The voltage v(t) shown in Fig. P7.85a is given by the graph shown in Fig. P7.85b. IfIL (0) = 0, what is i(t = 6 s)? Given L = 14 H and R = 10 Q.i(t)v(1)LHi(t)v(t)10 Vtu52-10 V(b)᾽R Ωt(s)

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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

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8. Evaluate the following convolution cos3t 8(t - 8) a. cos(24) b. 8cos (24) cos[3(t-8)] d. Sa(8w) 9. Evaluate the following convolution f(t) • 8(t−9) + f(t) • 8(t + 9) with F(w) is f(t)'s FT a. F(w+9)+F(@-9) b. f(t+9) + f(t-9) c. F(t+9) + F(t - 9) d. 2f(t-9)
The switch in the circuit in Fig. P7.4 has been in posi- tion 1 for a long time. At t = 0, the switch moves instantaneously to position 2. Find vo(t) for t≥ 0+. Figure P7.4 13Ω 1 12 Ω 36 mH w www t=0 2 (+ 90 V 3022 60% 802 Ω
Q6) a)Match the signals below with the candidate sketches below (i.e., you are supposed to choose which graph below corresponds to the signal below). Note: u(t) is the unit step function; i.e., u(t) = 1 for t>=0s and u(t) = 0 for t<0s. X₁(t) = u(t+1)-u(t-3) X₂(t) = u(t-3)-u(t-4) X3(t) = u(t+1)-u(t-1) d(t) = x₁(-t) e(t) = 2 x₁(t-1.5) f(t) = x₁(2t + 1.5) g(t) = x4(2t) h(t) = x3(t/2) i(t) = 2 x3(t) j(t) = -0.5 x₂(t) k(t) = x₁(1.5 t + 2.5) y(t) = 2 x₁(-t/2+1.5) 432 1 0 -1 4321OT 0 -1 -6-4-20 2 4 6 (1) 4321OT 0 -1 -6-4-20 246 (5) n -6-4-20 2 4 6 (9) 43210 3 -1 4321OT -1 43210 -6-4-20246 (2) -1 -6-4-20246 (6) 2 -6-4-20246 (10) 4321OT 0 4321OT -6-4-2 0 2 4 6 (3) -6-4-2 0 2 4 6 (7) 4321O -6-4-20246 (11) 4321OT 0 -1 -6-4-20 2 4 6 (4) 4321OT -6-4-20 2 4 6 (8) 43210 3 2 0 -6-4-20 246 (12)
7.2.3 Use the differential equation approach to find i,(t) for t> 0 in the network in Fig. P7.2.3. 12 V(+ 6 ΚΩ ww i(t) X=0 ΕΚΩ 100 μF 12 ΚΩ t = 0 FIGURE P7.2.3
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Using the step-by-step method, find the value of the output current, io(t), just before the switch closes (at t=0-) for the network in Fig. P7.81. Given lç, 11 mA, R1 = 3 kQ, R2 = 6 kQ, R3 = 6 k0, and R4 = 7 KQ. 1 Is mA R1 ΚΩ Figure P7.81 R3 ΚΩ R2K t = 0 Los i,(1) 2 mH SR4K! Notes on entering solution: • Enter your solution in milliAmps(mA) • Enter your solution to the nearest tenth of a mA . Do not include units in your answer .ex. 5mA is entered as 5.0 2iA
7.2.16 Use the step-by-step method to find uo(t) for t> 0 in the circuit in Fig. P7.2.16. + v(t) 18V ( Μ 3 ΚΩ 6ΚΩ 200 με FIGURE P7.2.16 ww 3 ΚΩ t = 0 +1 σν
7.5-3 Determine and sketch the capacitor voltage vc(t) for t> 0 in the circuit of Fig. P7.5-3. Answer: 10-5e-t/6 V. t=0 + 5A A ΖΩΣ 3F vc(t) <252 Figure P7.5-3

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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 = 14 H and R = 10 Q. i(t) v(1) LH i(t) v(t) 10 V tu 5 2 -10 V (b) ᾽R Ω t(s)