Question

A uniformly wound solenoid coil of self-inductance $1.8\times 10^{-4}H$ and resistance $6\Omega$ is broken up into two identical coils. These identical coils are then connected in parallel across a $12V$ battery of negligible resistance. The time constant for the current in the circuit is

• A. $0.1\times 10^{-4}$ $s$
• B. $0.2\times 10^{-4}s$
• C. $0.3\times 10^{-4}s$
• D. $0.4\times 10^{-4}s$

Answer 1

Answer: (C)

Given, self-inductance, $L=1.8\times 10^{-4}H$
Resistance, $R=6\Omega$
When self-inductance and resistance are broken up into identical coils.
Then, the self-inductance of each oil
$=\frac{1.8\times 10^{-4}}{2}H$
The resistance of each oil
$=\frac{6\Omega }{2}=3\Omega$
Coil are then connected in parallel
( $L_{{}_{eq}}=\frac{L_1{L}_2}{L_1+L_2}$ )
$\therefore L^{\prime }=\frac{\frac{1.8}{2}\times 10^{-4}\times \frac{1.8}{2}\times 10^{-4}}{\frac{1.8}{2}\times 10^{-4}+\frac{1.8}{2}\times 10^{-4}}$
$=0.45\times 10^{-4}H$
and $R^{\prime }=\frac{3\times 3}{3+3}=1.5\Omega$
( $R_{{}_{eq}}=\frac{R_1{R}_2}{R_1+R_2}$ )
Time constant $=\frac{L^{\prime }}{R\prime }$
$=\frac{0.45\times 10^{-4}}{1.5}=0.3\times 10^{-4}s$