Question

In the given figure, LCR circuit is shown. Voltage of the alternating current source is \( V = 100 \sin (500 t) \) volt. Calculate for the circuit:

Hint:

In an L-C-R circuit, the impedance depends on the resistance, inductive reactance, and capacitive reactance. The power factor is the cosine of the phase angle, and the peak current is determined by the peak voltage and total impedance.

Answer 1

(i) Total impedance
The total impedance \( Z \) of an L-C-R series circuit is given by: \[ Z = \sqrt{R^2 + \left(X_L - X_C \right)^2}, \] where: - \( R = 40 \, \Omega \) is the resistance,
- \( X_L = \omega L \) is the inductive reactance,
- \( X_C = \frac{1}{\omega C} \) is the capacitive reactance. Given: - \( L = 0.16 \, \text{H} \), - \( C = 40 \, \mu\text{F} = 40 \times 10^{-6} \, \text{F} \), - \( \omega = 500 \, \text{rad/s} \) (since the frequency \( f = \frac{500}{2\pi} \)). First, calculate \( X_L \) and \( X_C \): \[ X_L = \omega L = 500 \times 0.16 = 80 \, \Omega, \] \[ X_C = \frac{1}{\omega C} = \frac{1}{500 \times 40 \times 10^{-6}} = 50 \, \Omega. \] Now, substitute these values into the impedance formula: \[ Z = \sqrt{40^2 + (80 - 50)^2} = \sqrt{1600 + 900} = \sqrt{2500} = 50 \, \Omega. \] So, the total impedance is \( Z = 50 \, \Omega \). (ii) Power Factor:
The power factor (PF) in a series L-C-R circuit is given by: \[ \text{PF} = \cos \phi = \frac{R}{Z}, \] where: - \( R = 40 \, \Omega \), - \( Z = 50 \, \Omega \). So, the power factor is: \[ \text{PF} = \frac{40}{50} = 0.8. \] (iii) Peak Value of Current:
The peak current \( I_0 \) in the circuit is given by: \[ I_0 = \frac{V_0}{Z}, \] where: - \( V_0 = 100 \, \text{V} \) is the peak voltage,
- \( Z = 50 \, \Omega \) is the total impedance. So, the peak value of the current is: \[ I_0 = \frac{100}{50} = 2 \, \text{A}. \]