Question

A single phase wattmeter rated for 500 V; 5 A is having full scale deflection of 1000 watt. What is multiplying factor of the wattmeter?

Answer 1

To find the multiplying factor (MF) of a wattmeter, we need to understand that it helps scale the measured power reading from the wattmeter to match the actual power in the circuit. The multiplying factor is typically calculated by dividing the actual power (in watts) by the full-scale deflection (the maximum reading the wattmeter can show).

Given:
- Voltage rating \( V_{\text{rated}} = 500 \, \text{V} \)
- Current rating \( I_{\text{rated}} = 5 \, \text{A} \)
- Full-scale deflection \( P_{\text{full}} = 1000 \, \text{W} \)

The actual power at full load is calculated using the formula:

\[
P_{\text{actual}} = V_{\text{rated}} \times I_{\text{rated}}
\]

\[
P_{\text{actual}} = 500 \, \text{V} \times 5 \, \text{A} = 2500 \, \text{W}
\]

Now, the multiplying factor (MF) is:

\[
\text{MF} = \frac{P_{\text{actual}}}{P_{\text{full}}}
\]

\[
\text{MF} = \frac{2500 \, \text{W}}{1000 \, \text{W}} = 2.5
\]

Thus, the multiplying factor of the wattmeter is **2.5**.

Answer 2

To find the multiplying factor (MF) of the wattmeter, you need to use the following relationship:

\[ \text{Multiplying Factor (MF)} = \frac{\text{Full Scale Deflection Power}}{\text{Rated Power}} \]

Here’s how to calculate it:

1. **Full Scale Deflection Power (P\(_\text{fsd}\))**: The wattmeter is rated for full scale deflection of 1000 watts.

2. **Rated Power**: The rated voltage and current are 500 V and 5 A, respectively. Therefore, the rated power is:

   \[ \text{Rated Power} = \text{Rated Voltage} \times \text{Rated Current} = 500 \, \text{V} \times 5 \, \text{A} = 2500 \, \text{W} \]

3. **Multiplying Factor Calculation**:

   \[ \text{MF} = \frac{P_\text{fsd}}{\text{Rated Power}} = \frac{1000 \, \text{W}}{2500 \, \text{W}} = 0.4 \]

So, the multiplying factor of the wattmeter is 0.4.