Question:
A circuit connected to an ac source of emf \(e = e_0\sin(100t)\) with \(t\) in seconds, gives a phase difference of \(\pi/4\) between the emf \(e\) and current \(i\). Which of the following circuits will exhibit this ?
A circuit connected to an ac source of emf \(e = e_0\sin(100t)\) with \(t\) in seconds, gives a phase difference of \(\pi/4\) between the emf \(e\) and current \(i\). Which of the following circuits will exhibit this ?
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A phase difference of \(\pi/4\) needs \(\tan\phi = 1\), i.e. reactance equal to resistance, with \(\omega = 100\ \mathrm{rad/s}\).
Updated On: Nov 5, 2026
- RC circuit with \(R = 1\ \mathrm{k\Omega}\) and \(C = 1\ \mu\mathrm{F}\)
- RL circuit with \(R = 1\ \mathrm{k\Omega}\) and \(L = 1\ \mathrm{mH}\)
- RC circuit with \(R = 1\ \mathrm{k\Omega}\) and \(C = 10\ \mu\mathrm{F}\)
- RL circuit with \(R = 1\ \mathrm{k\Omega}\) and \(L = 10\ \mathrm{mH}\)
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The Correct Option is C
Solution and Explanation
From \(e = e_0\sin(100t)\): \(\omega = 100\ \mathrm{rad\,s^{-1}}\).
For a series RC or RL circuit, \(\tan\phi = \frac{X}{R}\). \(\phi = \pi/4\) means \(X = R = 1000\ \Omega\).
RL circuits: \(X_L = \omega L = 100 \times 1\times10^{-3} = 0.1\ \Omega\) and \(100 \times 10\times10^{-3} = 1\ \Omega\). Neither equals 1000 \(\Omega\), so (2) and (4) are wrong.
RC circuits: \(X_C = \frac{1}{\omega C}\). For \(C = 1\ \mu\mathrm{F}\): \(X_C = \frac{1}{100\times10^{-6}} = 10^4\ \Omega \ne R\). For \(C = 10\ \mu\mathrm{F}\): \(X_C = \frac{1}{100\times10\times10^{-6}} = 10^3\ \Omega = R\).
So the RC circuit with \(R = 1\ \mathrm{k\Omega}\), \(C = 10\ \mu\mathrm{F}\) gives \(\phi = \pi/4\).
Answer: (3) RC circuit with \(R = 1\ \mathrm{k\Omega}\) and \(C = 10\ \mu\mathrm{F}\)
For a series RC or RL circuit, \(\tan\phi = \frac{X}{R}\). \(\phi = \pi/4\) means \(X = R = 1000\ \Omega\).
RL circuits: \(X_L = \omega L = 100 \times 1\times10^{-3} = 0.1\ \Omega\) and \(100 \times 10\times10^{-3} = 1\ \Omega\). Neither equals 1000 \(\Omega\), so (2) and (4) are wrong.
RC circuits: \(X_C = \frac{1}{\omega C}\). For \(C = 1\ \mu\mathrm{F}\): \(X_C = \frac{1}{100\times10^{-6}} = 10^4\ \Omega \ne R\). For \(C = 10\ \mu\mathrm{F}\): \(X_C = \frac{1}{100\times10\times10^{-6}} = 10^3\ \Omega = R\).
So the RC circuit with \(R = 1\ \mathrm{k\Omega}\), \(C = 10\ \mu\mathrm{F}\) gives \(\phi = \pi/4\).
Answer: (3) RC circuit with \(R = 1\ \mathrm{k\Omega}\) and \(C = 10\ \mu\mathrm{F}\)
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