In a series LCR circuit, the inductive reactance (XL) is 10 Ω and the capacitive reactance (XC) is 4 Ω. The resistance (R) in the circuit is 6 Ω. The power factor of the circuit is: (a) 1/2 (b) √(3)/2 (c) 1/√(2) (d) (1)/(2√(2))

Answer: (c). Z = √(R² + (XL - XC)²) = √(6² + 6²) = 6√(2) Ω. cos φ = R/Z = (6)/(6√(2)) = 1/√(2). So the answer is (c) 1/√(2).

In a series LCR circuit, the inductive reactance (XL) is 10 Ω and the… | JEE Main 2021 (18 Mar, Shift 2)

Q56JEE Main 2021 · 18 MAR · SHIFT 2Easy

Power, Power Factor & Wattless Current

In a series LCR circuit, the inductive reactance (XL) is 10 $\Omega$ and the capacitive reactance (XC) is 4 $\Omega$ . The resistance (R) in the circuit is 6 $\Omega$ . The power factor of the circuit is:

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Answer: (c)

$Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{6^2 + 6^2} = 6\sqrt{2}\ \Omega$ .
$\cos\phi = \frac{R}{Z} = \frac{6}{6\sqrt{2}} = \frac{1}{\sqrt{2}}$ .
So the answer is (c) $\frac{1}{\sqrt{2}}$ .

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