Question

A coil of inductance $5\text{mH}$ and negligible resistance is connected to an oscillator giving an output voltage $E=10\sin \omega t$. The peak currents in the circuit for $\omega =10{\text{s}}^{-1},$$100{\text{s}}^{-1}$ and $500{\text{s}}^{-1}$ are $I_1,I_2$ and $I_3$ respectively. Then the ratio $I_1:I_2:I_3$ is:

• A. $50:5:1$
• B. $1:5:50$
• C. $5:1:50$
• D. $1:1:1$

Answer 1

The correct option is A $50:5:1$

The reactance of the circuit is:

$X_L=\omega L$

The peak value of current is:

$I_0=\frac{E_0}{X_L}=\frac{E_0}{\omega L}$

Here, $E_0$ and $L$ are constant.

$\Rightarrow I_p\propto \frac{1}{\omega }$

Therefore, the ratio of peak currents for
${\omega }_1=10\text{s}{}^{-1},{\omega }_2=100\text{s}{}^{-1}\text{and}{\omega }_3=500\text{s}{}^{-1}$ is,

$I_1:I_2:I_3=\frac{1}{{\omega }_1}:\frac{1}{{\omega }_2}:\frac{1}{{\omega }_3}$

$=\frac{1}{10}:\frac{1}{100}:\frac{1}{500}=50:5:1$

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