In a series L, C and R circuit with an ac source of frequency \( \omega \), the current leads the voltage by \( \pi/4 \). The value of capacitance C will be
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( (\omega^{2}L + \omega R)^{-1} \)
In a series LCR circuit the phase angle between the current and the voltage is given by
\( \displaystyle \tan\phi = \frac{X_C - X_L}{R} \)
where the current leads the voltage when \( X_C \gt X_L \) (the circuit is capacitive).
\( \phi = \pi/4 \Rightarrow \tan\phi = 1 \)
\( X_C - X_L = R \)
\( \displaystyle \frac{1}{\omega C} - \omega L = R \;\Rightarrow\; \frac{1}{\omega C} = \omega L + R \)
\( \displaystyle C = \frac{1}{\omega(\omega L + R)} = (\omega^{2}L + \omega R)^{-1} \)