When the frequency of AC is doubled, the impedance of an RC circuit is

Options

A. doubled
B. halved
C. increases
D. decreases

Answer

To solve the problem of how the impedance of an RC circuit changes when the frequency of AC is doubled, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Impedance Formula**:
The impedance \( Z \) of an RC circuit is given by the formula:
\[ Z = \sqrt{R^2 + X_C^2} \]
where \( R \) is the resistance and \( X_C \) is the capacitive reactance.

2. **Identify the Capacitive Reactance**:
The capacitive reactance \( X_C \) is defined as:
\[ X_C = \frac{1}{\omega C} \]
where \( \omega = 2\pi f \) is the angular frequency and \( C \) is the capacitance.

3. **Substituting for Angular Frequency**:
If the frequency \( f \) is doubled, the new frequency \( f' \) becomes:
\[ f' = 2f \]
Consequently, the new angular frequency \( \omega' \) will be:
\[ \omega' = 2\omega = 4\pi f \]

4. **Calculate New Capacitive Reactance**:
The new capacitive reactance \( X_C' \) at the doubled frequency is:
\[ X_C' = \frac{1}{\omega' C} = \frac{1}{4\pi f C} = \frac{1}{4} X_C \]

5. **Substituting into the Impedance Formula**:
Now, substituting \( X_C' \) into the impedance formula gives:
\[ Z' = \sqrt{R^2 + (X_C')^2} = \sqrt{R^2 + \left(\frac{1}{4} X_C\right)^2} \]
Simplifying this:
\[ Z' = \sqrt{R^2 + \frac{1}{16} X_C^2} \]

6. **Comparing New Impedance with Original Impedance**:
The original impedance \( Z \) was:
\[ Z = \sqrt{R^2 + X_C^2} \]
Since \( \frac{1}{16} X_C^2 < X_C^2 \), we can conclude that:
\[ Z' < Z \]

7. **Conclusion**:
Therefore, when the frequency of AC is doubled, the impedance of the RC circuit decreases.

### Final Answer:
The impedance of the RC circuit decreases when the frequency of AC is doubled.

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