The instantaneous current and volatage of an AC circuit are given by
`i = 10 sin (314 t) A and V = 100 sin (314 t) V`
What is the power dissipation in the circuit?

Options

A. 1000 W
B. 500 W
C. 300 W
D. 200 W

Answer

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

### Step 1: Identify the given values
The instantaneous current and voltage are given as:
- Current: \( i = 10 \sin(314t) \) A
- Voltage: \( V = 100 \sin(314t) \) V

### Step 2: Determine the peak values
From the equations, we can identify:
- Peak current, \( I_0 = 10 \) A
- Peak voltage, \( V_0 = 100 \) V

### Step 3: Calculate the RMS values
The root mean square (RMS) values for current and voltage are given by:
\[ I_{\text{rms}} = \frac{I_0}{\sqrt{2}} = \frac{10}{\sqrt{2}} \quad \text{and} \quad V_{\text{rms}} = \frac{V_0}{\sqrt{2}} = \frac{100}{\sqrt{2}} \]

### Step 4: Calculate the RMS values
Calculating the RMS values:
\[ I_{\text{rms}} = \frac{10}{\sqrt{2}} \approx 7.07 \, \text{A} \]
\[ V_{\text{rms}} = \frac{100}{\sqrt{2}} \approx 70.71 \, \text{V} \]

### Step 5: Determine the phase difference
In this case, the current and voltage are in phase since they have the same sine function. Therefore, the phase difference \( \phi = 0 \).

### Step 6: Calculate the power factor
The power factor \( \cos \phi \) is:
\[ \cos(0) = 1 \]

### Step 7: Calculate the average power
The average power \( P \) in an AC circuit is given by:
\[ P = V_{\text{rms}} \times I_{\text{rms}} \times \cos \phi \]
Substituting the values:
\[ P = \left(\frac{100}{\sqrt{2}}\right) \times \left(\frac{10}{\sqrt{2}}\right) \times 1 \]
\[ P = \frac{1000}{2} = 500 \, \text{W} \]

### Conclusion
The power dissipation in the circuit is \( 500 \, \text{W} \).

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