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)`W
B. 10 W
C. 2.5 W
D. 5 W

Explanation

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

### Step 1: Identify the given values
We have the voltage and current equations:
- Voltage: \( V = 100 \sin(100t) \) volts
- Current: \( I = 100 \sin(100t + \frac{\pi}{3}) \) mA

From the voltage equation:
- \( V_0 = 100 \) volts (amplitude of voltage)
- \( \omega = 100 \) rad/s (angular frequency)

From the current equation:
- \( I_0 = 100 \) mA = \( 100 \times 10^{-3} \) A (amplitude of current)
- Phase angle \( \phi = \frac{\pi}{3} \) radians

### Step 2: Calculate RMS values
The RMS (Root Mean Square) values for voltage and current are given by:
- \( V_{rms} = \frac{V_0}{\sqrt{2}} = \frac{100}{\sqrt{2}} \)
- \( I_{rms} = \frac{I_0}{\sqrt{2}} = \frac{100 \times 10^{-3}}{\sqrt{2}} \)

### Step 3: Calculate the power factor
The power factor \( \cos(\phi) \) can be calculated using the phase angle:
- \( \phi = \frac{\pi}{3} \)
- \( \cos(\phi) = \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \)

### Step 4: Calculate the power
The average power \( P \) in an AC circuit is given by:
\[ P = V_{rms} \times I_{rms} \times \cos(\phi) \]
Substituting the values we found:
\[ P = \left(\frac{100}{\sqrt{2}}\right) \times \left(\frac{100 \times 10^{-3}}{\sqrt{2}}\right) \times \frac{1}{2} \]

### Step 5: Simplify the expression
Calculating the above expression:
\[ P = \frac{100 \times 100 \times 10^{-3}}{2 \times 2} = \frac{10000 \times 10^{-3}}{4} = \frac{10000}{4} \times 10^{-3} = 2500 \times 10^{-3} = 2.5 \text{ watts} \]

### Final Answer
The power dissipated in the circuit is \( 2.5 \) watts.

---