Question

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

• A. $ 0.4\times 10^{-4}s $
• B. $ 0.2\times 10^{-4}\,s $
• C. $ 0.5\times 10^{-4}\,s $
• D. $ 0.1\times 10^{-4}\,s $

Answer 1

The Correct Option is A

Time constant $=\frac{L}{R}=\frac{1.2 \times 10^{-4}}{3}$
$=0.4 \times 10^{-1} s$