Question

Two inductor coils of self inductance $3H$ and $6H$ respectively are connected with a resistance $10\Omega$ and a battery $10V$ as shown in figure. The ratio of total energy stored at steady state in the inductors to that of heat developed in resistance in $10$ seconds at the steady is (neglect mutual inductance between $L_1$ and $L_2$)

• A. $\frac{1}{10}$
• B. $\frac{1}{100}$
• C. $\frac{1}{1000}$
• D. $1$

Answer 1

The correct option is B $\frac{1}{100}$

$\frac{1}{L}=\frac{1}{L_1}+\frac{1}{L_2}$

$L=\frac{6\times 3}{6+3}=2H$

A steady state inductor will offer no resistance.

$\therefore I=\frac{10}{10}=1A$

$\therefore H_R=I^{2}Rt=1^2\times 10\times 10=100$

$E_L=\frac{1}{2}L{I}^2=\frac{1}{2}\times 2\times 1^2=1$

$\therefore \text{ratio}=1:100$