CSC 32 Class exercise 1 – DC Circuit analysis
1. For the ladder network in the following figure, find I and Req.
Solution
Req = 3 + 4 (2 + 6 3) = 3 + 2 = 5Ω
Ω
I=
10
10
=
= 2A
Re q 5
2. Find i, v, and the power dissipated in the 6-Ω resistor in the following figure.
8Ω
Solution
i1
i
9A
+
6Ω
v
-
-1-
4Ω
By current division, i =
12
(9) = 6 A
6 + 12
i1 = 9 − 6 = 3A, v = 4i 1 = 4 x 3 = 12 V
p6 = i2R = 36 x 6 = 216 W
3. Use superposition to find i in the following figure.
Let
i
= i1 + i2 ,
where i1 is due to the 12-V source and i2 is due to the 4-A source.
For i1, consider Fig. (a).
2||3
= 2x3/5 = 6/5,
io = 12/(6 + 6/5) = 10/6
i1 = [3/(2 + 3)]io = (3/5)x(10/6) = 1 A
For i2, consider Fig. (b), 6||3 = 2 ohm, i2 = 4/2 = 2 A
i
= 1+2
-2-
= 3A
4. Solve for the current I in the circuit of the following figure using Thevenin’s
theorem.
Solution
Remove the 30-V voltage source and the 12-ohm resistor.
RTh
10Ω
Ω
a
10Ω
Ω
a
+
+
40Ω
Ω
−
b
50V
b
(a)
From Fig. (a),
From Fig. (b),
40Ω
Ω
VTh
(b)
RTh = 10||40 = 8 ohms
VTh = (40/(10 + 40))50 = 40V
8Ω
i
a
12 Ω
+
−
40V
+
−
30V
b
(c)
The equivalent circuit of the original circuit is shown in Fig. (c).
30 – 40 + (8 + 12)i
= 0, which leads to
<END>
-3-
Applying KVL,
i
= 500mA
CSC 32 Class exercise 2 – DC Circuit Analysis
1. Apply mesh analysis to find i in the following circuit.
Solution
For loop 1,
6 = 12i1 – 2i2
3 = 6i1 – i2
(1)
For loop 2,
-8 = 7i2 – 2i1 – i3
(2)
For loop 3,
-8 + 6 + 6i3 – i2 = 0
2 = 6i3 – i2
(3)
We put (1), (2), and (3) in matrix form,
6 − 1 0  i1   3
 2 − 7 1  i  =  8 

 2   
0 − 1 6 i 3  2
6
−1 0
6 3 0
∆ = 2 − 7 1 = −234,
0 −1 6
6
∆ 2 = 2 8 1 = −240
0 2 6
−1 3
∆ 3 = 2 − 7 8 = −38
0 −1 2
At node 0, i + i2 = i3 or i = i3 – i2 =
∆3 − ∆2
− 38 − 240
=
= 1.188 A
∆
− 234
-4-
2. Use nodal analysis to find currents i1 and i2 in the circuit of the following figure.
At node 1,
24 − v 1
v − v 2 v1 − 0
= 1
+
10
20
40
v1 − v 2 v 2
=
20
50
Solving (1) and (2) gives,
At node 2, 5 +
96 = 7v1 - 2v2
500 = -5v1 + 7v2
(2)
v1 = 42.87 V, v2 = 102.05 V
i1 =
v1
= 1.072 A,
40
v2 =
v2
= 2.041 A
50
3. For the bridge network in the following figure, find i0.
Assume all currents are in mA and apply mesh analysis for mesh 1.
30 = 12i1 – 6i2 – 4i3
15 = 6i1 – 3i2 – 2i3
(1)
for mesh 2,
0 = - 6i1 + 14i2 – 2i3
0 = -3i1 + 7i2 – i3
(2)
for mesh 2,
0 = -4i1 – 2i2 + 10i3
0 = -2i1 – i2 + 5i3
(3)
Solving (1), (2), and (3), we obtain,
io = i1 = 4.286 mA.
-5-
(1)
4.
What value of R in the circuit of the following figure would cause the current
source to deliver 800mV to the resistors?
Using R ∆ = 3RY = 3R, we obtain the equivalent circuit shown below:
30mA
3R R =
3R
3R
3R
R
30mA 3R
R
3RxR 3
= R
4R
4
3R (3RxR ) /(4R ) = 3 /( 4R )
3
3Rx R
3 
3
3
2 =R
3R  R + R  = 3R R =
3
4 
2
4
3R + R
2
P = I2R
800 x 10-3 = (30 x 10-3)2 R
R = 889 Ω
<END>
-6-
3R/2
CSC 23 Class exercise 3 –First Order Circuit
1. The switch in the following circuit has been closed for a long time. At t=0, the
switch is opened. Calculate i(t) for t>0.
Solution
When t < 0, the switch is closed and the inductor acts like a short circuit to dc.
short-circuited so that the resulting circuit is as shown in Fig. (a).
3Ω
12 V
i(0-)
+
4Ω
2H
−
(a)
i (0 − ) =
(b)
12
=4A
3
Since the current through an inductor cannot change abruptly,
i ( 0) = i ( 0 − ) = i ( 0 + ) = 4 A
When t > 0, the voltage source is cut off and we have the RL circuit in Fig. (b).
τ=
L 2
= = 0 .5
R 4
Hence,
i( t ) = i(0) e - t τ = 4 e -2t A
-7-
The 4 Ω resistor is
2. In the following circuit, v(0)=20V. Find v(t) for t>0.
Solution
v ( t ) = v ( 0) e - t τ ,
τ = R eq C
R eq = 2 + 8 || 8 + 6 || 3 = 2 + 4 + 2 = 8 Ω
τ = R eq C = (0.25)(8) = 2
v( t ) = 20 e -t 2 V
3.
The switch in the following figure has been position a for a long time. At t=0, it
moves to position b. Calculate I(t) for all t>0.
Solution
R eq = 6 || 3 = 2 Ω , τ = RC = 4
v( t ) = v(∞) + [ v(0) − v(∞)] e -t τ
Using voltage division,
-8-
v ( 0) =
3
3
(30) = 10 V , v(∞) =
(12) = 4 V
3+ 6
3+ 6
Thus,
v( t ) = 4 + (10 − 4) e - t 4 = 4 + 6 e - t 4
i( t ) = C
 - 1
dv
= (2)(6)   e -t 4 = - 3 e -0.25t A
4
dt
4. Find v(t) for t<0 and t>0 in the circuit in the following figure.
For t < 0, consider the circuit shown in Fig. (a).
3i o + 24 − 4i o = 0 
→ i o = 24
v( t ) = 4i o = 96 V
i=
v
= 48 A
2
For t > 0, consider the circuit in Fig. (b).
i( t ) = i(∞) + [ i(0) − i(∞)] e - t τ
i(0) = 48 ,
i(∞) =
R th = 2 + 8 = 10 Ω ,
20
=2A
8+ 2
τ=
L
0 .5 1
=
=
R th 10 20
i( t ) = 2 + ( 48 − 2) e -20t = 2 + 46 e -20t
v( t ) = 2 i( t ) = 4 + 92 e -20t V
<END>
-9-
CSC 23 Class exercise 4 –Second Order Circuit
Q1
Find v(t) for t>0 if v(0)=6V and i(0)=2A in the circuit shown in following
figure.
Solution
This is a series,
source-free circuit.
60||30
α = R/(2L) = 20/(2x2) = 5
ωo
and
v(0)
= Ldi/dt
= 6
ωo =
1
LC
=
1
0.04
= α leads to critical damping
i(t) = [(A + Bt)e-5t],
v
= 20 ohms
i(0) = 2
= 2{[Be-5t] + [-5(A + Bt)e-5t]}
= 2B – 10A = 2B – 20
Therefore,
= A
or B = 13.
i(t) = [(2 + 13t)e-5t] A
Q2 In the circuit in the following figure, calculate i(t) and v(t) for t>0.
Solution
-10-
= 5
At t = 0-, v(0) = (8/(2 + 8)(30) = 24
For t > 0, we have a source-free parallel RLC circuit.
α = 1/(2RC) = ¼
ωo = 1/ LC = 1 / 1x 1 4 = 2
Since α is less than ωo, we have an under-damped response.
ωd = ω o2 − α 2 = 4 − (1 / 16) = 1.9843
vo(t) = (A1cosωdt + A2sinωdt)e-αt
v(0) = 24 = A1 and i(t) = C(dv/dt) = 0 when t = 0.
dv/dt = -α(A1cosωdt + A2sinωdt)e-αt + (-ωdA1sinωdt + ωdA2cosωdt)e-αt
at t = 0, we get dv(0)/dt = 0 = -αA1 + ωdA2
Thus, A2 = (α/ωd)A1 = (1/4)(24)/1.9843
= 3.024
v(t) = (24cosω
ωdt + 3.024sinω
ωdt)e-t/4 volts
d 2i
di
3. The step response of a series RLC circuit is given by
+
2
+ 5i = 10
dt 2
dt
Given i(0) = 2 and di(0)/dt = 4, solve for i(t).
Solution
s2 + 2s + 5
= 0,
− 2 ± 4 − 20
= -1±j4
2
i(t) = Is + [(A1cos4t + A2sin4t)e-t], Is = 10/5 = 2
which leads to
s1,2
i(0) = 2
di/dt
-t
=
= = 2 + A1,
-t
= [(A2cos4t)e ] + [(-A2sin4t)e ]
or A1
= 4
-t
i(t) = 2 + sin4te A
-11-
= 0
= 4A2,
or A2
= 1
4. In the following circuit, find v(t) and i(t) for t > 0. Assume v(0) = 0V and i(0) =
1A.
t=0
i
2Ω
4A
1H
v
0.5F
Solution
ωo = 1/ LC = 1/ 1x 0.5 =
2
α = 1/(2RL) = (1)/(2x2x0.5) = 0.5
Since α < ωo, we have an underdamped response.
s1,2
=
− α ± α 2 − ω 2o =
i(t) = Is + [(Acos1.323t + Bsin1.323t)e-0.5t],
Thus,
i(0) = 1
v
di/dt
= vC
Is = 4
= 4 + A or A = -3
= vL = Ldi(0)/dt
= 0
= [1.323(-Asin1.323t + Bcos1.323t)e-0.5t] + [-0.5(Acos1.323t + Bsin1.323t)e-0.5t]
di(0)/dt
Thus,
-0.5 ± j1.323
= 0
= 1.323B – 0.5A or B = 0.5(-3)/1.323
i(t) =
{4 – [(3cos1.323t + 1.134sin1.323t)e-0.5t]} A
<END>
-12-
= -1.134
CSC 23 Class exercise 5 – Steady State AC Circuit Analysis
Q1 A sinusoidal voltage is give by the expression
v = 100 cos(240πt + 45o )mV .
Find
(a) f in hertz; (b) T in milliseconds; (c) Vm (d) v(0) (e) φ in degrees and radians;
(f) the smallest positive value of t at which v=0
(g) the smallest positive value of t at which dv/dt=0.
2. Find the impedance Zab in the circuit seen in following figure. Express Zab in both
polar and rectangular form.
-13-
3. Find the steady-state expression for i0(t) in the circuit in the following figure if
vs=750cos5000t mV.
4. The circuit in the following figure is operating in the sinusoidal steady state. Find
the steady-state expression for v0(t) if vg=64cos8000t V.
-14-
<END>
-15-
CSC 23 Class exercise 6 – Steady State AC Circuit Analysis (II)
Q1 Use nodal analysis, find i0(t) in the following circuit.
Solution
cos(2 t ) 
→ 1∠0°, ω = 2
8 sin(2 t + 30°) 
→ 8∠ - 60°
→
1H 
jωL = j2
12F 
→
1
1
=
= -j
jωC j (2)(1 2)
2H 
→
jωL = j4
14F 
→
1
1
=
= - j2
jωC j (2)(1 4)
-j2 Ω
2Ω
j4 Ω
j2 Ω
V1
Io
8∠-60° V
+
−
-j Ω
V2
1∠ 0° A
Consider the circuit below.
At node 1,
(8∠ - 60°) − V1 V1 V1 − V2
=
+
2
-j
j2
8∠ - 60° = (1 + j) V1 + j V2
(1)
-16-
At node 2,
1+
V1 − V2 (8∠ - 60°) − V2
+
=0
j2
j4 − j2
V2 = 4∠ - 60° + j + 0.5 V1
(2)
Substituting (2) into (1),
1 + 8∠ - 60° − 4∠30° = (1 + j1.5) V1
Therefore,
V1 =
1 + 8∠ - 60° − 4 ∠30°
1 + j1.5
Io =
V1 1 + 8∠ - 60° − 4 ∠30°
=
= 5.024 ∠ - 46.55°
-j
1 .5 − j
i o ( t ) = 5.024 cos(2t – 46.55°°) A
Q2 Determine V0 and I0 in the following circuit using mesh analysis.
Solution
Consider the circuit below.
For mesh 1,
(2 + j4) I 1 − 2 (4 ∠ - 30°) + 3 Vo = 0
where
Vo = 2 (4 ∠ - 30° − I 1 )
Hence,
(2 + j4) I 1 − 8∠ - 30° + 6 (4∠ - 30° − I 1 ) = 0
4∠ - 30° = (1 − j) I 1
or
I 1 = 2 2 ∠15°
-17-
Io =
3 Vo
3
=
(2)(4 ∠ - 30° − I 1 )
- j2 - j2
I o = j3 (4 ∠ - 30° − 2 2 ∠15°)
I o = 8.485∠
∠15°° A
Vo =
- j2 I o
∠-75°° V
= 5.657∠
3
Q3 Solve for v0(t) in the circuit of the following figure using the superposition principle.
Solution
Let v o = v 1 + v 2 + v 3 , where v1 , v 2 , and v 3 are respectively due to the 10-V dc source, the ac
current source, and the ac voltage source.
For v1 consider the circuit in Fig. (a).
The capacitor is open to dc, while the inductor is a short circuit.
v1 = 10 V
Hence,
For v 2 , consider the circuit in Fig. (b).
ω= 2
2H 
→
jωL = j4
1
F 
→
12
6Ω
-j6 Ω
1
1
=
= - j6
jωC j (2)(1 / 12)
+
4∠ 0° A
V2
(b)
Applying nodal analysis,
-18-
j4 Ω
4=
V2 V2 V2  1 j j 
+
+
=  + − V
6 - j6 j4  6 6 4  2
24
= 21.45∠26.56°
1 − j0.5
V2 =
v 2 = 21.45 sin(2t + 26.56°) V
Hence,
For v 3 , consider the circuit in Fig. (c).
ω=3
2H 
→
jωL = j6
6Ω
12∠0° V
+
−
j6 Ω
+
-j4 Ω
V3
(c)
1
F 
→
12
1
1
=
= - j4
jωC j (3)(1 / 12)
At the non-reference node,
12 − V3 V3 V3
=
+
6
- j4 j6
V3 =
12
= 10.73∠ - 26.56°
1 + j0.5
Hence,
v 3 = 10.73 cos(3t − 26.56°) V
Therefore,
v o = 10 + 21.45 sin(2t + 26.56°) + 10.73 cos(3t – 26.56°) V
<END>
-19-