Question

A Wattmeter has a current coil of 0.03 $\Omega$ resistance and a pressure coil of 6000 $\Omega$ resistance connected such that the current coil is on the load side. If the load take 20 A at a voltage of 220 V and 0.6 power factor then the percentage error in the reading of wattmeter is

• A. 0.45%
• B. 0.30%
• C. 0.15%
• D. 0.90%

Answer 1

The correct option is A 0.45%

Power consumed by load,

$P_{\tau }=VI\cos \varphi =220\times 20\times 0.6=2640W$
Now given that,

$R_C=0.03\Omega ,R_P=6000\Omega$
Power indicated by wattmeter,

$P_W=$ = Power consumed by load + power loss in current coil

$=P_{\tau }+I^2{R}_C=2640+{\left(20\right)}^2\left(0.03\right)=2652W$

$\%Error=\frac{P_W-P_T}{P_T}\times 100=\frac{2652-2640}{2640}\times 100$

$\%Error=0.45$%