Question

A $100\Omega$ resistance, a $0.1\mu \text{F}$ capacitor and an inductor are connected in series across a $250\text{V}$ supply at variable frequency. Calculate the value of inductance of inductor at which resonance will occur. Given that the resonant frequency is $60\text{Hz}.$

• A. $7.03\times {10}^{-5}\text{H}$
• B. $0.70\text{H}$
• C. $70.4\text{H}$
• D. $70.4\text{mH}$

Answer 1

The correct option is C $70.4\text{H}$

Given, $C=0.1\mu \text{F}={10}^{-7}\text{F}$
Resonant frequency $=60\text{Hz}$

We know that, at resonance
${\omega }_0=\frac{1}{\sqrt{LC}}$
$\Rightarrow 2\pi f_0=\frac{1}{\sqrt{LC}}\Rightarrow L=\frac{1}{4{\pi }^2{f}_0^{2}C}$
by putting values
$L=\frac{1}{4\times (3.14{)}^2\times {60}^2\times {10}^{-7}}=\frac{{10}^7}{4\times (3.14{)}^2\times 3600}$

$\Rightarrow L=70.43\text{H}$

Thus, option $\left(\text{D}\right)$ is correct.