Question

In a circuit \(\mathrm{L}, \mathrm{C}\) and \(\mathrm{R}\) are connected in series with an alternating voltage source of frequency \(\mathrm{f}\). The current leads the voltage by \(45^{\circ}\). The value of \(c\) is, (a) \([1 /\\{2 \pi \mathrm{f}(2 \pi \mathrm{fL}+\mathrm{R})\\}]\) (b) \([1 /\\{\pi f(2 \pi \mathrm{fL}+R)\\}]\) (c) \([1 /\\{2 \pi f(2 \pi f L-R)\\}]\) (d) \([1 /\\{\pi \mathrm{f}(2 \pi \mathrm{fL}-\mathrm{R})\\}]\)

Short answer

The value of capacitance (C) in the given circuit is \(C = [1 / \{2 \pi f(2 \pi f L-R)\}]\), which corresponds to option (c).

Step-by-step solution

Phase Angle of the Circuit

In a series RLC circuit, the phase angle (ϕ) determines how much the current leads or lags the voltage. It can be calculated using the formula: \(ϕ = arctan(\dfrac{X_L - X_C}{R})\) We know the current leads the voltage by 45°, so the phase angle is: \(ϕ = +45^{\circ}\)

Convert Degrees to Radians

To continue calculations, we need to convert the phase angle from degrees to radians: \(ϕ = +45^{\circ} \times \dfrac{\pi}{180^\circ} = \dfrac{\pi}{4}\) radians

Calculate Reactances of Inductor and Capacitor

In an RLC circuit, the reactances of the inductor (L) and capacitor (C) are given by: \(X_L = 2 \pi f L\) \(X_C = \dfrac{1}{2 \pi f C}\)

Plug the Values into Phase Angle Formula

Use the given phase angle formula and the ϕ and reactances we calculated: \(\dfrac{\pi}{4} = arctan(\dfrac{2 \pi f L - \dfrac{1}{2 \pi f C}}{R})\)

Solve for Capacitance (C)

To find the value of C, we need to solve the above equation. First, we use the tan function to get rid of arctan: \(tan(\dfrac{\pi}{4}) = \dfrac{2 \pi f L - \dfrac{1}{2 \pi f C}}{R}\) Since \(tan(\dfrac{\pi}{4}) = 1\), we have: \(1 = \dfrac{2 \pi f L - \dfrac{1}{2 \pi f C}}{R}\) Now solve for C: \(2 \pi f C(2 \pi f L - \dfrac{1}{2 \pi f C}) = R(2 \pi f C)\) \(4 \pi^2 f^2 L C^2 - 1 = 2 \pi f CR\) We want the value of C, divide by \(2 \pi f(2 \pi f L - R)\) to get: \(C = [1 / \{2 \pi f(2 \pi f L-R)\}]\) The answer is option (c).