A group of electric lamps having total power rating of 600 W, 200 V is supplied by an AC voltage `V=169sin(314t+60^(@))`. The rms vlaue of the current is

Options

A. 10A
B. 9.04 A
C. 1.48 A
D. 8 mA

Explanation

To find the RMS value of the current supplied to a group of electric lamps with a total power rating of 600 W and a voltage supply of \( V = 169 \sin(314t + 60^\circ) \), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the RMS Voltage**:
The given voltage is in the form \( V = V_m \sin(\omega t + \phi) \), where \( V_m = 169 \, \text{V} \). The RMS voltage \( V_{\text{rms}} \) can be calculated using the formula:
\[ V_{\text{rms}} = \frac{V_m}{\sqrt{2}} = \frac{169}{\sqrt{2}} \approx 119.79 \, \text{V} \]

2. **Use the Power Formula**:
The total power \( P \) in an AC circuit is given by:
\[ P = V_{\text{rms}} \cdot I_{\text{rms}} \cdot \cos(\phi) \]
Since the power factor \( \cos(\phi) \) is not provided, we will assume it to be 1 for maximum power transfer (purely resistive load).

3. **Rearranging the Power Formula**:
Rearranging the formula to find the RMS current \( I_{\text{rms}} \):
\[ I_{\text{rms}} = \frac{P}{V_{\text{rms}}} \]

4. **Substituting the Values**:
Now substitute \( P = 600 \, \text{W} \) and \( V_{\text{rms}} \approx 119.79 \, \text{V} \):
\[ I_{\text{rms}} = \frac{600}{119.79} \approx 5.00 \, \text{A} \]

5. **Final Result**:
The RMS value of the current is approximately:
\[ I_{\text{rms}} \approx 5.00 \, \text{A} \]