A 12.00-V battery is connected through a switch to two identical resistors and a

Question

A 12.00-V battery is connected through a switch to two identical resistors and an ideal inductor, as shown in the figure. Each of the resistors has a resistance of 100 ohm, and the inductor has an inductance of 4.00 H. The switch is initially open. Immediately after the switch is closed, what is the current in resistor R1 and in resistor R2? At 50.0 ms after the switch is closed, what is the current in resistor R1 and in resistor R2? At 500. ms after the switch is closed, what is the current in resistor R1 and in resistor R2?

Explanation / Answer

a]

When the switch closed,

current in R1= I1= V/R1= 12 V/100 ohms= 0.12 A

Current in R2= I2= V/R2[ 1- e[-R2t/L]]

b]

At t= 50 ms=0.05 s

current in R1= I1= V/R1= 12 V/100 ohms= 0.12 A

current in R1=0.12 A

Current in R2= I2= V/R2[ 1- e[-R2t/L]]

I2= 12/100* [1- e[-100*0.05/4]]

I2= 0.12*[1- e[-1.25]] =0.12 [ 1-0.287]= 0.086 A

Current in R2= 0.086 A

c]

At t= 500 ms=0.5 s

current in R1= I1= V/R1= 12 V/100 ohms= 0.12 A

current in R1=0.12 A

Current in R2= I2= V/R2[ 1- e[-R2t/L]]

I2= 12/100* [1- e[-100*0.5/4]]

I2= 0.12*[1- e[-12.5]] =0.12 [ 1-3.7*10-6]= 0.12 A

Current in R2= 0.12 A

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