Discussion
In Hay's bridge \( Q = 1/(\omega C_4 R_4) \). With \( C_4 = 0.5\ \mu \)F, \( R_4 = 100\ \Omega \) and a frequency of 1 kHz, the quality factor of the coil is about
Answer: A
The denominator is \( 2\pi \times 1000 \times 0.5 \times 10^{-6} \times 100 = 0.314 \), and its reciprocal is 3.18. Since Q here is only about three, Hay's bridge would introduce an appreciable error in the inductance and another form would be preferred.
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