Q1 (a) Use mesh-current analysis to find the currents i1, i2, and i3 in the circuit shown in Figure Q1(a).
i. Write down the 3 voltage equations for the 3 meshes.
[7 marks]
ii. Solve for i1, i2, and i3.
[3 marks]
Figure Q1(a)
Solution:
i. Voltage/mesh equations:
Mesh 1: − i1 − 1 − 2(i1 − i2 ) − (i1 − i3 ) = 0
⇒ 4i1 − 2i2 − i3 = −1
...(1)
Mesh 2: 3 − 3i2 − 2i2 − (i2 − i3 ) + 2(i1 − i2 ) = 0
⇒ 2i1 − 8i2 + i3 = −3
...(2)
Mesh 3: (i1 − i3 ) + (i2 − i3 ) − 3 − 2i3 = 0
⇒ i1 + i2 − 4i3 = 3
...(3)
ii. Consolidate the mesh equations in the matrix form,
⎡ 4 −2 −1⎤ ⎡ i1 ⎤ ⎡ −1⎤
⎢ 2 −8 1 ⎥ ⎢i ⎥ = ⎢ −3⎥
⎢
⎥⎢ 2⎥ ⎢ ⎥
⎢⎣ 1 1 −4 ⎥⎦ ⎢⎣ i3 ⎥⎦ ⎢⎣ 3 ⎥⎦
Hence, by Cramer’s Rule,
−1
−3
3
i1 =
4
2
1
−2
−8
1
−2
−8
1
−1
1
−4
17
= − = −0.3542 (A)
−1
48
1
−4
Page 1
4
2
1
i2 =
4
2
1
−1
−3
3
−2
−8
1
−1
1
−4
3
= = 0.1875 (A)
−1 16
1
−4
4
2
1
i3 =
4
2
1
−2
−8
1
−2
−8
1
−1
−3
3
19
= − = −0.7917 (A)
−1
24
1
−4
(b) For the circuit shown in Figure Q1(b),
i. Calculate the open-circuit voltage at terminal a-b.
[3 marks]
ii. Calculate the short-circuit current at terminal a-b.
[3 marks]
iii. Hence, or otherwise, determine the Thévenin equivalent circuit at terminal a-b.
[2 marks]
iv. If a 1Ω resistor is connected to terminal a-b, what is the power delivered by the 10 V source?
[2 marks]
Figure Q1(b)
Solution:
i.
10 V
V1
V2
Apply KCL at node V1 and V2:
V1 − 10 V1 V1 − V2
+ +
=0
1
1
1
⇒ 3V1 − V2 = 10
...(1)
V2 − V1 V2
+ =0
1
1
⇒ −V1 + 2V2 = 0
...(2)
2
Solving (1) and (2) gives
V1 = 4 (V)
V2 = 2 (V)
Therefore, the open-circuit voltage at terminal a-b is
Vab ,oc = V2 = 2 (V)
ii. When the terminal a-b is short-circuited, V2 = 0. Hence
V1 − 10 V1 V1
+ + =0
1
1 1
⇒ V1 =
10
(V)
3
0 − V1
+ I ab , sc = 0
1
⇒ I ab , sc =
10
(A)
3
iii. The Thevenin’s resistance RTH is given by
RTH =
Vab ,oc
I ab , sc
=
2
= 0.6 (Ω)
10 3
VTH = Vab ,oc = 2 (V)
iv. If 1-Ω is connected at terminal a-b,
V1 − 10 V1 V1 − V2
+ +
=0
1
1
1
⇒ 3V1 − V2 = 10
...(1)
V2 − V1 V2 V2
+ + =0
1
1 1
⇒ −V1 + 3V2 = 0
...(2)
Solving (1) and (2) gives
V1 = 3.75 (V)
V2 = 1.25 (V)
Current supplied by the 10-V source:
IS =
Hence,
10 − V1 10
+ = 16.25 (A)
1
1
P10V = (10 )(16.25 ) = 162.5 (W)
(The same answer is obtained by finding the equivalent resistance, Req =
3
8
(Ω) ).
13
Q2 (a) Find the currents IA, IB, IC, and IN in the AC circuit shown in Figure Q2(a).
[10 marks]
Figure Q2(a)
(b) For the circuit shown in Figure Q2(b), find I1 and I 2 .
i. Write down the 2 voltage equations for the 2 meshes.
ii. Solve for I1 and I 2 .
Figure Q2(b)
4
[8 marks]
[2 marks]
Solution:
(a) Consider the per-phase circuit,
220∠0°
= 1∠90° (A)
− j 220
220∠ − 120°
IB =
= 2∠ − 30° (A)
− j110
220∠120°
IC =
= 4∠ − 150° (A)
− j 55
I N = I A + I B + I C = 2.646∠ − 130.9° (A)
IA =
(b) i. Mesh equations:
381.051∠30° = − j 220 I1 − j110 ( I1 − I 2 )
−381.051∠ − 90° = − j110 ( I 2 − I1 ) − j 55I 2
ii. Solving (1) and (2) gives
I1 = 2.157∠143.4° (A)
I 2 = 3.569∠166.1° (A)
5
...(1)
...(2)
Q3 (a) Assuming continuous capacitor voltage and inductor current, state which of the following
concerning a first-order circuit is/are correct:
i. If vC (0 − ) = 10 V and vC (∞) = 16 V , then vC (t ) = 10 + 6 1 − e −t / τ .
(
−
ii. If vC (0 ) = 1 V and vC (∞) = 5 V , then vC (t ) = 1 + 5e
−t / τ
+
−t / τ
−
−t / τ
iii. If v R (0 ) = 10 V and v R (∞) = 0 V , then v R (t ) = 10e
iv. If v R (0 ) = 10 V and v R (∞) = 0 V , then v R (t ) = 10e
−
(
+
−t / τ
v. If i L (0 ) = 5 A and i L (∞) = 9 A , then i L (t ) = 5 + 4 1 − e
vi. If i L (0 ) = 5 A and i L (∞) = 1 A , then i L (t ) = 1 + 4e
)
.
.
.
−t /τ
).
.
[6 marks]
(b) For the circuit shown in Figure Q3(b), assume that the circuit has been in a steady state for t < 0,
and that the switch is closed at t = 0.
i. Find the value of the inductor current iL at t = 0+ and t = ∞.
[4 marks]
ii. Calculate the time constant and write down the expression for the inductor current for t > 0.
[6 marks]
iii. Also write down the expression for the inductor voltage vL for t > 0.
[4 marks]
Where t = 0− is the time instance before which the switch is opened or closed, t = 0+ the time
instance after which the switch is opened or closed, and t = ∞ is the time at which the switch has
been opened or closed for a long period.
VX
Figure Q3(b)
Solution:
(a) i.
ii.
iii.
iv.
v.
vi.
Correct
Incorrect
Correct
Incorrect
Correct
Correct
(b) i. Apply KCL at the node X for t = 0−,
VX ( 0− ) − 10 VX ( 0− )
+
=0
2
3
VX ( 0− ) = 6 (V)
6
iL ( 0+ ) = iL ( 0− )
=
VX ( 0 − )
3
= 2 (A)
Apply KCL at the node X for t = ∞,
VX ( ∞ ) − 10 VX ( ∞ ) − 24 VX ( ∞ )
+
+
=0
2
6
3
VX ( ∞ ) = 9 (V)
iL ( ∞ ) =
VX ( ∞ )
3
= 3 (A)
ii.
Req = 2 || 6 + 3 = 4.5 (Ω)
τ=
L
2
4
=
= (s)
Req 4.5 9
iL ( t ) = 2 + 1 ⎡⎣1 − exp ( −2.25t ) ⎤⎦
= 3 − exp ( −2.25t ) (A)
iii.
diL
dt
= ( 2 )( 2.25 ) exp ( −2.25t )
vL ( t ) = L
= 4.5exp ( −2.25t ) (V)
7