Question

A $0.07\text{H}$ inductor and a $12\Omega$ resistor are connected in series to a $220\text{V},50\text{Hz}$ ac source. The approximate current in the circuit and the phase angle between current and source voltage are respectively. [Take $\pi$ as $\frac{22}{7}$]

• A. $0.88\text{A}$ and ${\tan }^{-1}\left(\frac{11}{6}\right)$
• B. $88\text{A}$ and ${\tan }^{-1}\left(\frac{11}{6}\right)$
• C. $8.8\text{A}$ and ${\tan }^{-1}\left(\frac{11}{6}\right)$
• D. $8.8\text{A}$ and ${\tan }^{-1}\left(\frac{6}{11}\right)$

Answer 1

The correct option is C $8.8\text{A}$ and ${\tan }^{-1}\left(\frac{11}{6}\right)$

We know that,
$X_L=\omega L=2\pi fL$
$X_L=2\times \frac{22}{7}\times 50\times 0.07=22\Omega$

So, phase angle in $LR$ circuit is given by

$\varphi ={\tan }^{-1}\left(\frac{X_L}{R}\right)$

$\varphi ={\tan }^{-1}\left(\frac{22}{12}\right)={\tan }^{-1}\left(\frac{11}{6}\right)$

The required impedance of the circuit is,

$Z=\sqrt{X_L^2+R^2}=25.059$

$\therefore I=\frac{V}{Z}=\frac{220}{25.059}=8.77\text{A}$

Hence, option $\left(\text{A}\right)$ is correct.