The self inductance of the motor of an electric fan is 10 H. In order to impart maximum power at 50 Hz, it should be connected to a capacitance of (a) 8 (b) 4 (c) 2 (d) 1

Answer: (d). Maximum power at resonance: C = (1)/(4π² f² L). C = (1)/(4×9.87×2500×10) = 19.87×10⁵ ≈ 1×10⁻⁶ F So the answer is (d) 1 μ F.

The self inductance of the motor of an electric fan is 10 H. In order… | JEE Main 2005 (AIEEE)

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Resonance, Bandwidth & Q Factor

The self inductance of the motor of an electric fan is $10\text{ H}$ . In order to impart maximum power at $50\text{ Hz}$ , it should be connected to a capacitance of

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Answer: (d)

Maximum power at resonance: $C = \frac{1}{4\pi^2 f^2 L}$ .
$C = \frac{1}{4\times9.87\times2500\times10} = \frac{1}{9.87\times10^{5}} \approx 1\times10^{-6}$ F
So the answer is (d) $1\ \mu$ F.

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