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Question

A coil of inductance 300mH and resistance 2Ω is connected to a source of voltage 2v. The current reaches half of its steady-state value in____seconds.


A

0.1

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B

0.3

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C

0.05

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D

0.105

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Solution

The correct option is A

0.1


Step 1: Given

Inductance, L=300mH=300×10-3H

Resistance, R= 2Ω

Voltage, V=2V

Step 2: Use the formula of growth of current

Formula for the growth of current in inductor, I=I01-e-RtL

Also, I=I02 because the time requires is for the half of current steady-state value.

Now,

I02=I01-e-RtLI02=I0-I0e-RtLI0e-RtL=I0-I02I0e-RtL=I02e-RtL=12

taking log on both sides-

-RtL=-log2RtL=log2

Step 3: Calculate the time to reach half of its steady-state value

t=LRlog2

Substitute values-

t=300×10-32×0.693t=150×10-3×0.693t=0.10395s0.1s

Thus, the time to reach half of its steady-state value is 0.10395s0.1s.

Hence, Option A is correct.


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