Question

(1 point)

If Vs(t)=91e^(−42t)mV and is(0)= -5 mA, find is(t) for t>0

8 mF 20 mH 24 mF

is(t) =  mA

Answer 1

Given that From the circuit diagram C = 8 times 24/8 + 24 = 192/32 = 6 and also L = 20 mH = 20 times 10^-3 H Given data V_s (t) = 91 e^-42 t mv and i_s (0) = -5 mA \r\n i_s = i_c + i_L (1) = c dV_s/dt + 1/L integral V_s dt V_s = 91 e^-42 t mv i_c = c dv_s/dt = 6 times d/dt (91 e^-42 t) mu A = 6 times 91 times -42 times e^-42 t mu A = 6 times 91 times -42 times e^-42 t mu A = -22.93 e^-42 t mA i_L = 1/L integral V_s dt = 1/20 times 10^-3 integral 91 e^-42 t dt = 91/20 times 10^-3 integral e^-42 t dt = 4,550 times 1/-42 times e^-42 t + k = -108.33 e^-42 t + k \r\n Therefore i_s (0) = -5 mA Therefore From (1) -5 mA

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