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Physics Question on Electromagnetic induction

A uniformly wound coil of self-inductance 1.2×10−4H1.2\times 10^{-4}H and resistance 3 Ω3\,\Omega . is broken up into two identical coils. These coils are then connected parallel across a 6 V6\,V battery of negligible resistance. The time constant for the current in the circuit is (neglect mutual inductance)

A

0.4×10−4s0.4\times 10^{-4}s

B

0.2×10−4 s0.2\times 10^{-4}\,s

C

0.5×10−4 s0.5\times 10^{-4}\,s

D

0.1×10−4 s0.1\times 10^{-4}\,s

Answer

0.4×10−4s0.4\times 10^{-4}s

Explanation

Solution

Time constant =LR=1.2×10−43=\frac{L}{R}=\frac{1.2 \times 10^{-4}}{3}
=0.4×10−1s=0.4 \times 10^{-1} s