Question

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

Answer 1

Refer IMAGE $01$

$X_L=$ Inductive resistance

$X_C=$ Capacitive resistance

Phase angle $={35}^o$

Refer IMAGE $02$

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

$\tan {35}^o=\frac{X_L-X_C}{10}$

$0.7\times 10=X_L-X_C$

$X_L-X_C=7$ (equation $1$)

given that, $X_L+X_C=17$ (equation $2$)

From equation $1$ and $2$

$X_L=12$ $\Omega$

And,

$L=\frac{X_L}{2\pi f}=\frac{12}{2\pi \times 100}$

$L=1.9\times {10}^{-2}$ $H$