The current and voltage functions in an `AC` circuit are
`i=100 sin 100 tmA, V=100sin(100t+pi/3)V`
The power disspitated in the circuit is

Options

A. `10 W`
B. `2.5W`
C. `5W`
D. `5kW`

Explanation

To find the power dissipated in the AC circuit given the current and voltage functions, we can follow these steps:

### Step 1: Identify the peak current and voltage
The given current function is:
\[ i = 100 \sin(100t) \, \text{mA} \]
The peak current \( I_0 \) is:
\[ I_0 = 100 \, \text{mA} = 0.1 \, \text{A} \]

The given voltage function is:
\[ V = 100 \sin(100t + \frac{\pi}{3}) \, \text{V} \]
The peak voltage \( V_0 \) is:
\[ V_0 = 100 \, \text{V} \]

### Step 2: Calculate the RMS values of current and voltage
The RMS (Root Mean Square) values are calculated as follows:
\[ I_{\text{rms}} = \frac{I_0}{\sqrt{2}} = \frac{0.1}{\sqrt{2}} = \frac{0.1}{1.414} \approx 0.0707 \, \text{A} \]
\[ V_{\text{rms}} = \frac{V_0}{\sqrt{2}} = \frac{100}{\sqrt{2}} = \frac{100}{1.414} \approx 70.7 \, \text{V} \]

### Step 3: Determine the phase difference
From the voltage function, we see that the phase difference \( \phi \) is:
\[ \phi = \frac{\pi}{3} \, \text{radians} \]
To convert this to degrees:
\[ \phi = \frac{180}{3} = 60^\circ \]

### Step 4: Calculate the power factor
The power factor is given by:
\[ \cos(\phi) = \cos(60^\circ) = \frac{1}{2} \]

### Step 5: Calculate the average power dissipated
The average power \( P \) dissipated in the circuit can be calculated using the formula:
\[ P = V_{\text{rms}} \times I_{\text{rms}} \times \cos(\phi) \]
Substituting the values:
\[ P = (70.7 \, \text{V}) \times (0.0707 \, \text{A}) \times \left(\frac{1}{2}\right) \]
\[ P = 70.7 \times 0.0707 \times 0.5 \]
\[ P \approx 2.5 \, \text{W} \]

Thus, the power dissipated in the circuit is approximately:
\[ \boxed{2.5 \, \text{W}} \]