Question

Two coils have self-inductance $L_1=4mH$ and $L_2=1mH$ respectively. The currents in the two coils are increased at the same rate. At a certain instant of time, both coils are given the same power. If $i_1$and $i_2$ are the currents in the two coils, at that instant of time respectively, then the value of ($\frac{i_1}{i_2}$) is?

• A. $0$
• B. $1$
• C. $\frac{1}{2}$
• D. $\frac{1}{4}$

Answer 1

The correct option is D

$\frac{1}{4}$

The explanation for the correct option:

Step 1: Given Data

Two coils have self-inductance,

$$\begin{gathered} L_1=4mH \\ {L}_2=1mH \end{gathered}$$

Both coils are given the same power (i.e. $P_1=P_2$), at a certain instant of time after increasing the current.

Current flowing in the coils are $i_1$ and $i_2$.

Step 2: Formula used

The rate of change of current is given by $emf$(or $V$).

Mathematically,

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

where,

$V$ is the induced emf or voltage in the coil due to a change of current

$L$ is the inductance of the coil

$\frac{di}{dt}$is the rate of change of current in the coil

Now for two inductors, the emf induced is given by,

$V_1=L_1\frac{d{i}_1}{dt}$ and $V_2=L_2\frac{d{i}_2}{dt}.....\left(1\right)$

As the power in both the coils is equal so,

$\begin{array}{rcl}& & \\ & & \end{array}$$\begin{array}{rcl}& \Rightarrow & i_1{V}_1=i_2{V}_2\\ & \Rightarrow & \frac{i_1}{i_2}=\frac{V_2}{V_1}\cdots \cdots \left(2\right)\end{array}$

Step 3: Calculate the ratio between the currents in the two coils at the given instant of time.

Now from equation (1), we can write

$\Rightarrow \frac{i_1}{i_2}=L_2\frac{d{i}_2}{dt}\times \frac{1}{L_1{\displaystyle \frac{d{i}_1}{dt}}}......\left(3\right)$

Now the current in both coils is increased at the same rate i.e. $\frac{d{i}_1}{dt}=\frac{d{i}_2}{dt}$.

So equation (3) can be written as,

$\Rightarrow \frac{i_1}{i_2}=\frac{L_2}{L_1}$

Now putting the value of $L_1$and $L_2$, we got

$\frac{i_1}{i_2}=\frac{1}{4}$

Hence, option (D) is the correct answer.