Albaha University
Faculty of Engineering
Electrical Engineering Dept.
TAGSEER Scheme
Total Marks: 20
First Term — 1434/1435
Mid-Term Examination
Subject: Circuit Analysis 2
Date: 7/6/1435 H
Time Allowed: 90 Mins
Student Name: .......................
Academic Number: .......................
Answer The Following Questions:
20
1. Express the sinusoid v = −4 cos(3t + 40◦ ) as phasor.
[1 mark]
Answer: v = 4 cos(3t + 40◦ + 180◦ ) = 4 cos(3t + 220◦ ), thus V =4∠220◦ V
2. Find the sinusoid corresponding to the phasor I = j(5 − j12).
[1 mark]
√
5
Answer: I = 12 + j5 = 122 + 52 ∠ tan−1 ( 12 ) = 13∠22.62◦ , thus i(t) = 13 cos(ωt + 22.62◦ ) A
3. If v1 = 10 sin(ωt + 30◦ ) and v2 = 20 cos(ωt − 45◦ ), find V = v1 + v2 using phasors. [2 marks]
Answer: v1 = 10 cos(ωt+30◦ −90◦ ) = 10 cos(ωt−60◦ ), thus V = v1 +v2 = 10∠−60◦ +20∠−
45◦ = 10 cos(−60◦ ) + j10 sin(−60◦ ) + 20 cos(−45◦ ) + j10 sin(−45◦ ) = 5 − j8.7 + 14.1 − j14.1 =
19.1 − j22.8 = 29.7∠ − 50◦ , or V (t) = 29.7 cos(ωt − 50◦ ) V.
4. If voltage v = 6 cos(100t − 30◦ ) is applied to a 50 µF capacitor, calculate the current through
the capacitor.
[2 marks]
◦
= jωC ×6∠−30◦ = 100×50×10−6 ∠90◦ ×6∠−30◦ = 0.03∠60◦ A
Answer: IC = ZVC = 6∠−30
1/jωC
Thus iC = 0.03 cos(100t + 60◦ ) A
5. Determine v(t) and i(t) for the circuit in Fig. 1.
◦
[4 marks]
◦
Answer: VS = 5 cos(10t − 90 ) = 5∠ − 90 V
ZL = jωL = j10 × 0.2 = j2 Ω
Z = 4 + j2 = 4.47∠26.6◦ Ω
5∠−90◦
◦
I = VZS = 4.47∠26.6
◦ = 1.12∠ − 116.6
Thus i(t) = 1.12 cos(10t − 116.6◦ ) A
V = I×ZL = 1.12∠−116.6◦ ×2∠90◦ = 2.24∠−26.6◦ V
Thus v(t) = 2.24 cos(10t − 26.6◦ ) V
1
Fig. 1
6. In the circuit of Fig. 2, find i0 when ω = 10 rad/s.
[4 marks]
Answer: V = 4 cos(ωt) = 4∠0◦ V
ZC = jωC = −j/10 × 0.05 = −j2 Ω
ZL = jωL = j10 × 1 = j10 Ω
−j4(2+j2)
−j4
= (2−j2)(2+j2)
= 8−j8
=1−jΩ
Zk = 2 k −j2 = 2−j2
8
Zt = ZL + Zk = j10 + 1 − j = 1 + j9 = 9.06∠83.66◦ Ω
4∠0◦
◦
I0 = ZVt = 9.06∠83.66
◦ = 0.44∠ − 83.66 A.
◦
Thus i0 = 0.44 cos(10t − 83.66 ) A
Fig. 2
7. Using nodal analysis find v0 in the circuit of Fig. 3.
Answer: ω = 1 rad./s
v2 = 5 sin(t − 30◦ ) = 5 cos(t − 120◦ ) V
ZL = jωL = j1 × 1 = j = 1∠90◦ Ω
ZC = 1/jωC = 1/(j1 × 1) = −j = 1∠ − 90◦ Ω
Apply KCL at node v0
◦
10∠45◦ −V0
0
+ 5∠−120j −V0 = V
3
−j
◦
◦
10∠45 −V0
+ 5∠−120 j−V0 +V0 = 0, multiply by j3
3
j(10∠45◦ − V0 ) + 3(5∠ − 120◦ ) = 0
10∠135◦ − jV0 + 15∠ − 120◦ = 0
◦ +15∠−120◦
= 10∠45◦ + 15∠150◦ =
V0 = 10∠1351∠90
◦
7.07 + j7.07 − 12.99 + j7.5 = −5.92 + j14.57 =
15.73∠112.11◦ V, or v0 = 15.73 cos(t + 112.11◦ ) V
2
[6 marks]
Fig. 3