None The voltage v(t) shown in Fig. P7.85a is given by the graph shown in Fig. P7.85b. Ifio(0) = OA, how much energy is stored in the inductor at t = 3 s? Given L = 5 H and R =70.i(t)D(I) (+LHRO▼i₁(1)10 V-10 VD(1)521(S)(b)Figure P7.85Notes on entering solution:• Enter your solution in Joules(J)• Enter your solution to 2 decimal points• Do not include units in your answer• ex. 5 J is entered as 5.00

Step 1:Given information :Initial current is 0.i0=0AStep 2:First determine the current through the…

Answered: The voltage v(t) shown in Fig. P7.85a…

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

Section: Chapter Questions

Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...

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7.76 There is no energy stored in the capacitor in the PSPICE circuit in Fig. P7.76 when switch 1 closes at t = 0. MULTISIM Switch 2 closes 10 microseconds later. Find v.(t) for %3D t2 0. Figure P7.76 16 k2 1 = 0 t = 10 us 20 nF 5 mA 4 k2 30 V
3. The switch in Fig. P7.65 has been position a for a long time. At t = 0, it moves instantaneously to position b. For t20+, find i.(t), v1(t), v2(t), Vo(t). 2 kN 7.5 kN t = 0 + + 50 V 0.3 µF V1 100 V + Vo 0.9 µF - V2
I need the answer as soon as possible
The voltage v(t) shown in Fig. P7.85a is given by the graph shown in Fig. P7.85b. If iL(0) = OA, how much energy is stored in the inductor at t=3s? Given L=11H and R=170. D(1) (+ v(1) 10 V i(1) -10 V Figure P7.85 LH 2 (b) i₂(1) Notes on entering solution: RO [ (S) Enter your solution in Joules(J) • Enter your solution to 2 decimal points Do not include units in your answer • ex. 5 J is entered as 5.00
If you need use to Multisim.
The voltage v(t) shown in Fig. P7.85a is given by the graph shown in Fig. P7.85b. If IL (0) = 0A, how much energy is stored in the inductor at t = 3 s? Given L = 13 H and R - = 60. v(1) i(t) -10 V LH (a) i, (t) v(t) 10 V tu 5 2 3RΩ t (s)
In the circuit in Fig. P7.29 the voltage and current expressions are v = 72e 500 V, t≥ 0; i = 9e-500 mA, t = 0+. Find a) R. b) C. c) 7 (in milliseconds). d) the initial energy stored in the capacitor. how many microseconds it takes to dissipate 68% of the initial energy stored in the capacitor. Figure P7.29 - + v R R ww
P7.2-18+ The input to the circuit shown in Figure P 7.2-18 is the current i(t) = 3e-25 A fort > 0 The output is the voltage v(t) = 9.6e-25t+0.4 V fort > 0 The initial capacitor voltage is vc (0) = -2 V. Determine the values of the capacitance C and resistance R. Vc(t) www R + i(t) v(t) C
The voltage v(t) shown in Fig. P7.85a is given by the graph shown in Fig. P7.85b. If iL (0) = 0, what is i(t = 6 s)? Given L = 11 H and R = 50. D(1) i(t) LH Figure P7.85 (a) v(1) 10 V t 5 2 -10 V i, (1) (b) Notes on entering solution: • Enter your solution in Amperes(A) • Enter your solution to 2 decimal places • Do not include units in your answer .ex. 5 A is entered as 5.00 • Pay attention to signs RO r(s)
The switch in the circuit seen in fig. p7.52 has been in position a for a long time. At t=0, the switch moves instantaneously to the position b. For t >= 0+, find i0(t).
The voltage v(t) shown in Fig. P7.85a is given by the graph shown in Fig. P7.85b. If iL(0) = 0A, how much energy is stored in the inductor at t = 3 s? Given L = 7 H and R = 30. i(t) D(1) LH Figure P7.85 (a) D(1) 10 V tu 5 2 -10 V i, (t) (b) Notes on entering solution: • Enter your solution in Joules(J) • Enter your solution to 2 decimal points . Do not include units in your answer .ex. 5 J is entered as 5.00 R92 [(S)
7.62 For the RL circuit shown in the image below, i (0) = 0. If R1 = 1 S2 and R2 = 5 2, determine the current through the inductor, i, when t= 1.22 s. Please pay attention: the numbers may change since they are randomized. Your answer must include 3 places after the decimal point, and the proper Sl unit. R1 R2 u(t – 1) V(+ 2 H u(t) V Your Answer: Answer units ll

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