This question was previously asked in

PSPCL JE EE 2018 Previous Year Paper

Option 3 : 0.078%

**Concept:**

Reading of wattmeter = \(P = {V_{pc}}{I_{pc}}cos\alpha\)

Error due to current coil = loss in current coil \(= {I^2}{R_{cc}}\)

In this type of connection, error depends on load current. So this type of connection is mainly suitable for low load current value.

True power = \(P = VI\cos \alpha\)

Where V = Voltage across the load

I = current through the load

% Error \(= \frac{{loss\;in\;current\;coil}}{{true\;power}}\)

Reading of wattmeter = P = V_{pc}I_{cc }Cosα + \(\frac{{{V^2}}}{{{R_{pc}}}}\)

Error due to connection = loss in potential coil = \(\frac{{{V^2}}}{{{R_{pc}}}}\)

Error in this type of connection is fixed

Suitable for higher load current

% error = \(\frac{{error}}{{reading\;of\;wattmeter}}\)

**Calculation:**

Given that pressure coil resistance \({R_{pc}}\) = 8000 ohm

Current coil \({R_{cc}} =\) 0.3 ohm

Power factor = 0.8

Error in wattmeter = \(\frac{{{v^2}}}{{{R_{pc}}}} = \frac{{{{\left( {200} \right)}^2}}}{{8000}} = 5\;watt\)

Reading \(= 200 \times 40 \times 0.8 + 5 = 6405\)

% error \(= \frac{5}{{6405}} \times 100 = 0.078\;\%\)