Question

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 times

I 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

Answer 1

Right choice is (d) P decreases 100 times and Q is increased 10 times

Explanation: ω2 L = 10 ω1 LR will remain constant

∴ Q2 = \(\frac{10 ω_1  L}{R}\) = 10 Q1

That is Q is increased 10 times.

Now, I1 = \(\frac{V}{ω_1  L}\)

For a high Q coil, ωL ≫> R,

I2 = \(\frac{V}{10 ω_1  L} = \frac{I_1}{10}\)

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

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