An alternating current (AC) circuit contains an inductor of inductance L = \frac{0.2}{\pi^… — JAMB UTME

An alternating current (AC) circuit contains an inductor of inductance $L = \frac{0.2}{\pi^2}\ \text{H}$ , a resistor of resistance $R = 40\ \Omega$ , and a variable capacitor $C$ connected in series across a $50\ \text{Hz}$ voltage supply. What capacitance $C$ , in microfarads ( $\mu\text{F}$ ), is required for the circuit to operate at electrical resonance?

Answer: 500 μF

Answer

The capacitance required to achieve electrical resonance is 500 μF.

At electrical resonance in a series RLC circuit, the inductive reactance (

X_L

) equals the capacitive reactance (

X_C

). Setting

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

yields

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

. Substituting

f_0 = 50\ \text{Hz}

and

L = \frac{0.2}{\pi^2}\ \text{H}

into this equation gives

C = 5 \times 10^{-4}\ \text{F}

, which equals

500\ \mu\text{F}

Step-by-Step Solution

Recall the resonant frequency formula for a series RLC circuit.

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

At resonance, inductive reactance equals capacitive reactance (

X_L = X_C

Substitute the given numerical parameters into the equation.

50 = \frac{1}{2\pi \sqrt{\frac{0.2}{\pi^2} \cdot C}}

Given frequency

f_0 = 50\ \text{Hz}

and inductance

L = \frac{0.2}{\pi^2}\ \text{H}

Isolate the square root term and simplify.

\sqrt{0.2 C} = 0.01

Simplifying

2\pi \cdot \frac{1}{\pi} = 2

and rearranging

2 \sqrt{0.2 C} = \frac{1}{50} = 0.02

Square both sides and solve for

C

in farads.

C = 5 \times 10^{-4}\ \text{F}

0.2 C = (0.01)^2 = 10^{-4}

, so

C = \frac{10^{-4}}{0.2} = 5 \times 10^{-4}\ \text{F}

Convert capacitance from farads to microfarads.

C = 500\ \mu\text{F}

Multiply farads by

10^6

to express the result in microfarads.

Key Concept

Resonant Frequency in Series AC Circuits

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