A 45 V battery with negligible internalresistance, a 50 resistor and a 1.0-mH in

Question

A 45 V battery with negligible internalresistance, a 50 resistor and a 1.0-mH inductor with negligible resistance areconnected in series with an open switch. The switch is closedat t=0.
(a) What is the maximum current in the circuit and when does itoccur? (Ans: 0.90 A ,after a long time) (b) What is the maximum energy stored in theinductor? (Ans: 4.1x10^-4 J) (c) At what time t will the current reach one-half of its maximum value? (Ans: 1.4x10^ -5 s) A 45 V battery with negligible internalresistance, a 50 resistor and a 1.0-mH inductor with negligible resistance areconnected in series with an open switch. The switch is closedat t=0.
(a) What is the maximum current in the circuit and when does itoccur? (Ans: 0.90 A ,after a long time) (b) What is the maximum energy stored in theinductor? (Ans: 4.1x10^-4 J) (c) At what time t will the current reach one-half of its maximum value? (Ans: 1.4x10^ -5 s) (c) At what time t will the current reach one-half of its maximum value? (Ans: 1.4x10^ -5 s)

Explanation / Answer

V = 45 V, R = 50 , L = 1 mH I = (V/R)(1 - e-Rt/L) a) t --> , I = Imax = V/R = 0.90 A b) LImax2/2 = 4.05*10-4 J c) I = Imax/2 = Imax(1 -e-Rt/L) e-Rt/L = 1/2 t = L*ln(2)/R = 1.39*10-5 s

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