Figure 15 4 11 The steady state output equals the average value of the input

Figure 15 4 11 the steady state output equals the

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Figure 15-4 11. The steady-state output equals the average value of the input which is 15 V with a small ripple. 12. See Figure 15-5. Figure 15-5 13. τ = (1.0 k Ω )(1 μ F) = 1 ms See Figure 15-6. Steady-state is reached in 5 ms . Figure 15-6 SECTION 15-4 Response of RC Differentiators to a Single Pulse Full file at
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154 14. τ = (1.0 k Ω )(1 μ F) = 1 ms See Figure 15-7. Figure 15-7 15. The output voltage is approximately the same wave shape as the input voltage but with an average value of 0 V . 16. τ = Ω k 10 mH 10 = 1 μ s 5 τ = 5 μ s V out ( max ) = 0.632(8 V) = 5.06 V See Figure 15-8. 17. τ = Ω k 0 . 1 mH 50 = 50 μ s 5 τ = 250 μ s V out (max) = 12 V See Figure 15-9. SECTION 15-5 Response of RC Differentiators to Repetitive Pulses SECTION 15-6 Response of RL Integrators to Pulse Inputs Figure 15-8 Figure 15-9 Full file at
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155 18. (a) τ = 100 H 2.2 k μ Ω = 45.4 ns (b) At end of pulse, v out = (10 V) e 2.2 = 1.11 V See Figure 15-10. 19. τ = 100 H 2.2 k μ Ω = 45.4 ns 5 τ = 227 ns See Figure 15-11. 20. τ = RC = (22 k Ω )(0.001 μ F) = 22 μ s v B = V F (1 e t/RC ) = 10 V(1 e 440 μ s/22 μ s ) = 10.0 V 21. The output of the integrator is ideally a dc level which equals the average value of the input signal, 6 V in this case. 22. τ = RC = (3.3 k Ω )(0.22 μ F) = 726 μ s 5 τ = 5 RC = 5(726 μ s) = 3.63 ms (b) Since the output looks like the input, the capacitor must be open or the resistor shorted because there is no charging time. (c) The zero output could be caused by an open resistor or a shorted capacitor. 23. τ = 726 μ s; 5 τ = 5(726 μ s) = 3.63 ms (b) No fault. (c) Open capacitor or shorted resistor. SECTION 15-7 Response of RL Differentiators to Pulse Inputs SECTION 15-8 Applications SECTION 15-9 Troubleshooting Figure 15-10 Figure 15-11 100 ns 327 ns 8.89 V 250 600 850 Full file at
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156 ADVANCED PROBLEMS 24. (a) Looking from the source and capacitor, R tot = Ω Ω + Ω Ω k 4.2 ) k 0 . 1 k 0 . 1 )( k 2 . 2 ( = 1.05 k Ω τ = R tot C = (1.05 k Ω )(560 pF) = 588 ns = 0.588 s (b) See Figure 15-12. 2.6 s/0.588 s 10 = 120 mV v e = Figure 15-12 25. (a) Looking from the capacitor, the Thevenin resistance is 5 k Ω . τ = (5 k Ω )(4.7 μ F) = 23.5 ms ; 5 τ = 5(23.5 ms) = 118 ms (b) See Figure 15-13. Figure 15-13 26. L tot = 8 μ H + 4 μ H = 12 μ H R tot = Ω Ω Ω k 7 . 24 ) k 7 . 14 )( k 10 ( = 5.95 k Ω τ = Ω μ = k 95 . 5 H 12 tot tot R L = 2.02 ns This circuit is an integrator. Full file at
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157 27. v = V F (1 e t / τ ) 2.5 = 5(1 e t / τ ) 2.5 = 5 5 e t / τ 5 e t / τ = 5 2.5 e t / τ = 5 5 . 2 = 0.5 ln e t / τ = ln 0.5 τ t = 0.693 τ = 693 . 0 s 1 693 . 0 = t = 1.44 s 28. The scope display is correct. See Figure 15-14. Figure 15-14 29. Capacitor is open. 30. R 2 is open. 31. No fault 32. L 1 is open. Multisim Troubleshooting Problems Full file at
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158 1. Two types of semiconductor materials are silicon and germanium .
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