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Q.

The figure below shows a battery with emf 15 V in a circuit with R1=30Ω,R2=10Ω,R3=20Ω and 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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a

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b

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c

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d

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answer is A.

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Detailed Solution

Initially no current in the inductor, so initial current

i1=5R1+R2+R3=0.25A

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

i2=15R1=0.5A

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