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 $12 \, V$ 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
( $L_{_{e q}} = \frac{L_{1} L_{2}}{L_{1} + L_{2}}$ )
$\therefore \, L^{′}=\frac{\frac{1.8}{2} \times 10^{- 4} \times \frac{1.8}{2} \times 10^{- 4}}{\frac{1.8}{2} \times 10^{- 4} + \frac{1.8}{2} \times 10^{- 4}}$
$=0.45\times 10^{- 4 \, }H$
and $ \, \, R^{′}=\frac{3 \times 3}{3 + 3}=1.5\Omega $
( $R_{_{e q}} = \frac{R_{1} R_{2}}{R_{1} + R_{2}}$ )
Time constant $ \, =\frac{L^{′}}{R ′}$
$= \, \frac{0.45 \times 10^{- 4}}{1.5}=0.3\times 10^{- 4 \, }s$