Problem 1: [12 points] (CIRCLE THE CORRECT ANSWER. NO CREDIT FOR MULTIPLE ANSWERS.)
1) Determine the value of Zeq between a and b if ω is 104 rad/s.
0.1 uF
k
a) 2200 + j600 Ω
b) 2200 – j600 Ω
a
c) -2200 – j600 Ω
k
50 mH
d) -2200 + j600 Ω
b
2) Find is(t) for i1(t) = 2cos(1000t + 45) A.
a) 1.414cos(1000t + 45) A
b) 1.414cos(1000t – 45) A
c) 1.414cos(1000t ) A
d) None of the above
3) The RMS (effective) value for the periodic current shown in amperes is
a) 2
b)
2
3
2
3
2
c)
d) None of the above
4) Find the power dissipated by the 15 Ω resistor.
a) 21.6 W
b) 12 W
c) 15.5 W
d) None of the above
10
10 V
5) Find the Norton equivalent circuit between terminals a and b.
a) Isc = 0.75 A, Req = 15 Ω
b) Isc = 1.5 A, Req = 25 Ω
c) Isc = 2 A, Req = 15 Ω
3A
10
d) Isc = 1 A, Req = 20 Ω
2A
15
10 V
10
10
a
20
6) Find the value of the inductor L, if vs(t) = 10cos10t V and the average power dissipated by the
circuit is 4 W.
a) 0.25 H
b) 0.5 H
c) 1 H
d) 2 H
b
Problem 2:
a) [4 points] For the circuit shown find the equivalent resistance seen between terminals a and b.
SHOW ALL STEPS
Step
8Ω || 8Ω = 4Ω
4Ω || 4Ω = 2Ω
2Ω || 2Ω = 1Ω
1Ω + 2Ω = 3Ω
Req = 3
b) [6 points] Use node-voltage method to find the voltage v(t) in the circuit shown
4sin(4t ) V
v(t) =
Problem 3:
[10 points] Find v(t) in the following circuit.
5cos(2t+30) V
1
3
v(t)
1
F
8
10cos(4t) A
v(t) =
Solution:
Due to 5cos(2t+30)
Due to 10cos(4t)
Change to Phasor Domain
Change to Phasor Domain
+j2
3
V1
-j4
Combine two impedances in parallel
Using VDR V1= [j4/(3+j4)]*5<30
Combine two impedances in parallel
Using CDR I2=[3/(3-j4)]*10 = 3*10/5<-53.13
8V
=4<90*5<30/5<53.13 = 4<66.87
I2= 6<+53.13
V2=I2*-j4= 24<-36.87
v1(t)=4cos(2t+66.87)
v2(t)=24cos(4t-36.87)
v(t) = v1(t) + v2(t) +v3(t)
Due to 8V DC source
= 4cos(2t+66.87)+24cos(4t+36.87)+8
3
v3(t)
v3(t) = 8V
8V
Problem 4:
In the circuit shown, load 1 has impedance Z1 and load 2 has impedance Z2
Z1 absorbs 10 Watt (average power) and -15 var (Reactive power)
Z2 absorbs 20 Watt (average power) and 30 var (Reactive power)
10 cos(300t − 45 ) A
Find the following:
a) [2 points] The apparent power delivered to load Z1
Answer =
b) [2 points] The apparent power delivered to load Z2?
Answer =
c) [2 points] The power factor of load Z1 and indicate whether it is leading or lagging? pf1 =
d) [2 points] The power factor of load Z2 and indicate whether it is leading or lagging?
(e) [2 points] Find the voltage v(t)?
a)
b)
v(t) =
P1 = 10 − j15 watts
P1 = 18.03 VA
P1 = 20 + j 30 watts
P1 = 36.06 VA
c) pf1 = 10 /18.03 = 0.5546 leading
d) pf1 = 20 / 36.06 = 0.5546 lagging
e) Ptotal = 30 + j15 VA
V = 2 Ptotal / I *
= 2(30 + j15) /10∠45
= 6.7082∠ − 18.4349 = 6.3640 − j 2.1213 V
v(t ) = 6.7082 cos(300t − 18.4349) V
pf2 =
Problem 5:
[8 points] Three-phase power is supplied to a balanced, wye-connected load. The line-to-line voltage
is 208 Vrms, and the load consumes a total power of 16 kwatt at a lagging power factor of 0.9.
Determine:
a) The magnitude of the transmission-line currents
b) The individual load impedance ZL.
The magnitude of the line current is=
ZL =
Solution:
The transmission-line current (the load current) can be calculated as
( )
16000 = 3 VL I L cos θ Z L ;
IL =
The phase voltage Vp=208/
16000
= 49.346 A
3 (208)(0.9 )
=120 V.
So, the magnitude of the individual load impedance can be calculated as
ZL=Vp/IL=120/49.346=2.432 Ω.
Since the power factor is lagging, θ Z = cos −1 (0.9 ) = 25.842 .
Hence,
ZL= 2.432∠25.842 = 2.189+j 1.061 Ω.