Question

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

• A. $7.03 \times 10^{-5}$ H
• B. 70.3 H
• C. 0.70 H
• D. 70.3 mH

Hint:

The resonance condition in a series LCR circuit, $\omega_0 = 1/\sqrt{LC}$, is fundamental. Note that the resistance value affects the sharpness (Q-factor) of the resonance peak but not the resonant frequency itself.

Answer 1

The Correct Option is B

In a series LCR circuit, resonance occurs when the inductive reactance ($X_L$) equals the capacitive reactance ($X_C$).
$X_L = X_C$.
The formulas for reactances are $X_L = \omega L$ and $X_C = \frac{1}{\omega C}$, where $\omega$ is the angular frequency.
At the resonant angular frequency, $\omega_0$:
$\omega_0 L = \frac{1}{\omega_0 C}$.
We are given the resonant frequency $f_0 = 60$ Hz. The angular frequency is $\omega_0 = 2\pi f_0$.
$(2\pi f_0) L = \frac{1}{(2\pi f_0) C}$.
Rearranging to solve for the inductance, L:
$L = \frac{1}{(2\pi f_0)^2 C}$.
Now, substitute the given values. Note that the resistance (100 $\Omega$) and supply voltage (250 V) are not needed for this calculation.
Capacitance $C = 0.1 \mu\text{F} = 0.1 \times 10^{-6} \text{ F} = 10^{-7} \text{ F}$.
Frequency $f_0 = 60$ Hz.
$L = \frac{1}{(2\pi \times 60)^2 \times 10^{-7}}$.
$L = \frac{1}{(120\pi)^2 \times 10^{-7}} = \frac{1}{14400 \pi^2 \times 10^{-7}}$.
$L = \frac{10^7}{14400 \pi^2}$.
Using the approximation $\pi^2 \approx 9.87$:
$L \approx \frac{10^7}{14400 \times 9.87} \approx \frac{10^7}{142128}$.
$L \approx 70.36$ H.
This value is approximately 70.3 H.