Question

A $12\Omega$ resistor and inductance of $0.21H$ are connected in series to an AC source operating at $20V$, $50cycles/s$. The phase angle between the current and the source voltage is

Answer 1

Step 1: Given Data

Resistance of the resistor, $R=12\Omega$

The inductance of the inductor, $L=0.21H$

The voltage of the AC source $V_{rms}=20V$

Frequency of the AC source $f=50cycles/s=50Hz$

Let the phase angle between the current and the source voltage be $\varphi$.

Step 2: Establish the equation of $\varphi$

If $X_L$ is the reactance through the inductor, $Z$ is the impedance of the circuit, then the impedance triangle is given as,

From this triangle we get

$\tan \varphi =\frac{X_L}{R}$

Step 3: Calculate $\varphi$

We know that,

$X_L=\omega L$

where $\omega =2\mathrm{\pi f}$ is the angular frequency.

$\therefore X_L=2\mathrm{\pi f}L$

$=2\times 3.14\times 50\times 0.21$

$\tan \varphi =\frac{2\times 3.14\times 50\times 0.21}{12}$

$=5.5$

$\therefore \varphi ={\tan }^{-1}5.5$

$=80°$

Hence, the phase angle between the current and the source voltage is $80°$.