Electric Circuits

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vsff

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everything is explane on the pics. if you need any informtion text me

there are two home work.

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Exam #4 – Extra credit • Falstad Simulation Problem #3 http://www.falstad.com/circuit/ • Use a SPDT switch .] . Use Stacked “View in Scope” for Ve(t) and Ic(t) • Falstad Simulation Problem #4 http://www.falstad.com/circuit/ . . Use a switch • Use Stacked “View in Scope” for Il(t) and Vl(t) Exam #1 - Problem #3 3. (25 pts. total) Using any method a. Find the voltage Vc(t) across the capacitor for t> 0 (i.e. the switch moved to terminal B @ t=0) (hint: First find the initial condition at t < 0 for Vc(t) with the switch at terminal A) (15.0 pts) b. Sketch Vc(t), show t = 0, t = T, and t = 51 ( 200 ). Label your axes. ( 2.5 pts.) ( 5.0 pts.) c. Find the capacitor current Ic(t) for t > 0. d. Sketch Ic(t), show t= 0, t = t, and t = 5t ( 200 ). Label your axes. ( 2.5 pts.) B t=0 592 592 W + 10V .02 F Vc(t) 1 Ic(t) Exam #1 - Problem #4 ( 20 pts.) 4. (30 pts. total) Using the Differential Approach a. Find the current I (t) for t> 0.(i.e. the switch opens @t=0) (hint: First find the initial condition at t< 0 for iz(t) with the switch closed) b. Sketch Il(t), show t= 0, t = T, and t = 5t ( 200). Label your axes. c. Find the voltage V.(t) across the inductor for t > 0 d. Sketch V (t), show t = 0, t = t, and t = 51 ( 200 ). Label your axes. (2.5 pts.) ( 5.0 pts.) (2.5 pts.) 4 k22 M 4 k22 M iz(t) 8 V + + Switch Opens t=0 .5 H V_(t) Initial Condition: IL(0-) = i_(0+) = 0 All the current flows through the switch, no current reaches the inductor Exam #2 - Prob. #4 Solutions 1 4a. Find the equation for Vo in terms of Vin for the amplifier circuit below. (10 pts ) (assume ideal properties for the Op-Amp & the rules when there is feedback at the inverting terminal) 4b. If VIN = 2V, solve for Vo. ( 5 pts ) 4c. Determine the gain of the amplifier circuit below. ( 5 pts ) 2 k92 M If you simulate 1 ks 2 this circuit for M extra credit note VIN the polarity on the OpAmp inputs +10V V 1 I O 1 -10V + V Vo w 1 k 2 1 k2 ble KCL @ V+ 1. + 12 = 14 = 0 KCL @ V Iz = 14 + + = 0 Vo-V 2K VIN-V+ IK Vo-V. IK V-0 IK Vo-V + 2 VIN - 2 V = 0 Vo-V. = V-0 Vo = 3 V+ - 2 VIN (1) .5 Vo = V. (2) (a) Vo = 3 ( .5 Vo ) - 2 VIN -.5 Vo = - 2 VIN V =V+ (b) Vo = 4 VIN Vo = 4 (2V) (c) Gain = Vo/Vin = 4 Gain 4 Vo = 8V Vo = 4 VIN >
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