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Question

Two inductors L1 (Inductance 1 mH, internal resistance 2 Ω) and L2 (inductance 2 mH, internal resistance 4 Ω) and a resistor R (resistance12 Ω) are all connected in parallel across a 5 V battery. The circuit is switched on at time t=0. The ratio of the maximum to the minimum current (Imax/Imin) drawn from the battery is


A
4:1
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B
8:1
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C
6:1
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D
16:1
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Solution

The correct option is B 8:1
At t=0 instant, all the inductors will act like an open branch. So no flow of current across them

imin=VR

Given:-
R1=3 Ω L1=1 mH

R2=4 Ω L2=2 mH

R3=12 Ω

imin=VR3=512 A

At steady state, all the inductors will act as perfect conductor. So

imax=VR1+VR2+VR3

=5[13+14+112]=5(4+3+1)12

imax=4012 A

imaximin=4012×125=8:1

Hence, option (B) is correct.

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