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In the circuit below, the voltage source is a 5 V power supply. The inductor is

ID: 1430997 • Letter: I

Question

In the circuit below, the voltage source is a 5 V power supply. The inductor is considered to be ideal (it has no internal resistance) and large (say, 1 henry). R_1= 5 Ohm and R_2^=10 Ohm Assume the switch had been closed for a long time. Use Kirchhoff s rules to determine the voltage drop across R_2. At the instant just after the switch was opened, what is the current though the inductor? At the instant just after the switch was opened, what is the current though R_2? At the instant just after the switch was opened, what the current though R_1? At the instant just after the switch was opened, what the magnitude of the voltage drop across R_1? Suppose R_1 were replaced with a 50 Ohm resistor and the switch had been closed for a long time. What would the magnitude of the voltage drop across R_1 be just after the instant in which the switch was opened?

Explanation / Answer

V2 = V*(1-e^-tR/L)


t = infinity

V2 = V = 5 V

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IL = V/R2 = 5/10 = 0.5 A

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I2 = V/(R1+R2) = 5/15 = 1/3 = 0.33 A


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I1 = I1 = 0.33 A

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V1 = I1*R1 = 1.67 V

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V1 = I1*R1 = (V/(R1+R2))*R1 = (5/15)*50 = 16.67 V