WBJEE · Physics · Alternating Current
When a \(60 \mathrm{mH}\) inductor and a resistor are connected in series with an AC voltage source, the voltage leads the current by \(60^{\circ}\). If the inductor is replaced by a \(0.5 \mu \mathrm{F}\) capacitor, the voltage lags behind the current by \(30^{\circ} .\) What is the frequency of the AC supply?
- A \(\frac{1}{2 \pi} \times 10^{4} \mathrm{Hz}\)
- B \(\frac{1}{\pi} \times 10^{4} \mathrm{Hz}\)
- C \(\frac{3}{2 \pi} \times 10^{4} \mathrm{Hz}\)
- D \(\frac{1}{2 \pi} \times 10^{8} \mathrm{Hz}\)
Answer & Solution
Correct Answer
(A) \(\frac{1}{2 \pi} \times 10^{4} \mathrm{Hz}\)
Step-by-step Solution
Detailed explanation
Given, inductance of inductor, \(L=60 \mathrm{mH}\) \[ =60 \times 10^{-3} \mathrm{H} \] Phase difference between voltage and current in \(L-R\) circuit, \(\theta_{1}=60^{\circ}\) Capacitance of capacitor, \(C=0.5 \mu \mathrm{F}=0.5 \times 10^{-6} \mathrm{F}\) Phase difference…
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