A 174-V battery, an inductor, and a resistor are connected in series as shown in

Question

A 174-V battery, an inductor, and a resistor are connected in series as shown in the diagram below. A two-way switch makes it possible to include or exclude the battery. The switch that had been in position 1 for a long time is suddenly moved to position 2. (Enter your answers to at least two decimal places.)

(a) What is the voltage across the resistor at the end of four time constants? 3.19 V is the correct answer

Now I need help with B

(b) At this time, what is the voltage across the inductor? V (What is the total emf in the circuit when the switch is in position 2? )

Explanation / Answer

1)

when the switch is position 1 for a long time , the current through the inductor is in steady state

when the switch is moved to position 2 ,

the initial current in the inductor , Io = 174/R

let the time constant of the circuit is T

Now, for time t = 4 *T

current through the inductor ,

I = Io * e^(-t/T)

I = (174/R )* e^(-4)

voltage across the resistance = I*R

voltage across the resistance = (174/R )* e^(-4) * R

voltage across the resistance = 174/e^4

voltage across the resistance = 3.19 V

voltage across the resistance after 4 time constants is 3.19 V

2)

Using KVL in position 2

VL - I*R = 0

VL - VR = 0

VL = VR

Now, as voltage across the inductor = voltage across the resistance

voltage across the inductor = 3.19 V

voltage across the inductor is 3.19 V

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