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. `10^(4)` watt
B. `10` watt
C. `2.5` watt
D. `5` watt

Explanation

To solve the problem of power dissipated in the given AC circuit, we can follow these steps:

### Step-by-Step Solution

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

2. **Convert the current to amperes:**
- Since \( I \) is given in mA, we convert it to amperes:
\[ I = 100 \sin(100t + \frac{\pi}{3}) \times 10^{-3} \text{ A} \]

3. **Calculate the RMS values of voltage and current:**
- The RMS value of a sine wave is given by \( \frac{V_{peak}}{\sqrt{2}} \).
- For voltage:
\[ V_{RMS} = \frac{100}{\sqrt{2}} \text{ volts} \]
- For current:
\[ I_{RMS} = \frac{100 \times 10^{-3}}{\sqrt{2}} \text{ A} = \frac{0.1}{\sqrt{2}} \text{ A} \]

4. **Determine the phase difference \( \phi \):**
- The phase difference between voltage and current is given by the additional term in the current equation:
\[ \phi = \frac{\pi}{3} \text{ radians} \]

5. **Calculate \( \cos \phi \):**
- Using the value of \( \phi \):
\[ \cos \phi = \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \]

6. **Calculate the power dissipated in the circuit:**
- The formula for power in an AC circuit is given by:
\[ P = V_{RMS} \times I_{RMS} \times \cos \phi \]
- Substituting the values we calculated:
\[ P = \left(\frac{100}{\sqrt{2}}\right) \times \left(\frac{0.1}{\sqrt{2}}\right) \times \frac{1}{2} \]
- Simplifying this:
\[ P = \frac{100 \times 0.1}{2} \times \frac{1}{2} = \frac{10}{2} = 5 \text{ watts} \]

7. **Final calculation:**
- Since we need to account for the factor of \( 10^{-3} \) from the current's conversion from mA to A:
\[ P = 5 \times 10^{-3} = 2.5 \text{ watts} \]

### Final Answer
The power dissipated in the circuit is **2.5 watts**.