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

Question

A 12.00-V battery is connected through a snitch 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

It needs to understood here that parts of the circuit connected in parallel to each other have the same potential difference across them. Also, the amount of current flowing through an RL circuit at any time t is given as:

I = (V/R)[1 - e-Rt/L]

We will make use of the above to answer the problems given:

Part a.) As the voltage drop across R1 and the combination of R2 and L is same and each equal to V = 12 Volts

The current through R1 = V / R1 = 12 / 100 = 0.12 amperes

Now, the moment the current is switched, the inductor doesn't allow any current to flow through it, hence the current flowing through the resistor R2 would be zero.

Part b.) We know that the current flowing through an RL circuit is given as:

I = (V/R)[1 - e-Rt/L]

Now for t = 50 ms, the current would be = (12/100)[1 - e-100x0.05/4] = 0.10706 amperes

For the resistor R1, current would remain same as before, that is, 0.12 amperes.

Part c.) For t = 500 ms, the current in the inductor would be given as:

I = 0.12 x [1 - e-100x0.5/4] = 0.119999 amperes

For R1, the current would remain same as before, that is, 0.12 amperes.

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