Download Exam #2 Solutions

California State Polytechnic University, Pomona
Electrical & Computer Eng.
Dr. Zekeriya Aliyazicioglu
ECE 209 - 01 Network Analysis II
Exam #2 (Solution)
Name:__________________________
October 29, 2001
1. The average power in the 60Ω resistor is 240W. Find the complex power of each branch of the circuits.
What is the source voltage Vg and overall complex power of the circuit?
(35 point)
20Ω
60Ω
30Ω
Vg
I2
I1
-j20Ω
240
=2A
60
I1( rms ) =
j20Ω
2
O1 = I1( rms ) 20 = 80VAR
S1 = P1 + jQ1 = 240 + j80 VA
V( rms ) = (60 + j 20)2 = 120 + j 40 V
I 2( rms ) =
V( rms )
30 − j 20
=
120 + j 40
= 2.154 + j 2.769 = 3.508∠52.12! A
30 − j 20
S2 = V( rms ) I*2( rms ) = (120 + j 40)(2.154 − j 2.769)
S 2 = 369.24 − j 246.12
I( rms ) = I1( rms ) + I 2( rms ) = 2 + (2.154 + j 2.769) = 4.154 + j 2.769 A
2
P20 Ω = R I( rms ) = 498.46 W
S = P20Ω + S1 + S 2 = 498.46 + 240 + j80 + 369.24 − j 246.12
S = 1107.7 − j166.12 VAR
Z eq = 20 + Z 2 || Z1 = 20 + 24.444 − j 6.667 = 44.444 − j 6.667
Vg = I( rms ) Z eq = (4.154 + j 2.769)(44.444 − j 6.667) = 203.081 + j 95.371 V
S = Vg I*(rms ) = (203.081 + j 95.371) (4.154 − j 2.769)
S = 1107.68 − j166.162 VAR
By: Z. Aliyazicioglu
ECE209-2-1
Name:__________________________
2.
If
vg (t ) = 150 cos10000t V in the following circuit:
a. Determine the load impedance that absorbs the maximum average power.
b. Find the maximum average power transferred to load.
c. What is the overall average power in the circuit?
(35 point)
6H
2.5nF
60kΩ
Vg
ZL
jω L = j104 (6) = j 60 K Ω
1
109
=
= − j 40 K Ω
jω C j104 (2.5)
ZTh =
60( j 60)
− j 40 = 30 + j 30 − j 40 = 30 − j10 K Ω
60 + j 60
VTh =
150(60)
= 75 + j 75V
60 + j 60
*
Z L = ZTh
= 30 + j10 K Ω
Z = ZTh + Z L = 30 − j10 + 30 + j10 = 60 K Ω
ITh =
VTh 75 + j 75
=
= 1.25 + j1.25 mA
Z
60
1
1
Stotal = VTh I*Th = (75 + j 75)(1.25 − j1.25) = 93.75 mW
2
2
2
1
1
Sload = RL ITh = 30(1.767 ) 2 = 46.84 mW
2
2
By: Z. Aliyazicioglu
ECE209-2-2
Name:__________________________
3. For the following circuit:
a. Find the average and reactive powers at the terminals of the voltage sources.
b. Find the average and reactive powers associated with each impedance branch in the circuit.
c. Check the balance between delivered and absorbed average power.
d. Check the balance between delivered and absorbed reactive power.
(30 point)
40Ω
50Ω
Vg
j60Ω
-j80Ω
220∠0°V(rms)
Z eq =50+(-j80) || (40 + j 60 ) = 178 − j16 Ω
Vg
Ig =
Z eq
=
220
= 1226 + j 0.1102 A
178 − j16
S = Vg I*g = (178 − j16)(1226 − j 0.1102) = 269.72 − j 24.244 VA
P50Ω = R I g
2
= 50 1226 + j 0.1102 = 75.761 W
2
V50Ω = I g 50 = (1.226 + j 0.1102)50 = 61.3 + j 5.51V
Vp =Vg -V50 Ω = 220 − (61.3 + j 5.51) = 158.7 − j5.51V
Q j 80 Ω =
S=
Vg
2
−80
Vp
= −314.82 VAR
2
40 − j 60
= 193.97 + j 290.95 VA
ST = P50 Ω + Q80 + S 2 = 75.76 − j 314.82 + 193.97 + j 290.95 = 269.73 − j 23.86 VA
pf = cos(θ v − θ i ) = cos(0 − 5.136) = 0.996 ( loading)
By: Z. Aliyazicioglu
ECE209-2-3
Name:__________________________
4. Find currents I1 and I2 in the following circuit
(25 point)
By: Z. Aliyazicioglu
ECE209-2-4