In an a.c. circuit, voltage and current are given by :
`V = 100 sin (100 t) V` and
`I = 100 sin (100 t + (pi)/3) mA` respectively. The average power dissipated in one cycle is :

Options

A. `5W`
B. `10 W`
C. `2.5 W`
D. `25 W`

Answer

To find the average power dissipated in one cycle of the given AC circuit, we will use the formula for average power in an AC circuit:

\[ P = V_{rms} \cdot I_{rms} \cdot \cos(\phi) \]

where:
- \( V_{rms} \) is the root mean square voltage,
- \( I_{rms} \) is the root mean square current,
- \( \phi \) is the phase difference between the voltage and current.

### Step 1: Find \( V_{rms} \)

The given voltage is:

\[ V = 100 \sin(100t) \, \text{V} \]

The peak voltage \( V_0 \) is 100 V. The root mean square voltage \( V_{rms} \) is given by:

\[ V_{rms} = \frac{V_0}{\sqrt{2}} = \frac{100}{\sqrt{2}} \]

### Step 2: Find \( I_{rms} \)

The given current is:

\[ I = 100 \sin(100t + \frac{\pi}{3}) \, \text{mA} \]

The peak current \( I_0 \) is 100 mA. The root mean square current \( I_{rms} \) is given by:

\[ I_{rms} = \frac{I_0}{\sqrt{2}} = \frac{100 \times 10^{-3}}{\sqrt{2}} \]

### Step 3: Find the phase difference \( \phi \)

The phase difference \( \phi \) between the voltage and current is given as \( \frac{\pi}{3} \) radians.

### Step 4: Calculate \( \cos(\phi) \)

Now, we need to find \( \cos(\phi) \):

\[ \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \]

### Step 5: Substitute values into the power formula

Now we can substitute all the values into the power formula:

\[ P = V_{rms} \cdot I_{rms} \cdot \cos(\phi) \]

Substituting the values we found:

\[ P = \left(\frac{100}{\sqrt{2}}\right) \cdot \left(\frac{100 \times 10^{-3}}{\sqrt{2}}\right) \cdot \frac{1}{2} \]

### Step 6: Simplify the expression

Calculating this step by step:

1. Multiply the \( V_{rms} \) and \( I_{rms} \):

\[ P = \frac{100 \cdot 100 \times 10^{-3}}{2} \cdot \frac{1}{2} \]

2. Combine the terms:

\[ P = \frac{10000 \times 10^{-3}}{2 \cdot 2} = \frac{10}{4} = 2.5 \, \text{W} \]

### Final Answer

The average power dissipated in one cycle is:

\[ \boxed{2.5 \, \text{W}} \]