A series LCR circuit with L = (100)/(π) mH, C = 10⁻³π F and R = 10 Ω, is connected across an ac source of 220 V, 50 Hz supply. The power factor of the circuit would be ________. Numerical answer type. Work it out, then open the answer.

Answer: 1. ω = 2π f = 100π rad/s. XL = ω L = 100π × (0.1)/(π) = 10 Ω. XC = (1)/(ω C) = 1100π × 10⁻³/π = 10 Ω. XL = XC (resonance), so Z = R and cos φ = R/Z = 1. Power factor = 1.

A series LCR circuit with L = (100)/(π) mH, C = 10⁻³π F and R = 10 Ω… | JEE Main 2024 (27 Jan, Shift 2)

Q34JEE Main 2024 · 27 JAN · SHIFT 2Easy

Power, Power Factor & Wattless Current

A series LCR circuit with $L = \frac{100}{\pi}\text{ mH}, C = \frac{10^{-3}}{\pi}\text{ F}$ and $R = 10\ \Omega$ , is connected across an ac source of $220\text{ V}, 50\text{ Hz}$ supply. The power factor of the circuit would be ________.

Numerical answer type. Work it out, then open the answer.

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Answer: 1

$\omega = 2\pi f = 100\pi$ rad/s.
$X_L = \omega L = 100\pi \times \frac{0.1}{\pi} = 10\ \Omega$ .
$X_C = \frac{1}{\omega C} = \frac{1}{100\pi \times 10^{-3}/\pi} = 10\ \Omega$ .
$X_L = X_C$ (resonance), so $Z = R$ and $\cos\phi = \frac{R}{Z} = 1$ .
Power factor $= 1$ .

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