In an AC circuit the instantaneous values of emf and current are
`e=200sin300t` volt and `i=2sin(300t+(pi)/(3))` amp The average power consumed (in watts) is

Options

A. 200
B. 100
C. 50
D. 400

Answer

To solve the problem, we need to calculate the average power consumed in the given AC circuit using the provided instantaneous values of emf (e) and current (i).

### Step-by-Step Solution:

1. **Identify the Given Values:**
- The instantaneous emf is given by:
\[ e = 200 \sin(300t) \text{ volts} \]
- The instantaneous current is given by:
\[ i = 2 \sin\left(300t + \frac{\pi}{3}\right) \text{ amperes} \]

2. **Determine the Maximum Values:**
- From the equation for emf, the maximum voltage \( V_{\text{max}} \) is:
\[ V_{\text{max}} = 200 \text{ volts} \]
- From the equation for current, the maximum current \( I_{\text{max}} \) is:
\[ I_{\text{max}} = 2 \text{ amperes} \]

3. **Calculate the RMS Values:**
- The RMS (Root Mean Square) value of voltage \( V_{\text{rms}} \) is calculated as:
\[ V_{\text{rms}} = \frac{V_{\text{max}}}{\sqrt{2}} = \frac{200}{\sqrt{2}} = 100\sqrt{2} \text{ volts} \]
- The RMS value of current \( I_{\text{rms}} \) is calculated as:
\[ I_{\text{rms}} = \frac{I_{\text{max}}}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2} \text{ amperes} \]

4. **Determine the Phase Difference:**
- The phase difference \( \phi \) between the voltage and current is given by the argument of the sine function in the current equation. Here, it is:
\[ \phi = \frac{\pi}{3} \]

5. **Calculate the Cosine of the Phase Difference:**
- We need to find \( \cos(\phi) \):
\[ \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \]

6. **Calculate the Average Power:**
- The average power \( P \) consumed in the circuit is given by the formula:
\[ P = V_{\text{rms}} \cdot I_{\text{rms}} \cdot \cos(\phi) \]
- Substituting the values we calculated:
\[ P = (100\sqrt{2}) \cdot (\sqrt{2}) \cdot \left(\frac{1}{2}\right) \]
- Simplifying this:
\[ P = 100 \cdot 2 \cdot \frac{1}{2} = 100 \text{ watts} \]

### Final Answer:
The average power consumed in the circuit is \( \boxed{100} \) watts.