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Question

A constant voltage is applied to a series RL circuit by closing the switch. The voltage across inductor (L=2H) is 20V at t=0 and drops to 5Vat 20ms. The value of R in Ω


A

100ln2Ω

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B

100(1-ln2)Ω

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C

100ln4Ω

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D

1001-ln4Ω

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Solution

The correct option is C

100ln4Ω


Step 1:Given

At time(t)=0

voltage across inductor=20V

inductance of inductor=2H

At time(t)=20milliseconds

Voltage after decreasing=5V

Step 2:Formulae applied

Voltage becomes 14th in 20ms

which is equal to 2 half-lives

Thus, two half-lives equal to 20ms.

So, one half-life is 10ms.

Step 3:Calculation

t12=ln2LR

R=ln2Lt12

R=ln2210×10-3

R=100ln4Ω

Hence option C is the correct option.


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