Question

A coil has a resistance of 10 Ω and inductance of 0.4 Henry. It is connected to an AC source of 6.5 V, 30 Hz. Find the average power consumed in the circuit.

Hint:

In an R-L circuit, the average power consumed is given by \( P_{\text{avg}} = I_{\text{rms}}^2 R \), where \( I_{\text{rms}} \) is the RMS current.

Answer 1

Given:
- Resistance \( R = 10 \, \Omega \)
- Inductance \( L = 0.4 \, \text{H} \)
- AC Voltage \( V = 6.5 \, \text{V} \)
- Frequency \( f = 30 \, \text{Hz} \) The inductive reactance \( X_L \) is given by: \[ X_L = 2\pi f L = 2\pi \times 30 \times 0.4 = 75.4 \, \Omega. \] The impedance \( Z \) of the coil is: \[ Z = \sqrt{R^2 + X_L^2} = \sqrt{10^2 + 75.4^2} = \sqrt{100 + 5684.16} = 75.7 \, \Omega. \] The RMS value of the current \( I_{\text{rms}} \) is: \[ I_{\text{rms}} = \frac{V}{Z} = \frac{6.5}{75.7} = 0.086 \, \text{A}. \] The average power consumed is: \[ P_{\text{avg}} = I_{\text{rms}}^2 R = (0.086)^2 \times 10 = 0.074 \, \text{W}. \]