Question

The circuit in figure is used to measure the power consumed by the load. The current coil and the voltage coil of the wattmeter have $0.02\Omega and1000\Omega$ resistances respectively. The measured power compared to the load power will be

• A. 0.4% less
• B. 0.2% less
• C. 0.2% more
• D. 0.4% more

Answer 1

The correct option is C 0.2% more

$Loadpower\left(truepower\right)=VIcos\varphi$
$=200\times 20\times 1=4000W$

$Resistanceofcurrentcoil{R}_{cc}=0.02$

$CurrentthoughCC=I_c=20A$

Power consumed by current coil
$=I_C^2{R}_{cc}$
$={20}^2\times 0.02=8W$

Measured power= Power consumed by load + Power consumed by current coil

Measured power
$=4000+8=4008W$

$error\%=\frac{Measuredpower-Truepower}{Truepower}\times 100$
$=\frac{4008-4000}{4000}\times 100=0.2\%\left(more\right)$