Question

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 $\frac{\pi}{4}$ ahead, 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. $R= 1 k \Omega, \, L = 1 H$

Answer 1

Answer: (A)

$R= 1 k \Omega, \, C = 10 \mu F$

Answer 2

As the current i leads the emf e by $\frac{\pi}{4}$, it is an R-C circuit
$tan \phi = \frac{X_c}{R} \, \, or \, \, tan \frac{\pi}{4} = \frac{\frac{1}{? C}}{R}$
$\therefore \, \, \, \, \, \, \, \, \, \, \, \, \, ? CR = 1$
$As \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, ? 100 rad/s$
The product of C-R should be $\frac{1}{100}s^{-1}$
$\therefore$ Correct answer is (a).