Looking for classes? Ksquare Career Institute, Bengaluru →
In a circuit L, C and R are connected in series with an alternating voltage source of frequency $f$. The current leads the voltage by $45^\circ$. The value of C is -
A
$\dfrac{1}{2\pi f(2\pi fL - R)}$
B
$\dfrac{1}{2\pi f(2\pi fL + R)}$
C
$\dfrac{1}{\pi f(2\pi fL - R)}$
D
$\dfrac{1}{\pi f(2\pi fL + R)}$
Detailed Solution
In a series LCR circuit, $\tan\phi = \dfrac{X_C - X_L}{R}$, where $\phi$ is the angle by which the current leads the voltage.
The current leads the voltage, so the circuit is capacitive: $X_C \gt X_L$.
Given $\phi = 45^\circ$: $\tan 45^\circ = 1 = \dfrac{X_C - X_L}{R}$
Here $X_C - X_L = R$
$\Rightarrow \dfrac{1}{2\pi fC} = R + 2\pi fL$
$\Rightarrow C = \dfrac{1}{2\pi f(2\pi fL + R)}$
The current leads the voltage, so the circuit is capacitive: $X_C \gt X_L$.
Given $\phi = 45^\circ$: $\tan 45^\circ = 1 = \dfrac{X_C - X_L}{R}$
Here $X_C - X_L = R$
$\Rightarrow \dfrac{1}{2\pi fC} = R + 2\pi fL$
$\Rightarrow C = \dfrac{1}{2\pi f(2\pi fL + R)}$