Q.

An inductor (L = 0.03 H) and a resistance R=0.15kΩ are connected in series to a battery of 15V emf in a circuit shown below. The key K1 has been kept closed for a long time. Then at t=0, K1 is opened and key K2 is closed simultaneously. At t=1ms, the current in the circuit will be ( Given e5150)

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a

6.7 mA

b

100 mA

c

67 mA

d

0.67 mA

answer is B.

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Detailed Solution

Decay of current in L-R circuit

it=I0eRt/L

I0=ER=150.15×103=15×10015×1000=0.1A

it=0.1e150×1030.03=0.1e5

=0.1150=0.67mA

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