What is the frequency of power in an AC circuit connected to a capacitor. Given f = 50 Hz, V = 220 V

Options

A. 50 Hz
B. 100 Hz
C. Zero
D. 60 Hz

Answer

To find the frequency of power in an AC circuit connected to a capacitor, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Given Information**:
- Frequency of the AC source, \( f = 50 \, \text{Hz} \)
- Voltage applied, \( V = 220 \, \text{V} \)

2. **Write the Voltage Equation**:
- The voltage in an AC circuit can be expressed as:
\[ V(t) = V_0 \sin(\omega t) \]
- Here, \( V_0 \) is the peak voltage, and \( \omega \) (angular frequency) is given by:
\[ \omega = 2\pi f \]

3. **Determine the Current through the Capacitor**:
- In a capacitor, the current leads the voltage by 90 degrees. Therefore, the current can be expressed as:
\[ I(t) = I_0 \sin\left(\omega t + \frac{\pi}{2}\right) \]
- This can also be rewritten using the cosine function:
\[ I(t) = I_0 \cos(\omega t) \]

4. **Calculate Power in the Circuit**:
- The instantaneous power \( P(t) \) in the circuit is given by:
\[ P(t) = V(t) \cdot I(t) \]
- Substituting the expressions for voltage and current:
\[ P(t) = V_0 \sin(\omega t) \cdot I_0 \cos(\omega t) \]

5. **Use Trigonometric Identity**:
- Using the identity \( \sin(x) \cos(x) = \frac{1}{2} \sin(2x) \):
\[ P(t) = \frac{V_0 I_0}{2} \sin(2\omega t) \]

6. **Determine the Frequency of Power**:
- The frequency of the power \( f' \) is related to \( 2\omega \):
\[ 2\omega = 2(2\pi f) = 4\pi f \]
- Therefore, the frequency of power is:
\[ f' = 2f \]

7. **Substituting the Given Frequency**:
- Now substituting the value of \( f \):
\[ f' = 2 \times 50 \, \text{Hz} = 100 \, \text{Hz} \]

### Final Answer:
The frequency of power in the AC circuit connected to a capacitor is \( \mathbf{100 \, Hz} \).

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