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The figure below shown a battery with em...

The figure below shown a battery with emf 15 V in a circuit with `R_(1) = 30 (Omega), R_(3) = 20 (Omega 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

A

B

C

D

Text Solution

Verified by Experts

The correct Answer is:
A
Initially no current in the inductor, so initial current
`(i_1)=(15)/(R_(1)+R_(2)+R_(3))=0.25A`
In steady state, no current in `R_(2) and R_(3)`. So final current through battery: `I_(2)=(15)/(R_1)=0.5A`.
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