AP EAMCET · PHYSICS · Alternating Current
For the AC circuit shown below, phase difference between emf and current is \(\frac{\pi}{4}\) radian as shown in the graph. If the impedance of the circuit is \(1414 \Omega\), then the values of \(P\) and \(Q\) are

- A \(1 \mathrm{k} \Omega, 10 \mu \mathrm{F}\)
- B \(1 \mathrm{k} \Omega, 1 \mu \mathrm{F}\)
- C \(1 \mathrm{k} \Omega, 10 \mathrm{mH}\)
- D \(1 \mathrm{k} \Omega, 1 \mathrm{mH}\)
Answer & Solution
Correct Answer
(A) \(1 \mathrm{k} \Omega, 10 \mu \mathrm{F}\)
Step-by-step Solution
Detailed explanation
In the shown figure, current is ahead of voltage, so its a \(R C\) circuit, so \(P\) is a resistor and \(Q\) is a capacitor. Now, in \(R C\) circuit, \[ Z=\sqrt{R^2+X_C^2} \] For simple solution, Given…
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