Question

A resistor, an inductor, and a capacitor are connected in series with a $120V,100Hz$ ac source. Voltage leads the current by ${35}^{\circ }$ in the circuit. If the resistance of the resistor is $10\Omega$ and the sum of inductive and capacitive reactance is $17\Omega$, calculate self-inductance of the inductor.

Answer 1

Given:-

phase difference $\varphi =35°,R=10\Omega$

Sum of reactance of inductor and capacitor $(X_L+X_C)=17⟶e{q}^n\left(1\right)$

Frequency of applied voltage, $f=100Hz$

So, angular frequency=$2\pi \times 100$

Now, for LCR circuit, we know

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

or $X_L-X_C=R\tan \varphi =10\tan 35°$

$X_L+X_C=17$ [Using eq.1]

$2X_L=17+(10\times 0.7)$

$X_L=12$

$\omega L=12$

L=$\frac{12}{2\pi 100}=1.9\times {10}^{-2}H$