AP EAMCET · PHYSICS · Alternating Current
A 50 Hz AC circuit has a 10 mH inductor and a \(2 \Omega\) resistor in series. The value of capacitance to be placed in series in the circuit to make the circuit power factor as unity is
- A \(1.014 \times 10^{-6} \mathrm{~F}\)
- B \(1.014 \times 10^{-3} \mathrm{~F}\)
- C \(2.6 \times 10^{-3} \mathrm{~F}\)
- D \(4.125 \times 10^{-3} \mathrm{~F}\)
Answer & Solution
Correct Answer
(B) \(1.014 \times 10^{-3} \mathrm{~F}\)
Step-by-step Solution
Detailed explanation
For power factor, \(\cos \phi=1\), the LCR circuit is in resonance.…
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