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A coil (which can be modelled as a series RL circuit) has been designed for a high Quality Factor (Q) performance. The voltage is rated at and a specified frequency. If the frequency of operation is increased 10 times and the coil is operated at the same rated voltage. The new value of Q factor and the active power P will be?(a) P is doubled and Q is halved(b) P is halved and Q is doubled(c) P remains constant and Q is doubled(d) P decreases 100 times and Q is increased 10 timesThe question was posed to me during an internship interview.This question is from Problems of Series Resonance Involving Quality Factor topic in division Resonance & Magnetically Coupled Circuit of Network Theory

Answer» CORRECT option is (d) P DECREASES 100 times and Q is increased 10 times

The best explanation: ω2 L = 10 ω1 LR will remain constant

Q2 = \(\frac{10 ω_1L}{R}\) = 10 Q1

That is Q is increased 10 times.

Now, I1 = \(\frac{V}{ω_1L} \)

For a high Q coil, ωL >> R,

I2 = \(\frac{V}{10 ω_1L} = \frac{I_1}{10}\)

∴ P2 = R \((\frac{I_1}{10})^2 = \frac{P_1}{100}\)

Thus, P decreases 100 times and Q is increased 10 times.


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