Electromagnetic Induction

JEE Main level, numerical answer, medium

148 of 153

Question 148 of 153

For the given circuit the current through battery of 6V6V just after closing the switch 'SS' will be _____ AA.
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Answer: 1

Just after closing the switch SS, inductor behaves like an open circuit, so no current will flow through 2Ω2\Omega resistors.
Thus, current through the circuit is I=62+4=1AI=\dfrac{6}{2+4}=1A (Here 2Ω2\Omega and 4Ω4\Omega resistors are in series).

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