In an AC circuit V and I are given by `V = 100 sin 100t V and I = 100 sin (100t + pi//3)` mA. The power dissipated in the circuit is

Options

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

Answer

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

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

### Step 2: Determine the peak values
From the equations, we can identify:
- The peak voltage \( V_0 = 100 \) V
- The peak current \( I_0 = 100 \) mA = \( 100 \times 10^{-3} \) A = \( 0.1 \) A

### Step 3: Calculate the RMS values
The RMS (Root Mean Square) values are calculated using the formula:
- \( V_{rms} = \frac{V_0}{\sqrt{2}} \)
- \( I_{rms} = \frac{I_0}{\sqrt{2}} \)

Calculating these:
- \( V_{rms} = \frac{100}{\sqrt{2}} \)
- \( I_{rms} = \frac{0.1}{\sqrt{2}} \)

### Step 4: Determine the phase difference
The phase difference \( \theta \) between the voltage and current is given as:
- \( \theta = \frac{\pi}{3} \)

### Step 5: Use the power formula
The average power \( P \) dissipated in the circuit can be calculated using the formula:
\[ P = V_{rms} \cdot I_{rms} \cdot \cos(\theta) \]

### Step 6: Substitute the values into the power formula
Substituting the RMS values and the cosine of the phase difference:
\[ P = \left(\frac{100}{\sqrt{2}}\right) \cdot \left(\frac{0.1}{\sqrt{2}}\right) \cdot \cos\left(\frac{\pi}{3}\right) \]

### Step 7: Calculate \( \cos\left(\frac{\pi}{3}\right) \)
We know that:
\[ \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \]

### Step 8: Simplify the power equation
Now substituting \( \cos\left(\frac{\pi}{3}\right) \):
\[ P = \left(\frac{100}{\sqrt{2}}\right) \cdot \left(\frac{0.1}{\sqrt{2}}\right) \cdot \frac{1}{2} \]
\[ P = \frac{100 \cdot 0.1}{2} \cdot \frac{1}{2} = \frac{10}{2} = 5 \text{ watts} \]

### Step 9: Final calculation
Thus, the power dissipated in the circuit is:
\[ P = 2.5 \text{ watts} \]