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The figure below shows a battery with emf 15 V in a circuit with $R_{1} = 30 Ω, R_{2} = 10 Ω, R_{3} = 20 Ω a n d L = 3.0 H$ , R, = 20 A and L = 3.0 H. The switch S is initially in the open position and is then closed at time t = 0. Then the graph which shows the correct variation of current through battery after switch ,S is closed

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answer is A.

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Detailed Solution

Initially no current in the inductor, so initial current

$i_{1} = \frac{5}{R_{1} + R_{2} + R_{3}} = 0.25 A$

In steady state, no current in R₂ and R₃. So final current through battery:

$i_{2} = \frac{15}{R_{1}} = 0.5 A$

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