In an AC circuit, a capacitor of capacitance `C=(25)/(pi) mu F` and a resistor of resistance `^(4) R=300 Omega` are connected in series with an `A C` source of `200 V` and `50 s^(-1)` frequency. The power dissipated (in watt) in the circuit will be

Answer

To solve the problem of finding the power dissipated in the given AC circuit with a capacitor and a resistor, we will follow these steps:

### Step 1: Identify the given values
- Capacitance, \( C = \frac{25}{\pi} \, \mu F = \frac{25 \times 10^{-6}}{\pi} \, F \)
- Resistance, \( R = 300 \, \Omega \)
- Voltage, \( V = 200 \, V \)
- Frequency, \( f = 50 \, s^{-1} \)

### Step 2: Calculate the capacitive reactance (\( X_C \))
The capacitive reactance is given by the formula:
\[ X_C = \frac{1}{2 \pi f C} \]
Substituting the values:
\[ X_C = \frac{1}{2 \pi (50) \left(\frac{25 \times 10^{-6}}{\pi}\right)} = \frac{1}{2 \times 50 \times 25 \times 10^{-6}} = \frac{1}{0.0025} = 400 \, \Omega \]

### Step 3: Calculate the impedance (\( Z \))
The impedance in an RC circuit is given by:
\[ Z = \sqrt{R^2 + X_C^2} \]
Substituting the values:
\[ Z = \sqrt{(300)^2 + (400)^2} = \sqrt{90000 + 160000} = \sqrt{250000} = 500 \, \Omega \]

### Step 4: Calculate the current (\( I \))
Using Ohm's law for AC circuits, the current can be calculated as:
\[ I = \frac{V}{Z} \]
Substituting the values:
\[ I = \frac{200}{500} = 0.4 \, A \]

### Step 5: Calculate the power dissipated (\( P \))
The power dissipated in an AC circuit with a resistor is given by:
\[ P = I^2 R \]
Substituting the values:
\[ P = (0.4)^2 \times 300 = 0.16 \times 300 = 48 \, W \]

### Final Answer
The power dissipated in the circuit is \( 48 \, W \).

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