In an `AC` circuit, `V` and `I` are given by `V=100sin(100t)volts, I=100sin(100t+(pi)/3)mA`. The power dissipated in circuit is

Options

A. 100 W
B. 10 W
C. 5 W
D. 2.5 W

Explanation

To find the power dissipated in the given AC circuit, we will follow these steps:

### Step-by-Step Solution:

1. **Identify the given values:**
- Voltage: \( V = 100 \sin(100t) \) volts
- Current: \( I = 100 \sin(100t + \frac{\pi}{3}) \) mA

2. **Determine the peak values:**
- The peak voltage \( V_0 \) is 100 volts.
- The peak current \( I_0 \) is 100 mA, which is \( 100 \times 10^{-3} \) A = 0.1 A.

3. **Calculate the RMS values:**
- The RMS value of voltage \( V_{rms} \) is given by:
\[ V_{rms} = \frac{V_0}{\sqrt{2}} = \frac{100}{\sqrt{2}} \text{ volts} \]
- The RMS value of current \( I_{rms} \) is given by:
\[ I_{rms} = \frac{I_0}{\sqrt{2}} = \frac{0.1}{\sqrt{2}} \text{ A} \]

4. **Calculate the phase difference \( \phi \):**
- The phase difference \( \phi \) is given as \( \frac{\pi}{3} \).

5. **Use the power formula:**
- The average power \( P \) dissipated in the circuit can be calculated using the formula:
\[ P = V_{rms} \times I_{rms} \times \cos(\phi) \]
- We need to find \( \cos\left(\frac{\pi}{3}\right) \):
\[ \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \]

6. **Substitute the values into the power formula:**
- Substitute \( V_{rms} \), \( I_{rms} \), and \( \cos(\phi) \):
\[ P = \left(\frac{100}{\sqrt{2}}\right) \times \left(\frac{0.1}{\sqrt{2}}\right) \times \frac{1}{2} \]
- Simplifying this:
\[ P = \frac{100 \times 0.1}{2} \times \frac{1}{2} = \frac{10}{2} = 5 \text{ watts} \]

7. **Final Calculation:**
- Therefore, the power dissipated in the circuit is:
\[ P = 2.5 \text{ watts} \]

### Conclusion:
The power dissipated in the circuit is **2.5 watts**.