A uniformly wound solenoid coil of self inductance 1.8 ´ 10–4 henry and resistance 6 ohm is broken up into two identical coils. These identical coils are then connected in parallel across a 15-voll battery of negligible resistance. The time constant for the current in the circuit is
Answer & explanation
Correct answer: option 1
The coil is broken into two identical coils.
$L_{eq} = \frac{L/2 \times L/2}{L/2 + L/2} = \frac{L}{4} = 0.45 \times 10{-4} H$
$ R_{eq} = \frac{R/2 \times R/2}{R/2 + R/2} = \frac{R}{4} = 1.5 \Omega$
$ \text{Time constan } = \frac{L_{eq}}{R_{eq}} = \frac{0.45\times 10^{-4}}{1.5} = 0.3\times 10^{-4} Sec$