Question

Values of v()/V = -1/ IR + (1)/V V R T 21 31 0.36788 0.13534 0.04979 0.01832 0.00674 ST Figure 1: A source free RC circuit. Table 1: values of the exponential term for different t Observe from the table 1 that the smaller the time constant, the more rapidly the voltage decreases, that is, the faster the response. This is illustrated in Fig. 2. A circuit with a small time constant gives a fast response in that it reaches the steady state (or final state) quickly due to quick dissipation of energy stored, whereas a circuit with a large time constant gives a slow response because it takes longer to reach steady state. At any rate, whether the time constant is small or large, the circuit reaches steady state in five-time constants. TOS O Figure 2: plot of v/V0 = e for various values of the time constant. 1. Suppose V= 1 V, R = 1 kN and L = 1 uF in the figure 1. Calculate v(t) for t>O if the switch is closed for a very long time. Then: a) Suppose switch closes at t=0 and opens at tel ms. Plot va(t) in terms of time for (<t<2 ms. b) Suppose switch closes at t=0 and opens at t=5 ms. Plot vat) in terms of time for 0<t<10 ms. c) Suppose switch closes at t=0 and opens at t=10 ms. Plot vat) in terms of time for (<t<20 ms. R t=0 + Us C VC Figure 3: Circuit # 1

Answer 1

Lir Ralke Velt) + C=1 Mf It=o switch closed at t=lms ve Lt) = vs (1-794) -ims = Vs (in e ) Ne (t) = Vs (1 e ims T=Rc = (1K2) (luf) =ims 0.632 v Ne= NS(1-0.3678) A Velt) 0 '632v .632v t> ims Ims at t=2ms -2ms & Ne (t) = Vs line ims - ) vs (1-22) tims ve Velt) = 0.632 =0.864 v = 0.632 e -0.08555 0.632v 0.0855V Ims 2ms

att 5ms b) Similarly, Velt) 111-ės) = lv Velt) = Ne R c = 100 . at t=1oms soms ty sms (.) Velt) Ne (1-etc) at t= loms Velt) le nou At t=20ms e the Velt) = veo 20ms - le tms Velt) W 2008 10m