A circuit connected to an ac source of emf \( e=e_0\sin(100t) \), with t in seconds, shows a phase difference of \( \pi/4 \) between the emf e and the current i. Which of the following circuits will show this?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → RC circuit with \( R=1\,\mathrm{k\Omega} \) and \( C=10\,\mu\mathrm{F} \)
In a series circuit the phase difference \( \phi \) between the emf and the current is given by \( \tan\phi=(X_C-X_L)/R \). With only one reactive element this becomes \( \tan\phi=X/R \), where \( X \) is the inductive reactance \( \omega L \) or the capacitive reactance \( 1/\omega C \).
A phase difference of \( \pi/4 \) needs \( \tan\phi=1 \), that is, a reactance equal to the resistance, \( R=1\,\mathrm{k\Omega}=1000\,\Omega \). From the emf, \( \omega=100\,\mathrm{rad\,s^{-1}} \).
Option (1) RL circuit with \( R=1\,\mathrm{k\Omega} \) and \( L=10\,\mathrm{mH} \) — Incorrect. Here \( X_L=100\times0.01\,\Omega \), only \( 1\,\Omega \). This is far below \( R \), so \( \tan\phi=0.001 \) and the phase difference is almost zero.
Option (2) RL circuit with \( R=1\,\mathrm{k\Omega} \) and \( L=1\,\mathrm{mH} \) — Incorrect. Here \( X_L=100\times0.001\,\Omega \), only \( 0.1\,\Omega \), so \( \tan\phi=10^{-4} \) and the current is almost in phase with the emf.
Option (3) RC circuit with \( R=1\,\mathrm{k\Omega} \) and \( C=1\,\mu\mathrm{F} \) — Incorrect. Here \( X_C=1/(100\times10^{-6})\,\Omega \), that is, \( 10^{4}\,\Omega \). This is ten times \( R \), so \( \tan\phi=10 \) and \( \phi \) is about \( 84^\circ \).
Option (4) RC circuit with \( R=1\,\mathrm{k\Omega} \) and \( C=10\,\mu\mathrm{F} \) — Correct. Here \( X_C=1/(100\times10^{-5})\,\Omega \), that is, \( 1000\,\Omega \), equal to the resistance. Hence \( \tan\phi=1 \) and \( \phi=\pi/4 \), with the current leading the emf.