An alternating supply of 220 V is applied across a circuit with resistance `22 Omega` and impedance `44 Omega`. The power dissipated in the circuit is

Options

A. 1100 W
B. 550 W
C. 2200 W
D. (2200/3) W

Explanation

To solve the problem of finding the power dissipated in a circuit with an alternating supply of 220 V, a resistance of 22 ohms, and an impedance of 44 ohms, we can follow these steps:

### Step 1: Identify the given values
- Voltage (V) = 220 V
- Resistance (R) = 22 Ω
- Impedance (Z) = 44 Ω

### Step 2: Calculate the current (I) using Ohm's Law
The current in an AC circuit can be calculated using the formula:
\[ I = \frac{V}{Z} \]

Substituting the known values:
\[ I = \frac{220 \text{ V}}{44 \text{ Ω}} \]
\[ I = 5 \text{ A} \]

### Step 3: Calculate the power dissipated (P)
The power dissipated in an AC circuit with resistance is given by the formula:
\[ P = I^2 \times R \]

Substituting the current and resistance:
\[ P = (5 \text{ A})^2 \times 22 \text{ Ω} \]
\[ P = 25 \text{ A}^2 \times 22 \text{ Ω} \]
\[ P = 550 \text{ W} \]

### Conclusion
The power dissipated in the circuit is **550 watts**.

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