Compute the reactive power entering the one-port AB. Sketch a phasor diagram.
iC (t )
A
C
B
i2(t)
vS (t )
iL (t )
L
R2
R1
Fig. 13
Data:
R1 = R2 = 10Ω, L = 0.02H, C = 2 ⋅ 10 −5 F .
Solution
Let us redraw the circuit using the phasor format as depicted in Fig. 14, where
Z AB
1 ⎞
⎛
jωL⎜ − j
⎟
ωC ⎠ j20(− j50)
⎝
=
=
= j33.3Ω .
1
− j30
jωL − j
ωC
A
Vs
ZAB
VAB
B
I2
R2
R1
Fig. 14
53
The phasor of the current traversing all the elements of the circuit shown in Fig. 14 is
I 2 = 1.5e j60° A .
To find Vs we compute the total impedance faced by the voltage source
Z = R1 + Z AB + R2 = (20 + j33.3) Ω
and expressed it in polar from
Z = 20 2 + (33.3) e
2
j tan −1
33.3
20
= 38.9e j59° Ω .
Ohm’s law in phasor from gives
VS = ZI 2 = 58 .3e j119 ° V.
Now we come from the phasor back to the sinusoidal function using the expression
(
)
v S (t ) = Re V S e jωt ,
where ω = 1000
rad
. As a result we obtain
s
v S (t ) = 58.3 cos (1000t + 119°) V .
The reactive power entering the one-port AB is specified by
2⎞
⎛1
⎛ 1
2⎞
PX = Im⎜ Z AB I 2 ⎟ = Im⎜ j 33.3 ⋅ (1.5) ⎟ = 37.5VAR .
⎝2
⎠
⎝ 2
⎠
A phasor diagram is shown in Fig. 15.
VAB = Z AB I 2
VS
(R1+R2)I2
IL
IC
I2
Fig. 15
54
Question 6
In the circuit shown in Fig. 16 the voltage vL(t) across the inductor L is given:
v L (t ) = 3 cos (1000t + 60°)V .
(i) Find the sinusoidal source voltage vs(t) using the phasor concept.
(ii) Compute the average and reactive power delivered to the branch AB consisting of L and R3.
(iii) Sketch a phasor diagram.
Data: R1 = 50Ω, R2 = 40Ω, R3 = 60Ω, L = 0.03H.
R1
A
L
VS (t)
vL(t)
R2
R3
IC (t)
B
Fig. 16
Solution
We redraw the circuit of Fig.16 in the phasor form (see Fig. 17).
I1
R1
A
IL
ZL
VS
V2
R2
R3
I2
B
Fig. 17
(i) We find the phasor of the voltage vL(t)
VL = 3e j60° V
55
and the inductor impedance
Z L = jωL = j ⋅ 1000 ⋅ 0.03 = j30 = 30e j90° Ω.
Then we compute the phasor IL
VL 3 ⋅ e j60°
IL =
=
= 0.1e − j30° = 0.1(cos 30° − j sin 30°) = (0.087 − j0.05) A
j90°
Z L 30e
and find the phasor of the voltage v2(t)
V2 = (R3 + Z L )I L = (60 + j30 )(0.087 − j0.05) = (6.72 − j0.39 ) V.
Having V2 we compute I2
I2 =
V2 6.72 − j0.39
=
= (0.17 − j0.01) A .
R2
40
Using KCL at the top node we obtain
I 1 = I 2 + I L = (0.26 − j0.06) A .
KVL applied to the loop consisting of VS, R1, R2 leads to the equation
VS = RI + V2 = 50(0.26 − j0.06 ) + 6.72 − j0.39 = (19.57 − j3.39 ) V =
= 19.57 2 + 3.39 2 e
− j tan −1
3.39
19.57
= 19.86e − j9.8° V .
Hence, we have
v S (t ) = Re (VS e jωt ) = 19.86 cos (1000t − 9.8°) V .
(ii) The complex power of the branch AB is given by
1
1
PAB = V2 I L∗ = (6.72 − j0.39 )(0.087 + j0.05) = (0.302 + j0.151) VA .
2
2
Thus, it holds
Pav = Re(PAB ) = 0.302 W ,
PX = Im(PAB ) = 0.151 VAR .
56
(i)
A phasor diagram is sketched in Fig. 18
VL
R 3+ IL
I2
V2
I1R 1
IL
I1
VS
Fig. 18
Question 7
In the circuit depicted in Fig. 19 determine the resonant angular frequency ωO of the branch AB
consisting of R1, L1, C1.
The circuit is driven by the voltage source
v S (t ) = 100 cos(ω O t + 45°) V .
(i) Find the capacitance C2 knowing that the current i (flowing through C2) leads voltage vS by 45˚.
(ii) Compute i(t).
Data: R1 = R2 = 1000Ω, L1 = 0.04H, C1 = 1μF.
i
A
C2
i1
vR
i2
R1
1
vs (t )
R2
L1
vL
1
C1
vC
1
B
Fig. 19
57
Solution
First we compute the resonant angular frequency
ωO =
1
L1C1
1
=
0.04 ⋅ 10
= 5000
−6
rad
s
and the voltage source
v S (t ) = 100 cos (5000t + 45°)V .
Since the branch AB is in resonance, its impedance equals R1. Consequently, the circuit can be
redrawn as shown in Fig. 20, where
R=
R1 R2
= 500Ω .
R1 + R2
C2
i
R
VS
Fig. 20
(i) Current I is given by the expression
I=
VS
1
R− j
ωOC2
,
where
V S = 100e j45° V .
Hence, we obtain
R− j
V
1
= S.
ωOC2
I
Since I leads VS by 45˚, its phase is 90˚.
Thus, it holds
I = I e j90°
and
58
VS 100 − j45°
=
e
.
I
I
Using the above results we write
⎛ −1
tan −1 ⎜⎜
⎝ Rω O C 2
⎞
⎟⎟ = −45°
⎠
or
1
= 1.
Rω O C 2
Solving for C2 we obtain
C2 =
1
1
=
= 0.4 ⋅ 10 −6 F .
Rω O 500 ⋅ 5000
(ii) To find the current i(t) we use equation
I=
100e j45°
=
500 − j500
100e j45°
500 + 500 e
2
2
− j 45°
= 1.414e j90° A
and go from the frequency domain to the time domain
(
)
i (t ) = Re Ie jωOt = 1.414 cos (5000t + 90°) A .
59
Three-phase systems
Question 1
In the three-phase system shown in Fig. 1 the terminals B’ and C’ of the load have been short
circuited. The system is supplied with a balanced three-phase generator. The amplitude of the
generator phase voltage is 100V, the impedance of each connecting wire is Z p = (1 + j)Ω and the
phase impedance of the balanced load is Z = (6 + j10)Ω . Find the line currents Ia, Ib, Ic and sketch a
phasor diagram.
Va
A
Zp
Ia
A’
Z
Vb
B
Zp
Ib
B’
Z
Vc
C
Zp
Ic
C’
Z
Fig. 1
Solution
We redraw the circuit as shown in Fig. 2.
Zp
Va
A
Vb
B
Zp
C
Zp
n
Vc
Z
Ia
Z
Ib
Z
n’
Vn
Fig. 2
Let
1
Z a = Z p + Z + Z = (10 + j16)Ω ,
2
Z b = Z p = (1 + j)Ω,
Z c = Z p = (1 + j)Ω,
then the circuit depicted in Fig. 2 can be simplified as shown in Fig. 3.
60
Va
A
Za
Ia
Va’
Vb
n
B
Zb
Ib
n’
Vb’
Vc
C
Zc
Ic
Vc’
Vn
Fig. 3
The generator phase voltages are as follows:
Va = 100V,
⎛ 1
Vb = ⎜⎜ − − j
⎝ 2
⎛ 1
Vc = ⎜⎜ − + j
⎝ 2
3⎞
⎟ ⋅100 = (− 50 − j86.6 ) V,
2 ⎟⎠
3⎞
⎟ ⋅100 = (− 50 + j86.6) V.
2 ⎟⎠
We find the voltage Vn
Va Vb Vc
+
+
Za Zb Zc
= (− 44.8 − j1.2) V
Vn =
1
1
1
+
+
Za Zb Zc
and next calculate the currents
Va − Vn
= (4.1 − j 6.5) A,
Za
V − Vn
Ib = b
= (− 45.3 − j 40.1) A,
Zb
V − Vn
Ic = c
= (41.2 + j 46.5) A.
Zc
Ia =
The phasor diagram is shown in Fig. 4.
(1)
(2)
(3)
(4)
61
Vc
V’c
Ic
Va
Vn
Ia
Ib
V’a
V’b
Vb
Fig. 4
Question 2
In the three-phase system shown in Fig. 5 the phase c is open circuited. The system is supplied with
a balanced three-phase generator. The amplitude of the generator phase voltage is 220V, the
impedance of each connecting wire is Zp = (3+j 4) Ω and the phase impedance of the balanced load
~
is Z = (27+j 36) Ω. Find I a , I b , Vo and sketch a phasor diagram.
Zp
Va
Z
Ia
Va’
Vb
Zp
Z
Ib
n
n’
Vb’
Zp
Vc
Z
Ic= 0
~
Vo
Vc’
Vn
Fig. 5
Solution
Since Ic = 0, then Ib = –Ia. We write KVL equation in the loop consisting of the phases a and b:
Va − Z p I a − ZI a + Z (− I a ) + Z p (− I a ) − Vb = 0.
Hence, we find
Ia =
The generator phase voltages are
Va − Vb
.
2Z + 2Z p
(5)
62
Va = 220 V,
⎛ 1
3⎞
Vb = Va ⎜⎜ − − j ⎟⎟ = (− 110 − j190.5) V,
2 ⎠
⎝ 2
⎛ 1
3⎞
Vc = Va ⎜⎜ − + j ⎟⎟ = (− 110 + j190.5) V.
2 ⎠
⎝ 2
Substituting these voltages into (5) yields
Ia =
~
To find Vo we apply KVL
330 + j190.5
= (3.5 − j1.6)A = − I b .
60 + j80
~
Vo − Vc + Vn = 0 ,
(6)
where
Va
Vb
V
+
+ c
Zp + Z Zp + Z Zc
.
Vn =
1
1
1
+
+
Zp + Z Zp + Z Zc
(7)
In equation (7) Z c → ∞, hence, we have
Vn =
Va + Vb
V
=− c.
2
2
(8)
Using (6) we find
V
3
~
Vo = Vc − Vn = Vc + c = Vc = (− 165 + j285.75)V.
2 2
The phasor diagram is shown in Fig. 6.
Vc
~
Vo
Va
Vn
Ib
Va-Vn
Ia
Vb
Vb - V n
Fig. 6
63
Question 3
In the three-phase system shown in Fig. 7, the phase c of the load is short circuited. The system is
supplied with a balanced three-phase generator. The amplitude of the generator phase voltage is
100 V, the impedance of the balanced load Z = 100Ω. Find the currents Ia, Ib, Ic and the voltages
Va' and Vb' .
Va
Z
Ia
Va’
Vb
Z
Ib
n
n’
Vb’
Ic
Vc
Z
Vn
Fig. 7
Solution
Using KVL we write
Vn = Vc .
(9)
Va' = Va − Vc ,
(10)
V = Vb − Vc .
(11)
Hence, we have
'
b
We find the source voltages
Va = 100V,
⎛ 1
3⎞
⎟ = (− 50 − j86.6 )V,
Vb = 100⎜⎜ − − j
⎟
2
2
⎠
⎝
⎛ 1
3⎞
⎟ = (− 50 + j86.6 ) V
Vc = 100⎜⎜ − + j
⎟
2
2
⎠
⎝
and substitute them into the equations (10) and (11)
Va' = Va − Vc = (150 − j86.6 )V ,
Vb' = Vb − Vc = − j173V.
Next we calculate the currents
Va'
= (1.5 − j 0.87 ) A,
Z
V'
I b = b = (− j1.73) A ,
Z
I c = −(I a + I b ) = (− 1.5 + j2.6 ) A.
Ia =
The phase diagram is shown in Fig. 8.
(12)
(13)
(14)
64
Ic
Ia
Va ’
Vc= Vn
Ib
Va
Vb’
Vb
Fig. 8
Question 4
In the three-phase system shown in Fig. 9 determine the indications of the wattmeters and the
average power consumed by the load. The system is supplied with a balanced three-phase generator.
The amplitude of the generator phase voltage is 324V, the impedances of the load are: Za = 100Ω,
Zb = (100 – j100)Ω, Zc = 200Ω.
Va
Ia
*
Za
*
W1
Vac
Vb
*
Ib
n
*
Vbc
Vc
Zb
W2
n’
Zc
Ic
Vn
Fig. 9
Solution
The indications of the wattmeters are
1
PW1 = Re(Vac I a∗ ),
2
1
PW2 = Re V bc I b∗ .
2
(
)
(15)
(16)
65
At first we find the phase voltages of the generator and the voltage Vn:
Va = 324V ,
⎛ 1
3⎞
Vb = 324⎜⎜ − − j ⎟⎟ = (− 162 − j280.3)V ,
2 ⎠
⎝ 2
⎛ 1
3⎞
Vc = 324⎜⎜ − + j ⎟⎟ = (− 162 + j280.3)V ,
2 ⎠
⎝ 2
Va Vb Vc
+
+
Za Zb Zc
Vn =
= (132.6 − j73.6)V.
1
1
1
+
+
Za Zb Zc
Next we compute the currents Ia and Ib:
Ia =
Ib =
Va − Vn 324 − (132.6 − j73.6 )
=
= (1.91 + j0.74 )A,
Za
100
Vb − Vn − 162 − j280.3 − (132.6 − j73.6 )
=
= (− 0.44 − j2.51)A
Zb
100 − j100
and apply the equations (15) and (16).
1
1
1
PW1 = Re((Va − Vc )(1.91 − j0.74 )) = Re((486 − j280.3)(1.91 − j0.74 )) = (486 ⋅1.91 − 280.3 ⋅ 0.74 ) = 360.4,
2
2
2
1
1
1
PW2 = Re((Vb − Vc )(− 0.44 + j2.51)) = Re((− j560.6 )(− 0.44 + j2.51)) = 560.6 ⋅ 2.51 = 703.6 .
2
2
2
To find the average power consumed by the load we use the formula
P = PW1 + PW2 = 1064 W = 1.064 kW.