A resistance is connected to a n AC source. If a capacitor is induced in the series cirucit, the average power absorbed by the resistance

Options

A. will increase
B. will decrease
C. may increase or decrease
D. will remain constant

Answer

To solve the problem, we need to analyze the effect of adding a capacitor in series with a resistor connected to an AC source on the average power absorbed by the resistor.

### Step-by-Step Solution:

1. **Understanding the Circuit**:
- We have a resistor \( R \) connected in series with a capacitor \( C \) and an AC source.

2. **Impedance Calculation**:
- The total impedance \( Z \) of the series circuit is given by:
\[ Z = \sqrt{R^2 + X_C^2} \]
where \( X_C \) is the capacitive reactance, defined as:
\[ X_C = \frac{1}{\omega C} \]
with \( \omega \) being the angular frequency of the AC source.

3. **Current Calculation**:
- The root mean square (RMS) current \( I_{\text{rms}} \) through the circuit can be expressed as:
\[ I_{\text{rms}} = \frac{V_{\text{rms}}}{Z} \]
where \( V_{\text{rms}} \) is the RMS voltage of the AC source.

4. **Average Power Calculation**:
- The average power \( P \) absorbed by the resistor is given by:
\[ P = I_{\text{rms}}^2 R \]
Substituting the expression for \( I_{\text{rms}} \):
\[ P = \left(\frac{V_{\text{rms}}}{Z}\right)^2 R \]

5. **Effect of Adding Capacitor**:
- When the capacitor is added, the impedance \( Z \) increases because \( X_C \) is positive. This leads to:
\[ Z = \sqrt{R^2 + \left(\frac{1}{\omega C}\right)^2} \]
- As \( Z \) increases, \( I_{\text{rms}} \) decreases since:
\[ I_{\text{rms}} = \frac{V_{\text{rms}}}{Z} \]

6. **Conclusion on Power**:
- Since \( P \) is proportional to \( I_{\text{rms}}^2 \), a decrease in \( I_{\text{rms}} \) will result in a decrease in the average power \( P \). Therefore, the average power absorbed by the resistance will decrease.

### Final Answer:
The average power absorbed by the resistance will **decrease**.

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