A series alternating current (AC) circuit contains an inductor of inductance L = \frac{1}{… — JAMB UTME

A series alternating current (AC) circuit contains an inductor of inductance $L = \frac{1}{\pi^2}\ \text{H}$ , a capacitor of capacitance $C = 25\ \mu\text{F}$ , and a resistor of resistance $R = 50\ \Omega$ . What is the resonant frequency of the circuit in hertz ( $\text{Hz}$ )?

Answer: 100 Hz

Answer

The resonant frequency of the circuit is

100\ \text{Hz}

At resonance, the inductive reactance

X_L = 2\pi f L

equals the capacitive reactance

X_C = \frac{1}{2\pi f C}

. Equating both yields

f_0 = \frac{1}{2\pi\sqrt{LC}}

. Substituting

L = \frac{1}{\pi^2}\ \text{H}

and

C = 25 \times 10^{-6}\ \text{F}

gives

\sqrt{LC} = \frac{5 \times 10^{-3}}{\pi}\ \text{s}

, leading to

f_0 = \frac{1}{2\pi \left(\frac{5 \times 10^{-3}}{\pi}\right)} = 100\ \text{Hz}

Step-by-Step Solution

Write down the formula for the resonant frequency of a series RLC circuit.

f_0 = \frac{1}{2\pi\sqrt{LC}}

Resonance occurs when the inductive reactance equals the capacitive reactance (

X_L = X_C

Substitute the values of inductance

L = \frac{1}{\pi^2}\ \text{H}

and capacitance

C = 25 \times 10^{-6}\ \text{F}

into

\sqrt{LC}

\sqrt{LC} = \sqrt{\frac{1}{\pi^2} \times 25 \times 10^{-6}} = \frac{5 \times 10^{-3}}{\pi}\ \text{s}

Simplifying the square root removes the fraction containing

\pi

Calculate the resonant frequency

f_0

f_0 = \frac{1}{2\pi \times \frac{5 \times 10^{-3}}{\pi}} = \frac{1}{10^{-2}} = 100\ \text{Hz}

The factor of

\pi

cancels out in the denominator, resulting in a whole number value.

Key Concept

Resonant Frequency in AC Series Circuits

Estimated Time:1m 30s

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