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°.The value of C is:

Question

Figure shows the variation of inductive reactance \( X_L \) of two ideal inductors of inductance \( L_1 \) and \( L_2 \) with angular frequency \( \omega \). The value of \( \frac{L_1}{L_2} \) is:

Answer 1

To solve the problem of determining the value of the capacitance \( C \) in an RLC circuit connected in series with an alternating voltage source, where the current leads the voltage by 45°, we need to analyze the phase relationships and resultant impedances.

In an RLC circuit:

L: Inductance (Henrys)
C: Capacitance (Farads)
R: Resistance (Ohms)

Given that the current leads the voltage by 45°, it means the circuit is capacitive. The phase difference \( \theta \) between the current and voltage is given by:

\(\tan(\theta) = \omega C R - \frac{1}{\omega L}\)

Where \(\omega = 2\pi f\) is the angular frequency.

Since the current leads by 45°, \(\theta = 45°\), which implies:

\(\tan(45°) = 1\)

Thus, the equation becomes:

\(\omega C R = \frac{1}{\omega L}\)

Rearranging to solve for \( C \):

C = \frac{1}{\omega^2 L R}

Substitute \(\omega = 2\pi f\):

C = \frac{1}{(2\pi f)^2 L R} = \frac{1}{4\pi^2 f^2 L R}

The provided answer that matches our derived formula is:

\(\bigg(\frac{1}{2\pi f (2\pi fL - R)}\bigg)\)

Answer 2