Question

Part A The action of the two switches in the circuit seen in the figure is as follows. For t<0, switch 1 is in position a and switch 2 is open. This state has existed for a long time. At t 0, switch 1 moves instantaneously from position a to position b, while switch 2 remains open. Ten milliseconds after switch 1 operates, switch 2 closes, remains closed for 10 ms and then opens. (Figure 1) Find vo(t) 44 ms after switch 1 moves to position b Express your answer with the appropriate units 0.6139 SubmitPre ious Ans Request Answen gure 1 of 1 X Incorrect; Try Again; 3 attempts remaining Provide Feedback 5 n 0+ 10 ms 15 A 1012 350 mH 20 2

Answer 1

Now, At t = 0^-: - the system is in steady state Therefore inductor acts as short circuited i_L (0^-) = I_s times 10/(10 + 5) = 10/15 times 15 = 10A (Using current divider Rule) As, [i_L (0^-) = i_L (0^+) = 15A] For t = 0 to 10 m sec Implies 's1' At position (b) It is a source free R-L circuit having time constant- tau = L_eq/R_eq, Here R_eq = 5 Ohm L_eq = 50 mH tau = 50 mH/5 Ohm = 0.01 sec. Therefore i_L (t) = i_L (0) e^-t/tau Implies 10 e^-100t Ampere 0 lessthanorequalto t (m sec) lessthanorequalto 10 Implies At t = 10 (m sec) Implies i_L (10) = 10 times e^-100 times 10/1000 = 10 times e^-1 = 0.3678 times 10 = 3.678 Ampere. \r\n Now, S2 closes for t > 10^+ 2 keeps conducts for next 10 seconds Now, new time constant Implies R_eq = R_1 Vertical-bar Vertical-bar R_2 = 5 Vertical-bar Vertical-bar 20 = 20 times 5/25 = 4 Ohm Therefore tau_new = L/R_eq = 50 mH/4 Ohm = 0.0125 sec tau^-1_new = 80 Now, [i (t) for (10 lessthanorequalto t