Question

Two different coils have self-inductance $L_1=8\text{mH}$ and $L_2=2\text{mH}$. The currents in both the coils is increased at a same constant rate. At a certain instant of time, the power given to both the coils is the same. At that time the current and the induced voltage in the first coil are, $i_1$ and $V_1$ respectively. Corresponding values for the second coil, at the same instant are, $i_2$ and $V_2$ respectively. Then -

• A. $\frac{i_1}{i_2}=\frac{1}{2}$
• B. $\frac{i_1}{i_2}=\frac{1}{4}$
• C. $\frac{i_1}{i_2}=2$
• D. $\frac{i_1}{i_2}=4$

Answer 1

The correct option is B $\frac{i_1}{i_2}=\frac{1}{4}$

The magnitude of induced emf is given by,

$E=L\frac{dI}{dt}$

$\therefore \frac{V_1}{V_2}=\frac{L_1}{L_2}=\frac{8}{2}=4[\because \text{Rate of current change}\text{is same in both the coils}]$

As the power is same at the given time instant,

$V_1{i}_1=V_2{i}_2$

$\therefore \frac{i_1}{i_2}=\frac{V_2}{V_1}=\frac{1}{4}$

Hence, option $\left(B\right)$ is the correct answer.