In an a.c circuit, `V` & I are given by `V = 100 sin (100 t)` volt.
`I = 100 sin (100 t + (pi)/(3)) mA`
The power dissipated in the circuit is:

Options

A. `10^(4)` watt
B. 10 watt
C. 2.5 watt
D. 5.0 watt

Answer

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

### Step 1: Identify the given values
The voltage \( V \) and current \( I \) are given as:
- \( V = 100 \sin(100t) \) volts
- \( I = 100 \sin(100t + \frac{\pi}{3}) \) mA

From this, we can extract:
- The peak voltage \( V_0 = 100 \) volts
- The peak current \( I_0 = 100 \) mA = \( 0.1 \) A

### Step 2: Determine the phase difference
The phase difference \( \phi \) between the voltage and current can be calculated as:
- \( \phi = \frac{\pi}{3} - 0 = \frac{\pi}{3} \) radians

### Step 3: Use the power formula
The average power \( P \) dissipated in an AC circuit can be calculated using the formula:
\[ P = \frac{V_0 I_0}{2} \cos(\phi) \]

### Step 4: Substitute the values into the power formula
Substituting the known values into the formula:
- \( V_0 = 100 \) volts
- \( I_0 = 0.1 \) A
- \( \phi = \frac{\pi}{3} \)

The power becomes:
\[ P = \frac{100 \times 0.1}{2} \cos\left(\frac{\pi}{3}\right) \]

### Step 5: Calculate \( \cos\left(\frac{\pi}{3}\right) \)
We know that:
\[ \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \]

### Step 6: Substitute and simplify
Now substituting \( \cos\left(\frac{\pi}{3}\right) \) into the power equation:
\[ P = \frac{100 \times 0.1}{2} \times \frac{1}{2} \]
\[ P = \frac{10}{2} \times \frac{1}{2} = 5 \times \frac{1}{2} = 2.5 \text{ watts} \]

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

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