Question

A wattmeter has current coil resistance, $R_C=0.01\Omega$ and pressure coil resistance, $R_P=8000\Omega$ connected such that current coil is on the load side. If the load takes 10 A at the voltage of 230 V at 0.9 power factor, the percentage error in reading will be

• A. 0.048%
• B. 0.020%
• C. 0.072%
• D. 0.036%

Answer 1

The correct option is A 0.048%

Power consumed by the load
= 230 × 10 × 0.9
= 2070 W

The wattmeter also measures loss in current coil
$=I^2{R}_c=(10{)}^2\times 0.01=1W$

$\therefore$ The percentage error

$=\frac{1}{2070}\times 100=0.048\%$