WBJEE · Physics · Alternating Current
When an \(A C\) source of emf \(E\) with frequency \(\omega=100 \mathrm{~Hz}\) is connected across a circuit, the phase difference between
\(E\) and current I in the circuit is observed to be \(\frac{\pi}{4}\) as shown in the figure. If the circuit consist of only \(\mathrm{RC}\) or \(\mathrm{RL}\) in series, then
- A \(\mathrm{R}=1 \mathrm{k} \Omega, \mathrm{C}=5 \mu \mathrm{F}\)
- B \(\mathrm{R}=1 \mathrm{k} \Omega, \mathrm{L}=10 \mathrm{H}\)
- C \(\mathrm{R}=1 \mathrm{k} \Omega, \mathrm{L}=1 \mathrm{H}\)
- D \(\mathrm{R}=1 \mathrm{k} \Omega, \mathrm{C}=10 \mu \mathrm{F}\)
Answer & Solution
Correct Answer
(D) \(\mathrm{R}=1 \mathrm{k} \Omega, \mathrm{C}=10 \mu \mathrm{F}\)
Step-by-step Solution
Detailed explanation
.. Since supply voltage lags the current \(\operatorname{By}(\phi=\pi / 4)\)…
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