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 $12V$ battery of negligible internal resistance.
Values of equivalent resistance and equivalent inductance are respectively :

• A. $1.5\Omega ,4.5\times {10}^{-5}H$
• B. $3\Omega ,9\times {10}^{-5}H$
• C. $6\Omega ,1.8\times {10}^{-4}H$
• D. $12\Omega ,3.6\times {10}^{-4}H$

Answer 1

The correct option is A $1.5\Omega ,4.5\times {10}^{-5}H$

For resistance, $6\Omega$ is broken into two identical ,

$R_1=R_2=3\Omega$

Equivalent resistance in parallel combination,

$\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}$

$\frac{1}{R_{eq}}=\frac{1}{3}+\frac{1}{3}=\frac{2}{3}$

$R_{eq}=\frac{3}{2}=1.5\Omega$

For self inductance, $1.8\times {10}^{-4}H$ is broken into two identical

$L_1=L_2=0.9\times {10}^{-4}H$

Equivalent self inductance in parallel combination,

$\frac{1}{L_{eq}}=\frac{1}{L_1}+\frac{1}{L_2}$

$\frac{1}{L_{eq}}=\frac{1}{0.9\times {10}^{-4}}+\frac{1}{0.9\times {10}^{-4}}=\frac{2}{0.9\times {10}^{-4}}$

$L_{eq}=\frac{0.9\times {10}^{-4}}{2}=4.5\times {10}^{-5}H$

The correct option is A.