This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: EE10 Sample Final Exam
UCLA
Department of Electrical Engineering Instructor: Prof. Gupta ’ Spring 2011
Q1. (5 + 5 points) (a) The linear circuit (black box) is connected to the load resistor R L at the terminals a and
b as shown on the diagram. If the RL = 2 k0 , then the current through the load equals  = 8 mA. If the RL = 5 k0 , then the current through the load equals l = 4 mA. Draw the Norton Equivalent of this circuit. (Linear
circuit} $272»— itzzﬁ. iii: Ellis/14A W Wm. wag; CK
if" “a
»« f7?“ .2? 3%;
AWE? \i 33 (b) Find the Thevenin equivalent of the simple transformer circuit below. Q2. (5 + 3 +2 points)
Assume w = 104 rad/sec. (a) Use source transformation to obtain the current IO. Express your answer in milliamps,
both in the phasor notation (rectangular form) and in the time domain. (b) Draw a phasor diagram showing the voltage across 30ohm resistor, 4mH inductor, 5uF
capacitor. (c) Graphically solve (roughly) for current through the 20 ohm resistor. You don’t need to
calculate the exact value of the current. Just show it graphically. 4*siuuat) V 3 ME 2 it}? ‘
law—T 23 Q;
C 03 65 V
icy10:) \ ‘5th
10003 4.
WSMGB—j l; 10
l+l5V
—1 +1" ‘
1Q: ﬁ’ 3 a 40 *AGQ WA e "56. éLlSSQMA : SAESM LiooootHBSQ)
‘EO‘JLci.
\xl 1;» farm,
r \T—W />
mm a ”W
\k— 1‘)“; VS“? 3"».
{We
um”
p3
O
t Q3. (8 + 2 points) Consider the circuit with the following parameters: VSl =12V,‘VSZ =1OV;RS =250,‘R=1000;C=32nF;L=500uH;RL =SOQ
Assume DC steady—state conditions at t <0. (a) Write and solve the differential equation for the capacitor voltage vc as a function of time at
t >0. (b) What is the energy stored in the inductor at steady state ? What about the capacitor ? VSZ W®1® (PVC +300000i99L613X1010V“ =6 X1”
cw dt' ' C" ”K 0 gun. “ *[5‘0000 fr; 3.000605 Vow (”‘33 ’7' 8,16% (kg (05 1000063 * {él 811/! 10000013) V6C13'3 Vamtt) rvcﬁm): e’lsmt ([4. (05 2.000002: Wls‘maooaoct) '1 0 Maw {41 —(0= ~18 Kw 8 M03 alt '2 " [50000 {g t‘ looooolﬁl “‘ 9000000 l42 ‘— lq » —~(
"Vctﬂ 2 € smooté —gco$moooo£ +i0tS‘nM1ooooot3‘lO / "C70
W w W (b) “Cr—00 ”é
i“ *x
, vc "TR *waho J VCCoﬁ32nlO m g ‘L ‘ E ) M be ‘3 3%”“3 ’1 %,%§lnxir3 {gem/m
“x” M a: 3L § “ ‘1 ,0 KL» ,“ :2. a.“ ’hé’é‘) a (:3 Q; Q4. (4+6 points) (a) Label the nodes in this circuit and write down the matrix equation to solve this circuit by
node method. You don’t need to solve the actual KCL equations. lo ”Lissa was —o.i o ”v“
o 2 was omv‘iecieb'ioxﬂl :5
0 soil q 166'] owls—041$ Va
0 o ~0.o°ll «0.05 u’lal’l 5 (b) Use source transformation and source shifting to contain minimum number of
independent sources with all of them being voltage sources. Then write the matrix equation for solving the circuit using loop method. You don’t need to actually solve the
KVL equations. Q5. (3 + 3 + 4 points) (a) What is the Norton Equivalent of this circuit at the input terminals ? The network N
obeys the vn(t) = 5in(t) 3. (b) You design a ”special” capacitor for which Q = CV15. Can you use superposition to solve
circuits containing this capacitor ? What about KCL and KVL ? (c) For a linear circuit without any independent sources, when v(t) = 105in(3t + pi/6) is
applied, a current i(t) = 270sin(3t + 3pi/8) results in steady state. What is the complex
impedance ofthis circuit? Give one R/L/C realization of this impedance. (at) W \ ”h w ' , \ ‘ imam".
“0) Y‘” can NOT use superpos‘rﬁon, Same A [WWWW “3 “0" \(GUK CA‘I’V M36 KCL. CLMCL KVL/o W) Va 10412; } lemoﬁg 3 mg
I at: L, a: 3?” «a» 3“ u;% at Q ,m i 15:“
Z?» 1: :1 3’1 iii/“t “" g} cm ”m if: :14 a» {9a 1% {Li/'31 J‘j
Q i a ,M
W: Wlﬁ .1) £5?” SLR 31"th {315294; ”(1“
EL. L“; :39: ...
View
Full Document
 Spring '07
 Chang

Click to edit the document details