Electric Circuits
ECSE 2010
Prof. Shayla Sawyer
HW11 solution
Fall 2014
Homework due: December 10, 2014
Grading: Pick part a or b in Units 1, 2, 3, 4 (Problems 2-5)
For Extra Credit on this Extra Credit 11: Do BOTH a and b
Score will replace lowest homework score
Problem 1) 3 phase
A 3-phase circuit is terminated with a balanced Wye-load. The voltage between line A and B, VAB, is 100
VRMS with 30 degree phase. The frequency of operation is 60 Hz. The load is a Wye type with a resistive 75Ω
load.
a. Determine the three line voltages, VAB, VBC, and VCA.
V AB = 100∠30° [VRMS]
V BC = 100∠ − 90° [VRMS]
VCA = 100∠150° [VRMS]
b. Determine the three phase voltages, VAN, VBN, and VCN.
Note to graders VA=VAN and VB=VBN and VC=VCN
Using the relationship |VAB|=sqrt(3)|VA| and considering a balanced system
where all sources have the same amplitude
100
VA =
∠0° [VRMS]
3
100
VB =
∠ − 120° [VRMS]
3
100
VC =
∠120° [VRMS]
3
c. Determine the three line currents, IA, IB, and IC.
Since the load is a Wye type and Ohm’s Law relates the phase voltages to the
line currents in a Wye type load
100
IA =
∠0° [IRMS]
75 3
100
IB =
∠ − 120° [IRMS]
75 3
100
IC =
∠120° [IRMS]
75 3
1 of 11
Electric Circuits
ECSE 2010
Prof. Shayla Sawyer
HW11 solution
Fall 2014
d. Determine the three phase currents, IAB, IBC, and ICA.
For balanced loads, the delta configuration and Wye configuration are related
by Z? = 3ZWye. Ohm’s Law relates the line voltage and phase current, giving
100
I AB =
∠30° [IRMS]
225
100
I BC =
∠ − 90° [IRMS]
225
100
I CA =
∠150° [IRMS]
225
e. Determine the power consumed through one of the loads.
Since the Wye load and an equivalent Delta load must result in the same
power
 100
 10000
*
S = V ABRMS I AB
= (100∠30° )
∠ − 30°  =
[W], real power only
RMS
225
 225

f. If a 0.2 H inductor is added in series to the resistive load, which of the above currents and voltages will
change? What is the new total power supplied?
The voltages are determined by the source generators and are therefore
unchanged. Changing the load results in a change of the currents. Both line
currents and equivalent phase currents will change.
ZWye = 75+j75.4
In one branch of the Wye circuit, the power is
10000
2
V ARMS
3
S=
=
= 22.1 + j 22.2 = 31.34∠45° [VA]
*
75 − j 75.4
Z Wye
2 of 11
Electric Circuits
ECSE 2010
Prof. Shayla Sawyer
HW11 solution
Fall 2014
Problem 2. Thevenin/Norton
R1
2k
R5
4k
V2
2
R3
4
4k
V1
R4
1k
R2
2k
I1
1E-3
RLoad
a. Find the Thevenin equivalent circuit.
VTH = 5.66 V
RTH = 2.33k?
IN = 2.43mA
b. Dependent sources.
R15
1k
2k
5V
I1
4k
10 Vdc
V3
0.001Vy
V2
+
Vy
-
1k
0Vdc
2000(Ix)
2k
a. Find the current through the 4k resistor.
3 of 11
Electric Circuits
ECSE 2010
Prof. Shayla Sawyer
HW11 solution
1
1
1
 2k + 1k + 4k

1

−
4k

1

1

4k

VA = -0.4545
VB = 4.091
Vy = VA
Ix = -1.1mA
1
4k
1
1
1
+
+
2k 4k 1k
0
1
−
4k
−
Fall 2014

− 2
V A   0 
    10 5 
− 0.001 − 2 V B  =  2k + 1k 
 V  

−1
0  y   0 
I
0
− 1   x   0 

0.001
Problem 3)
a. Differential equations
R2
R1
1
L1
1E-3
Vs
2
C1
1E-6
1. Determine the differential relationship for the voltage across C1.
d 2VC 1  R1
1  dVC1
1  R1 
1
+ +
+
+ 1VC 1 =
VS


2
dt
L1C1  R 2 
L1C1
 L1 R 2C1  dt
2. Determine values for R1 and R2 so that the circuit is underdamped.
For the circuit to be underdamped, we require α < ωo. Using the differential
expression above and again rearranging some terms
1  R1
1 
1  R1 
+ 1
 +
<

2  L R 2C1 
LC  R 2 
Pick any two values that meet this requirement. R1 = 20, R2 = 50 satisfies the
relationship
4 of 11
Electric Circuits
ECSE 2010
Prof. Shayla Sawyer
HW11 solution
Fall 2014
3. For those values, plot the voltage across C1 if the source is 10V for t<0 and turns off at t=0.
10V
5V
0V
-5V
-10V
0s
0.1ms
V(R2:1)
0.2ms
0.3ms
0.4ms
0.5ms
0.6ms
0.7ms
0.8ms
0.9ms
1.0ms
Time
b. Laplace
Vs
C2
U1
+
1
OUT
-
R3
1
C3
0.333
OPAMP
R4
2
2.5
L2
0.5
1
0
+
Vout
-
0
The source voltage is 5 V for t < 0 and becomes 10 V for t > 0.
5 of 11
Electric Circuits
ECSE 2010
Prof. Shayla Sawyer
HW11 solution
Fall 2014
1. Draw the s-equivalent circuit.
1/sC2
VC1(0-)/s=5/s
U2
1/sC3
+
1
10/s
VC2(0-)/s=0/s R4
OUT
0.333
-
Vs
OPAMP
R3
1
2
2.5
+
sL2
0.5 Vout
1
-
0
LI(0-)=0
0
2. Use circuit analysis and partial fraction expansion to find the voltage across the inductor for t>0.
Vout = 2.5 exp(− 1t ) − 20 exp(− 2t ) + 22.5 exp(− 3t ) [V]
Problem 4)
a. Phasors
R1
1k
VOFF = 0
VAMPL = 10
FREQ = 1590
V1
R2
1
2
L2
5E-2
L1
2
2E-2
C1
1
1k
1. Determine a value for C1 so that the current through the source is 5mA and has zero phase.
C1 = 5E-7 F
6 of 11
Electric Circuits
ECSE 2010
Prof. Shayla Sawyer
HW11 solution
Fall 2014
b. Frequency response
Vs
C2
U1
+
1E-7
OUT
-
R3
1E5
C3
1E-5
OPAMP
R4
U2
2
101
L3
1E-3
+
OUT
-
C4
+
1E-9
OPAMP
R5 Vout
1k
1
0
0
0
-
1. Sketch the Bode plots for the magnitude and phase, H(s)=Vout/Vin
7 of 11
Electric Circuits
ECSE 2010
Prof. Shayla Sawyer
HW11 solution
Fall 2014
Problem 5)
a. Transformers
1:2
R3
0.111
1:9
C1
1.074E-2
V1
1:2
FREQ = 60
VAMPL = 110
VOFF = 0
R2
8
1:5
R4
800
1:4 (top:bottom)
R1
160
1. Determine the phasor form of the current through the source.
IS = 3.0+j1.61
2. Determine the total power in the left branch, (R3 and C1)
S = 98.3-j177 [VA]
b. Power
8 of 11
Electric Circuits
ECSE 2010
Prof. Shayla Sawyer
HW11 solution
Fall 2014
R6
75
VOFF = 0
VAMPL = 2kVrms
FREQ = 60
V2
R5
50
R7
10
2
C2
1.06E-4
Zunknown
L1
0.106
1
1. Determine Zunknown such that the power produced is purely real.
2. Determine the power produced in each load and the source
Load 1: S = 80k
Load 2: S = 48k-j16k
Load 3: S = 23.5k+94.1k
Load 4: S = -j78.1k
Source: S = 151k
Zunknown: -j51.2?
9 of 11
Electric Circuits
ECSE 2010
Prof. Shayla Sawyer
HW11 solution
Fall 2014
10 of 11
Electric Circuits
ECSE 2010
Prof. Shayla Sawyer
HW11 solution
Fall 2014
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