Question

The adjoining diagram represents EMF $e$ and the current $I$ having a phase difference of $\frac{\pi }{4}$ in a circuit which has an AC source of emf $e=E_{0}sin\left(100t\right)V$ connected across it. If the circuit only possibly consists of RC or RL or LC in series, then find the relation between the two possible elements of the circuit.

• A. $R=1k\Omega ,C=10\mu F$
• B. $R=1k\Omega ,C=1\mu F$
• C. $R=1k\Omega ,L=10H$
• D. $R=1k\Omega ,L=1H$

Answer 1

Answer: (A)

The current is leading the voltage in the given graph and the phase difference is not equal to $90^0$ . So, we can conclude that resistor and capacitor can be the probable elements of the circuit.
If we take $R$ to be the resistance and $C$ to be the capacitance and $\varphi$ be the phase difference between current and applied voltage.
For RC circuit,
$tan\varphi =\frac{X_c}{R}$
For $\varphi =45^{{}^{\circ }}$ , $tan\varphi =1$
$\frac{X_c}{R}=1X_c=R\Rightarrow \frac{1}{\omega C}=R$
Given that, $R=1k\Omega ,\omega =100$
Substituting in the above equation, we get
$\frac{1}{100C}=1\times 10^{3}C=10\mu F$