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A coil (which can be modelled as a series RL circuit) has been designed for high 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 timesI had been asked this question in homework.Question is taken from Advanced Miscellaneous Problems on Measurement of Power in section Measurement of Power and Related Parameters of Electrical Measurements |
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Answer» RIGHT CHOICE is (d) P decreases 100 TIMES and Q is increased 10 times 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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