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1 Circuit Exam 2 Spring 206. /25 2. /30 3. /25 4. /20 Total /00 Name Pleae write your name at the top of every page! Note: ) If you are tuck on one part of the problem, chooe reaonable value on the following part to receive partial credit 2) You don t need to implify all your numerical calculation. For example, you can leave quare root term in radical form.

2 Name ) Short Anwer (25 point) Quetion (5 point) Which of the following circuit can have a econd order repone? Circle all that apply. R R2 R3 R R3 R4 V C C2 C3 V R2 R5 C 0 a) b) 0 R R3 R4 C V 0 R2 c) 2 L C U R OUT V in - - OPAMP 0 V o - Both op amp are powered. Relationhip between Vin and Vo. d) V in v - o - v o - a. c. e. Both op amp are powered. Relationhip between Vin and Vo. e) Jut for fun (and extra credit 7 pt): Derive the relationhip between Vin and Vo for e) to prove the order of the repone. There i no partial credit. d 2 V o dt 2 τ τ 2 dv 0 dt τ τ 2 V in V 0 R C R 2 C where τ n R n C n 2 2

3 Quetion 2 (5 point) Name R k TCLOSE 2E-4 2 U V C R2 In the above circuit, the voltage ource turn on at t0, V20u(t). At t 2E-4 (), witch U cloe. (Only the value of R i given but the variable C and R2 will be determined below). a) Determine C uch that the voltage acro C i 3.5 V when the witch cloe. 0 Before the witch cloe, the circuit i the familiar RC circuit with votlage acro the capacitor defnied a t V c ( t) A exp A E3C 2 The inital condition at t 0 i Vc(t) 0 AA2 The teady tate condition a t approache i Vc 20 0 A2 The two condition lead to A -20 and A2 20 t V c ( t) 20exp 20 E3C At t 0.2 m, Vc3.5V. Putting thoe value into the above equation exp C ln( 0.825) C F C [F] b) Determine R2 uch that the teady tate (t>>>0) voltage acro the capaictor i 7.5V. At teady tate, the capacitor i open and the circuit i a votlage divider R 2 V c V R2 V k R 2 R 2 600Ω R R 2 R2 [ohm] 3

4 Name Quetion 3 (5 point) The diagram to the hown the voltage pike of an RL circuit when meauring acro the inductor. a) Draw the correponding current through the inductor. In the diagram below it. (Note: Pay attention to time). b) What i the caue of the voltage pike when meauring the voltage acro the inductor in an RL circuit? It take time for current to flow due to back EMF effect. Therefore with little to no current the voltage pike or rie to a maximum value until back EMF reduce. - if they DON'T mention the magnific field around the changing current cauing EMF induced back into the inductor 4

5 Name 2) Firt order circuit (30 point) R 20k 50V V R2 0k R3 20k 2 U TCLOSE 0 I R3 C 0. uf The witch in the circuit in the above figure ha been open for a long time. It cloe at t 0. 2.: Find the current I R3 (t) fort > 0 uing differential equation (5 point). Thi can be olved two way.. Ue a ource converion and then find the current through R3 uing a current divider. 2. Solve for the voltage acro the capacitor C firt uing Vth. It i in parallel with R3 after the witch cloe. Then find the current uing ohm' law. Solving uing method #2. R Need initial condition : 20kΩ R 2 : 0kΩ R 3 : 20kΩ V : 50V C : 0.μF R 2 R 3 V C0- : V R R 2 R 30V 3 R R 3 V tht0 : V C0inf 25V R tht0 : R R 3 τ : R tht0 C m 0 3 τ V C ( t) A A 2 e 000t Final condition Vc( ) R 3 V C0inf : V R R 3 0kΩ 25 V 25 A A 2 0 A A A 2e V t 0 C ( t) 25 5e 000t Initial condition: Vc(0) 30 A A 2 I R3 ( t) V R3 ( t) e 000t ma 20kΩ 5

6 Name 2.2: Determine the -domain expreion for the current through R 3, I R3 (). Your expreion hould be in the form N()/D() where both the numerator and denominator expreion are polynomial. (The expreion hould be partial fraction expreion-ready!) Draw the circuit for full credit. (0 pt) Thi problem can alo be olved three way:. Source converion for V a. With initial condition voltage ource for C b. With initial condition current ource for C 2. Thevenin eq. find Vc then ohm' law S-domain circuit Source converion with initial condition voltage ource Source converion with intial condition current ource R 20k 50/ V R2 0k R3 20k 2 U TCLOSE 0 C /0.uF* 30/ I 2.5E-3/ R 20k R3 20k C E7/ Ic(0) 3E-6 I 2.5E-3/ 3E-6 R3 0k RC E7/(500) Thevenin eq. finding Vc (from part a) and then ue ohm' law Thi may be eaiet. Rth 0k 25/ C /0.E-6 E7/ 30/ 6

7 Uing ource converion with intial condition current ource V C ( t) V R3 ( t) Uing ohm' law 50 I V : R H I C0 : 30C F RC RC Ue current divider with norton circuit to find RC ( 500) ( 500) ( ) ( 000).50 3 ( 000) ( )

8 Thevenin eq. finding Vc (from part a) and then ue ohm' law To find Vc ue voltage divider and don't forget initial condition voltage ource! thi i Vc but you mut divide by R3 or 20k ( 000) : Ue partial fraction expanion to find the time domain expreion for the current through R 3 for t > 0 (5 pt). From ource converion with intial condition current ource A A For A.25 ( 000) 0.25A A For A e 000t ma 8

9 From Thevenin eq. finding Vc (from part a) and then ue ohm' law partial fraction expanion for 0.25 ( 000) A A A A For A For A e 000t ma Curiou, which method do you prefer for thi problem -domain or diff. eq. time domain? (0 pt) 9

10 Name 3) Second order circuit/differential Equation (25 pt) R R2 9V V 250 ohm 2 50 ohm U TCLOSE 0 L H C 4uF I w 2 I L - The witch in the figure above ha been open for a long time and i cloed at t 0. R p3 : 250Ω R 2p3 : 50Ω L p3 : H V p3 : 9V C p3 : 4μF 3.: Find the intial condition at t 0. (Should include both capacitor and inductor value). V C30- : 0 Recognize that the inductor i horted! V p3 I L0- : R p3 R 2p3 30mA R R2 and L are alo in erie with ource. V C (0) I L (0) [V] [A or ma] 3.2: Find the inductor current for t > 0. Thi i an RLC parallel circuit where the input current i 0! d 2 i L di L 2 2 dt 2 2α ω dt o il ω o in ω o : 500 L p3 C p3 α : 2R 2p3 C p R2 50 ohm 2 L H C 4uF 0

11 : 2 : α α 2 2 ω o 50.5 α α 2 2 ω o I L ( t) K e 50.5t K 2 e 4949t K 3 Find coefficient I L ( 0) di L ( 0) dt I L ( ) I L ( ) K 3 K 3 0mA there i no ource connected a t goe to infinity... I L ( 0) K K 2 30mA di L ( 0) 50.5K dt 4949K 2 0 M : C m : 0 M C m I L ( t) 30.3e 50.5t 0.309e 4949t ma check t check t goe to 0mA 3.3: Find the capacitor voltage for t > 0. The inductor and capactor are in parallel o VCVLLdiL/dt ( ) V C ( t) L dil.53e 50.5t.529e 4949t dt V C ( t).53e 50.5t.529e 4949t V mA.53A mA.529A

12 3.4: Find the current through the witch for t > 0. (Extra Credit 3 pt) Current through witch i the current coming down from R (250 Ω) on the left hand ide of the witch, minu the current coming up through R2 (50 Ω) on the right hand ide of the witch. I R I R2 I w I R2 I R I w 0 I R 0V I R2 9V 0V R p3 36mA I R 36mA V C ( t) 0 50 I w t 0.03e 50.5t 0.3e 4949t A I R2 ( ) e 50.5t 3e 4949t ma ) Second Order Laplace Circuit (20pt) R L 2 E-6 V C E-2 0 In the circuit above, the voltage ource i 5V for t<0 and 0 V for t >0. L : 0 6 H C : 0 2 F V t0 : 0V V t0- : 5V 2

13 4.: Draw the -domain equivalent circuit. Include the intial condition in your -domain circuit. Label your component value uing ymbolic notation (i.e. L). R L 2 Li(0-) V() /C Vc(0-)/ I L ( 0-) I L ( 0 ) 0 DC teady tate for component o capacitor i open o the current in erie circuit in 0. V C ( 0-) V C ( 0 ) 5V The voltage acro the capacitor mut be 5V, the value of the ource at t<0. R L 2 L*I(0-) 0 *E-6 C E2/ V 5/ 0 3

14 4.2: Uing impedance, determine the voltage acro C, VC(). Ue ymbolic notation in your expreion. (You do not need to inert number) V C ( ) C R C L V ( ) LI( 0-) V C ( 0-) Thi i a voltage divider. The ( voltage ) ource mut be accounted for. V C 0- V C ( ) L C 2 R L L C V ( ) LI( 0-) V C ( 0-) V C ( 0-) You can inert intial condition -domain value here. If we were looking for the V c(t), then you'd do partial fraction expanion and the invere laplace tranform after the following. V C ( ) L C 2 R L L C V C ( ) L C 2 R L L C : Determine the current through the capacitor, I C (). I C ( ) V C ( ) C V C ( 0-) Uing ohm' law where current i related to capacitor impedance Zc/C by Vc Zc Ic The voltage ource i ubtracted from Vc to meaure current through capacitor and ue ohm' law. 4

15 I C ( ) L 2 R L L C V ( ) L I( 0-) V C ( 0-) You can inert intial condition value here but it i not neceary to get full credit for the problem. I C ( ) L 2 R L L C I C ( ) L 2 R L L C 5 5 I C ( ) L 2 R L L C 4.4: Uing the equation from part c. for R 5kΩ, determine the current through the capacitor a a function of time for t>0. I C ( ) 50 6 ( ) R : 5kΩ L C L H R L 5

16 Partial fraction expanion 50 6 ( ) ( ) A ( ) A 2 ( ) For A For A ( ) ( ) I C ( ) (.0 3 ) ( ) (. 0 3 ) ( ) I C ( t).0 3 e t.0 3 e t 6

17 Name Survey for Undergraduate Studie Committee (3 pt extra credit jut for anwering) What i your major? CHECK or CIRCLE all that apply: Electrical Engineering Computer and Sytem Engineering Computer Science Applied Phyic Mechanical Engineering Biomedical Engineering Other Have you applied for an internhip thi ummer? YES NO Have you already accepted an internhip office for thi ummer? YES NO Pleae ue thi pace for any addition calcualtion. Label the problem clearly. 7

18 Name Pleae ue thi pace for any addition calcualtion. Label the problem clearly. 8