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 determine the average power dissipated by the circuit, we first calculate the total impedance \( Z \) of the series circuit. The inductor's reactance \( X_L \) is \( 100 \, \Omega \), the capacitor's reactance \( X_C \) is \( 50 \, \Omega \), and the resistor's resistance \( R \) is \( 50 \, \Omega \). The formula for total impedance in a series circuit is:

\( Z = \sqrt{R^2 + (X_L - X_C)^2} \)

Substituting the values, we compute:

\( Z = \sqrt{50^2 + (100 - 50)^2} = \sqrt{50^2 + 50^2} = \sqrt{2500 + 2500} = \sqrt{5000} \)

\( Z = 50\sqrt{2} \, \Omega \)

The voltage \( V \) supplied by the AC source is \( 10 \, V \). The root mean square (RMS) current \( I \) is given by \( I = \frac{V}{Z} \):

\( I = \frac{10}{50\sqrt{2}} = \frac{1}{5\sqrt{2}} \, \text{A} \)

The average power dissipated by a resistive element in an AC circuit is determined by the formula:

\( P = I^2 \cdot R \)

Plugging in the RMS current and resistance values, we calculate:

\( P = \left(\frac{1}{5\sqrt{2}}\right)^2 \cdot 50 = \frac{1}{50} \cdot 50 = 1 \, \text{W} \)

The computed average power dissipated, \( 1 \, \text{W} \), falls within the expected range of \( 1,1 \). Hence, the average power dissipated by the circuit is 1 W.

Answer 2

Step 1: Identify the given parameters.
Inductive reactance, \( X_L = 100 \, \Omega \) 
Capacitive reactance, \( X_C = 50 \, \Omega \) 
Resistance, \( R = 50 \, \Omega \) 
RMS voltage of the AC source, \( V_{rms} = 10 \, V \) 
Frequency of the AC source, \( f = 50 \, Hz \) 
Step 2: Calculate the impedance \( Z \) of the series LCR circuit.
The impedance of a series LCR circuit is given by: \[ Z = \sqrt{R^2 + (X_L - X_C)^2} \] Substitute the given values: \[ Z = \sqrt{(50 \, \Omega)^2 + (100 \, \Omega - 50 \, \Omega)^2} \] \[ Z = \sqrt{(50)^2 + (50)^2} = \sqrt{2500 + 2500} = \sqrt{5000} \, \Omega \] \[ Z = 50\sqrt{2} \, \Omega \] 
Step 3: Calculate the RMS current \( I_{rms} \) in the circuit.
Using Ohm's law for AC circuits: \[ I_{rms} = \frac{V_{rms}}{Z} \] Substitute the values of \( V_{rms} \) and \( Z \): \[ I_{rms} = \frac{10 \, V}{50\sqrt{2} \, \Omega} = \frac{1}{5\sqrt{2}} \, A = \frac{\sqrt{2}}{10} \, A \] 
Step 4: Calculate the average power \( P_{avg} \) dissipated by the circuit.
The average power dissipated in an AC circuit is only through the resistor and is given by: \[ P_{avg} = I_{rms}^2 R \] Substitute the values of \( I_{rms} \) and \( R \): \[ P_{avg} = \left(\frac{\sqrt{2}}{10} \, A\right)^2 \times 50 \, \Omega \] \[ P_{avg} = \left(\frac{2}{100}\right) \times 50 \, W \] \[ P_{avg} = \frac{1}{50} \times 50 \, W \] \[ P_{avg} = 1 \, W \] The average power dissipated by the circuit is 1 W.