Question

An inductor $(L=200\text{mH})$ is connected to an AC source of peak emf $210\text{V}$ and frequency $50\text{Hz}$. The value of peak current and instantaneous voltage of the source when the current is at its peak value is respectively :

• A. $3.3\text{A},210\text{V}$
• B. $6.6\text{A},0\text{V}$
• C. $3.3\text{A},0\text{V}$
• D. $6.6\text{A},210\text{V}$

Answer 1

The correct option is C $3.3\text{A},0\text{V}$

The reactance of the inductor is,

$X_L=\omega L$

$=(2\pi \times 50{\text{s}}^{-1})\times (200\times {10}^{-3}\text{H})$

$=62.8\Omega$

The peak current is,

$i_0=\frac{E_0}{X_L}=\frac{210}{62.8}=3.3\text{A}$

Phasor diagram:


As the current lags the voltage by $\pi /2$, the voltage is zero when the current has its peak value.

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