Question

An inductor of reactance $ 100 \, \Omega $, a capacitor of reactance $ 50 \, \Omega $, and a resistor of resistance $ 50 \, \Omega $ are connected in series with an AC source of $ 10 \, V $, $ 50 \, Hz $. Average power dissipated by the circuit is ____ W.

Hint:

In an AC circuit containing resistors, inductors, and capacitors, only the resistor dissipates average power. The inductor and capacitor store and release energy but do not dissipate it on average over a complete cycle. The power dissipated is calculated using the RMS current and the resistance.

Answer 1

Correct Answer: 1

To ascertain the circuit's average power dissipation, the total impedance \( Z \) of the series circuit is computed first. The inductive reactance \( X_L \) is \( 100 \, \Omega \), the capacitive reactance \( X_C \) is \( 50 \, \Omega \), and the resistance \( R \) is \( 50 \, \Omega \). The total impedance \( Z \) for a series circuit is calculated using the formula: \( Z = \sqrt{R^2 + (X_L - X_C)^2} \). Substituting the given values yields: \( Z = \sqrt{50^2 + (100 - 50)^2} = \sqrt{50^2 + 50^2} = \sqrt{2500 + 2500} = \sqrt{5000} = 50\sqrt{2} \, \Omega \). The AC source provides a voltage \( V \) of \( 10 \, V \). The root mean square (RMS) current \( I \) is determined by \( I = \frac{V}{Z} \). Therefore, \( I = \frac{10}{50\sqrt{2}} = \frac{1}{5\sqrt{2}} \, \text{A} \). The average power dissipated by a resistor in an AC circuit is given by \( P = I^2 \cdot R \). Using the calculated RMS current and resistance, the power is: \( P = \left(\frac{1}{5\sqrt{2}}\right)^2 \cdot 50 = \frac{1}{50} \cdot 50 = 1 \, \text{W} \). This computed average power of \( 1 \, \text{W} \) is within the expected range of \( 1,1 \). Consequently, the circuit's average power dissipation is 1 W.