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Question

Two resistors of 10Ω and 20Ω and an ideal inductor of 10 H are connected to a 2 V battery as shown in Fig. Key K is inserted at time t=0. The initial (t=0) and final (t) currents through the battery are:
120956_9629c0dd2dbb418a95ecc381acdea2dc.png

A
115A,110A
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B
110A,115A
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C
215A,110A
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D
115A,225A
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Solution

The correct option is A 115A,110A
Initially, inductor does not allow flux through it to change, therefore current passes through the parallel resistor. Therefor, at t=0, current through batteryI=2(10+20)=115A
Now at t= infinity, inductor acts as a short.
Current through battery I=220=110A

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