samplefinal_sol

samplefinal_sol - EE10 Sample Final Exam UCLA Department of...

Info icon This preview shows pages 1–7. Sign up to view the full content.

View Full Document Right Arrow Icon
Image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 2
Image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 4
Image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 6
Image of page 7
This is the end of the preview. Sign up to access the rest of the 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} $272»— itzzfi. iii: Ellis/14A W Wm. wag; CK if" “a »« f7?“ .2? 3%; AWE? \i 33 (b) Find the Thevenin equivalent of the simple transformer circuit below. Q2. (5 + 3 +2 points) Assume w = 104 rad/sec. (a) Use source transformation to obtain the current IO. Express your answer in milliamps, both in the phasor notation (rectangular form) and in the time domain. (b) Draw a phasor diagram showing the voltage across 30ohm resistor, 4mH inductor, 5uF capacitor. (c) Graphically solve (roughly) for current through the 20 ohm resistor. You don’t need to calculate the exact value of the current. Just show it graphically. 4*siuuat) V 3 ME 2 it}? ‘ law—T 23 Q; C 03 65 V icy-10:) \ ‘5th 10-0-03 4. WSMGB—j l; 10 l+l5V —1 +1" ‘ 1Q: fi’ 3 a 40 *AGQ WA e "56. éLlSSQMA : SAESM LiooootHBSQ) ‘EO‘JLci. \xl 1;» farm, r \T—W /> mm a ”W \k— 1‘)“; VS“? 3"». {We um” p3 O t Q3. (8 + 2 points) Consider the circuit with the following parameters: VSl =12V,‘VSZ =1OV;RS =250,‘R=1000;C=32nF;L=500uH;RL =SOQ Assume DC steady—state conditions at t <0. (a) Write and solve the differential equation for the capacitor voltage vc as a function of time at t >0. (b) What is the energy stored in the inductor at steady state ? What about the capacitor ? VSZ W®1® (PVC +300000i99-L613X1010V“ =6 X1” cw dt' ' C" ”K 0 gun. “- *[5‘0000 fr; 3.000605 Vow (”‘33 ’7' 8,16% (kg (05 1000063 *- {él 811/! 10000013) V6C13'3 Vamtt) rvcfim): e’lsmt ([4. (05 2.000002: Wls‘maooaoct) '1 0 Maw {41 —(0= ~18 Kw -8 M03 alt '2 " [50000 {g t‘ looooolfil “-‘ 9000000 l42 ‘— lq » —~( "Vctfl 2 € smooté —gco$moooo£ +i0tS‘nM1ooooot3‘lO / "C70 W w W (b) “Cr—00 ”é i“ *x , vc "TR *waho J VCCofi32-nlO 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'i-oxfll :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

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern