Question

Two different coils have self-inductance $L_1=6\text{mH}$ and $L_2=3\text{mH}$. The current in the first coil is increased at a constant rate. The current in the second coil is also increased at the same constant rate. At a certain instant of time, the power given to the two coils is the same. At that instant of time, the current, the induced voltage and the energy stored in the first coil are $i_1,V_1\text{and}{W}_1$ respectively. Corresponding values for the second coil at the same instant are $i_2,V_2\text{and}{W}_2$ respectively. Then

• A. $\frac{V_1}{V_2}=\frac{2}{1}$
• B. $\frac{i_1}{i_2}=\frac{1}{4}$
• C. $\frac{i_1}{i_2}=4$
• D. $\frac{W_1}{W_2}=\frac{1}{2}$

Answer 1

Answer: (A), (D)

$\frac{W_1}{W_2}=\frac{1}{2}$

Here, $L_1=6\text{mH}$ and $L_2=3\text{mH}$

Induced voltage in the coil is,

$V=-L\frac{di}{dt}$

$\because \frac{di}{dt}$ is constant and same for both coils.

$\therefore$ $V\propto L\Rightarrow \frac{V_1}{V_2}=\frac{L_1}{L_2}=\frac{6}{3}=\frac{2}{1}.....\left(1\right)$

Given that, power given to two coils is same,

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

$\Rightarrow \frac{i_1}{i_2}=\frac{V_2}{V_1}=\frac{1}{2}.....\left(2\right)\left[\text{Using 1}\right]$

Energy stored in the coil, $W=\frac{1}{2}L{i}^2$

$\frac{W_1}{W_2}=\left(\frac{L_1}{L_2}\right){\left(\frac{i_1}{i_2}\right)}^2$

Using $\left(1\right)$ and $\left(2\right)$,

$\frac{W_1}{W_2}=\left(\frac{2}{1}\right){\left(\frac{1}{2}\right)}^2=\frac{1}{2}$

Hence, $\left(\text{A}\right)$ and $\left(\text{D}\right)$ are the correct options.