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. 0.41666666666667
B. `5sqrt(2)`A
C. 20 A
D. `10sqrt(2)`A

Explanation

To solve the problem step-by-step, we will follow the given information and apply relevant formulas.

### Step 1: Identify the given values
- Total power rating (P) = 1000 watts
- AC voltage equation: \( E = 200 \sin(310t + 60^\circ) \)
- Amplitude of voltage (E₀) = 200 V
- Phase difference (φ) = 60°

### Step 2: Calculate the RMS voltage (E_rms)
The RMS (Root Mean Square) value of the voltage can be calculated using the formula:
\[ E_{rms} = \frac{E_0}{\sqrt{2}} \]
Substituting the value of E₀:
\[ E_{rms} = \frac{200}{\sqrt{2}} = 100\sqrt{2} \text{ V} \]

### Step 3: Use the power formula
The power in an AC circuit is given by:
\[ P = E_{rms} \cdot I_{rms} \cdot \cos(\phi) \]
Where:
- \( P \) = Power (1000 W)
- \( E_{rms} \) = RMS voltage (calculated above)
- \( I_{rms} \) = RMS current (unknown)
- \( \cos(\phi) \) = cosine of the phase difference (cos(60°) = 1/2)

### Step 4: Substitute the known values into the power formula
Substituting the known values into the power equation:
\[ 1000 = (100\sqrt{2}) \cdot I_{rms} \cdot \frac{1}{2} \]

### Step 5: Simplify the equation
Rearranging the equation to solve for \( I_{rms} \):
\[ 1000 = 50\sqrt{2} \cdot I_{rms} \]
\[ I_{rms} = \frac{1000}{50\sqrt{2}} = \frac{20}{\sqrt{2}} = 10\sqrt{2} \text{ A} \]

### Final Answer
The RMS value of the circuit current is:
\[ I_{rms} = 10\sqrt{2} \text{ A} \]

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