QJEEM24Ph41e8: JEE Main 2024 Physics - Class 12 JEEM24Ph41e8 Solution: A series LCR circuit with L = 100 π mH , C = 10 − 3 π F and R = …

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QJEEM24Ph41e8: JEE Main 2024 Physics - Class 12 JEEM24Ph41e8 Solution: A series LCR circuit with L = 100 π mH , C = 10 − 3 π F and R = …

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

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$\begin{aligned} X_{c} = \frac{1}{ω C} = \frac{π}{2 π \times 50 \times 10^{- 3}} = 10 Ω \\ X_{L} = ω L = 2 π \times 50 \times \frac{100}{π} \times 10^{- 3} \\ = 10 Ω \\ ∵ X_{C} = X_{L}, Hence, circuit is in resonance \\ ∴ power factor = \frac{R}{Z} = \frac{R}{R} = 1 \end{aligned}$

QJEEM24Ph41e8: JEE Main 2024 Physics - Class 12 JEEM24Ph41e8 Solution: A series LCR circuit with L = 100 π mH , C = 10 − 3 π F and R = …

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