Q.
When an AC source of emf $e = E_{0}\sin{(100t})$ is connected across a circuit, the phase difference between the emf e and the current i in the circuit is observed to be $\frac{\pi}{4}$ , as shown in the diagram. If the circuit consists possibly only of R – C or R – L or L – C in series, find the relationship between the two elements
A) $R = 1\ k\ \Omega,\ C = 10\ \mu F$
B) $R = 1\ k\ \Omega,\ C = 1\ \mu F$
C) $R = 1\ k\ \Omega,\ L = 10\ H$
D) $R = 1\ k\ \Omega,\ L = 1\ H$
Explanation
(a) As the current i leads the emf e by $\frac{\pi}{4}$, it is an R – C circuit
$\tan{\phi = \frac{X_{C}}{R}}$
or $\tan\frac{\pi}{4} = \frac{\frac{1}{\omega C}}{R}$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \omega CR = 1$
As $\omega = 100\ rads^{- 1}$
The product of C – R should be $\frac{1}{100}s^{- 1}$.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.