The elements in the circuit shown above have the following values: E = 102 V, L

Question

The elements in the circuit shown above have the following values: E = 102 V, L = 2 H, R1 = 14Ohm, R2 = 25 Ohm, and R3 = 30Ohm. After the switch has been open for a long time, it is closed at t = 0. Calculate the current through R2 at t= 0+. Calculate the potential drop across the inductor at t= 0+. At what rate is the current through the inductor increasing at t = 0+? After a long time, what is the current through R2? After being closed for a long time, the switch is opened at time T, What is the current through R2 immediately afterward.

Explanation / Answer

a) at t=0
inductor is opened.

Rnet = R1+R2 = 39 ohm

i = v/Rnet = 102/39 = 2.615 A

b)potential drop across inductor = potential drop across R2

= R2*V/(R1+R2) = 25*102/(39 = 65.38 volts

c) VL = L*di/dt

di/dt = 65.38/2 = 32.69 A/s

d) after a long time inductor is short circuited

Rnet = R1 + R2*R3/(R2+R3) = 14 + 25*30/(25+30) = 27.64 ohm

I = V/Rnet = 102/27.64 = 3.69 A

current through inductor = R2*I/(R2+R3) = 3.69*25/(25+30) = 1.678 A

e) zero

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