A group of electric lamps having a total power rating of `1000` watt is supplied by an `AC` voltage `E=200sin(310t+60^(@))`. Then the r.m.s value of the circuit current is

Options

A. `10 A`
B. `10 sqrt(2) A`
C. `20 A`
D. `20 sqrt(2) A`

Explanation

To find the r.m.s value of the circuit current supplied to a group of electric lamps with a total power rating of 1000 watts, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the given parameters:**
- Total power (P) = 1000 watts
- AC voltage equation: \( E(t) = 200 \sin(310t + 60^\circ) \)

2. **Calculate the r.m.s value of the voltage (E_rms):**
- The peak voltage (E_0) is given as 200 V.
- The r.m.s value of the voltage is calculated using the formula:
\[ E_{rms} = \frac{E_0}{\sqrt{2}} = \frac{200}{\sqrt{2}} = 100\sqrt{2} \, \text{V} \]

3. **Use the power formula to relate power, voltage, and current:**
- The power in an AC circuit is given by:
\[ P = V_{rms} \cdot I_{rms} \cdot \cos(\phi) \]
- Here, \(\phi\) is the phase difference, which is given as 60 degrees. Thus, \(\cos(60^\circ) = \frac{1}{2}\).

4. **Substitute the known values into the power formula:**
- Rearranging the power formula to find \(I_{rms}\):
\[ I_{rms} = \frac{P}{V_{rms} \cdot \cos(\phi)} \]
- Substituting the values:
\[ I_{rms} = \frac{1000}{100\sqrt{2} \cdot \frac{1}{2}} = \frac{1000 \cdot 2}{100\sqrt{2}} = \frac{2000}{100\sqrt{2}} = \frac{20}{\sqrt{2}} \, \text{A} \]

5. **Calculate the final r.m.s value of the current:**
- Simplifying further:
\[ I_{rms} = 10\sqrt{2} \, \text{A} \]

### Final Answer:
The r.m.s value of the circuit current is \(10\sqrt{2} \, \text{A}\).

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