EGR252 Solutions to Quiz #1
1. In Figure 1-1, what is the frequency of the sources?
Answer
A. 500 rad/sec
B. 500 Hz
C. 100 rad/sec
D. 100 Hz
2. What is i2(t) in the circuit of Figure 1-1?
Answer
A. 2 cos (500t)
B. 5 cos (500t)
C. 2 cos (500t+30o)
D. 5 cos
EGR252 Solutions to Quiz #2
1. Question What is the complex power being delivered to a load that draws 1000 W (P) at a lagging PF of
0.8 ?
Answer
A. 1000+j750
(VA)
B. 1000+j600
(VA)
C. 1000-j750
(VA)
D. 800-j1000
(VA)
2. Question A composite load consists
EGR252 Solutions to Quiz #3
Question A Y-connected power system is known to have V an =100<80 o V and V bn =100< -40 o V. Which of
the following statement is false? Answer
A. Vca=173.2 <130o V
B. Vbc=173.2 <-10o V
C. Vcn=100 <200o V
D. The phase sequence
EGR252 Solutions to Quiz #8
Question If a two-port network A and a two-port network B are in series, which of the following statement is
true? Answer
A. The series-ed network impedance matrix Z= Z A+ Z B .
B. The series-ed network admittance matrix Y= Y A
Chapter 14
14.1
(a) s =0;
(b) s = j9s 1;
(c) s =-8s-1;
(d) s =-1000 j1000s 1;
(e) v(t)=8+2cost mV cannot be attributed a single complex frequency. In a circuit
analysis problem, superposition will need to be invoked, where the original function v(t)
is ex
Chapter 12
12.5
V25=V24+V45=56.67 -11.52o V
V13=V12+V25+V53=189.8 34.95o V
12.7
(a) Van=400 -45o V ; Vbn=400 75o V ; Vcn=400 195o V
(b) The phase sequence is negative, since sequence is acbacb.
12.13 (a) C=40.85sin21.80o/(377440)=91.47 (F)
I
(b) AB =37791
Chapter 10
10.3
(a) f(t)=58.31cos(t+149.04 )
g(t)=57.01cos(t+15.255 )
Amplitude of f(t)=58.31 and Amplitude of g(t)=57.01
(b) f(t) leads g(t) by 133.8
10.11
-2x10-3cos5t+10i+Vc=0
dV
dV
Since i=ic=C c , then 30 c +Vc=2x10-2cos5t
dt
dt
Vc(t)=Acos(5t+)
We fi
Chapter 11
11.6
R
For t>0, i(t)= 8e L t = 8e 2 t .
(a)P(0+)=82x1=64 (W)
(b)at t=1 s, i=1.083 A; P(1)=i2R=1.723 W.
(c)at t=2 s, i=146.5 mA; P(2)=i2R=21.47 mW.
11.11
Zin=6.5+j5
s=100/(6.5+j5)
I
s,abs=(-1/2)x100x12.194cos37.57=-483.3 (W)
P
P4,abs=(1/2)(12
NET PRAVICY
Gramajo, R. 2007. Social Issues of Today. McGraw-Hill: New York.
[A]
The internet has brought mixed blessings to the
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EGR 252 Practice Midterm Exam
1. Let N1=100 turns and N2=200 turns in the ideal transformer shown below, find the average power (P) deliver
to ZL for (a) I2(max)
A;
o
230
(b)V1(rms)
V;
100 60o
(c) VS(max)
V
120 0o
2. Let k=0.5;
(a) Draw the circuit in fre
Chapter 13
13.3 1 and 3, 2 and 4
1 and 4, 2 and 3
3 and 1, 2 and 4
13.4 (a)
di1
di
+ M 2 = 2400 cos 80 t 1200 sin 80 t
dt
dt
1200
= 24002 + 12002 cos(80t tan 1
)
2400
= 2683 cos(80t 26.57)(V)
V1 = L1
(b)
di 2
di
+ M 1 = 7200 sin 80 t + 1200 cos 80t
dt
dt
EGR252 Solutions to Quiz #7
Question What is Q 0 for the circuit shown in Figure 7-1? Answer
A. Q0=2.12
B. Q0=0.47
C. Q0=3.52
D. Q0=1.35
Question For the circuit shown in Figure 7-2, which of the following statement is true ? Answer
A. Z in =10- j990 (Ohm
EGR252 Solutions to Quiz #6
Question What is Z in for the circuit of Figure 6-2 as a ratio of two polynomials in S if R=10 Ohm; L=1H;
C=2F? Answer
A. 10+ S/( 2 S 2+1)
B. 10+ 1/(2 S 2+1)
C. 10+ S/( S 2+2)
D. 10+ 1/( S 2+2)
Question What is Z in (j10) in re
EGR252 Solutions to Quiz #4
Question Two coupled inductors, L1=1 H and L2=100 H, have the coupling coefficient of 0.9. What is the value
of mutual inductance, M? Answer
A. 9 H
B. 90 H
C. 10 H
D. 100 H
Question In the circuit of Figure 13.38 (p521), let L1
EGR252 Solutions to Quiz #5
Question What is the Laplace Inverse of V( S)= (3 S-1) / [( S-3)( S+5)] ? Answer
A. v(t)=(e3t+2e-5t) u(t)
B. v(t)=(e-3t+2e5t) u(t)
C. v(t)=(2e-3t+e5t) u(t)
D. v(t)=(2e3t+e-5t) u(t)
Question What are the poles and zeros of V( S)
10.8
(p386)
The inductor is represented by a j(10 10 3) (1200) = j12 impedance and the capacitor
by a impedance.
(a)
The voltage across the 20 resistor is then
and the current through it is
Thus,
(by KCL)
= 2.33231.04 A
(b)
34.86 4.55 V
7
(c)
3.986 cos (1
EGR252 Midterm Exam Notes
Three
Three Phase
Phase Power
Power
* Ideal Transformers
Turn Ratioa=
N2
N1
Voltage Adjustment
V2
=a
V1
I2 1
Current Adjustment =
I1 a
Impedance MatchingZ =
ZL
a
2