110`underset(rms)(V)` is applied across a series circuit having resistance `11 Omega` and impedance `22 Omega` . The power consumed is

Options

A. 275 W
B. 366 W
C. 550 W
D. 1100 W

Explanation

To solve the problem of finding the power consumed in a series circuit with a given RMS voltage, resistance, and impedance, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values:**
- RMS Voltage (\( V_{rms} \)) = 110 V
- Resistance (\( R \)) = 11 Ω
- Impedance (\( Z \)) = 22 Ω

2. **Use the Power Formula:**
The power consumed in an AC circuit can be calculated using the formula:
\[ P = V_{rms} \cdot I_{rms} \cdot \cos(\phi) \]
where \( \cos(\phi) \) is the power factor.

3. **Relate \( I_{rms} \) to \( V_{rms} \) and \( Z \):**
The current \( I_{rms} \) can be expressed as:
\[ I_{rms} = \frac{V_{rms}}{Z} \]

4. **Determine the Power Factor \( \cos(\phi) \):**
The power factor \( \cos(\phi) \) can be calculated using:
\[ \cos(\phi) = \frac{R}{Z} \]

5. **Substituting Values:**
Now substitute the values into the power formula:
\[ P = V_{rms} \cdot \left(\frac{V_{rms}}{Z}\right) \cdot \left(\frac{R}{Z}\right) \]
This simplifies to:
\[ P = \frac{V_{rms}^2 \cdot R}{Z^2} \]

6. **Plug in the Values:**
Substitute \( V_{rms} = 110 \, V \), \( R = 11 \, \Omega \), and \( Z = 22 \, \Omega \):
\[ P = \frac{(110)^2 \cdot 11}{(22)^2} \]

7. **Calculate \( V_{rms}^2 \):**
\[ (110)^2 = 12100 \]

8. **Calculate \( Z^2 \):**
\[ (22)^2 = 484 \]

9. **Substituting Back:**
\[ P = \frac{12100 \cdot 11}{484} \]

10. **Perform the Multiplication:**
\[ 12100 \cdot 11 = 133100 \]

11. **Final Calculation:**
\[ P = \frac{133100}{484} \approx 275 \, W \]

### Final Answer:
The power consumed in the circuit is approximately **275 Watts**.