Question

A single-phase load is connected between R and Y terminals of a 415 V, symmetrical, 3-phase, 4 wire system with phase sequence RYB. A wattmeter is connected in the system as shown in the figure. The power factor of the load is 0.8 lagging. The wattmeter will read

• A. -795 W
• B. -597 W
• C. +597 W
• D. +795 W

Answer 1

The correct option is B -597 W

$Linetolinevoltage=415V$

$Phasetoneutralvoltage=\frac{415}{\sqrt{3}}V$

Taking $V_R$ as the reference,

$V_{RN}=\frac{415}{\sqrt{3}}\angle 0^{\circ }V$

$V_{YN}=\frac{415}{\sqrt{3}}\angle -{120}^{\circ }V$

$and{V}_{BN}=\frac{415}{\sqrt{3}}\angle -{240}^{\circ }V$

$V_{RY}=V_{RN}-V_{YN}$

$=\frac{415}{\sqrt{3}}\angle 0^{\circ }V-\frac{415}{\sqrt{3}}\angle -{120}^{\circ }$

$=415\angle {30}^{\circ }V$

Load current,

$I_L=\frac{V_{RY}}{Z}=\frac{415\angle {30}^{\circ }}{100\angle 36.87}$

$=4.15\angle -{6.87}^{\circ }A$

Current though current coil,

$I_{cc}=I_L=4.15\angle -{6.87}^{\circ }A$

Voltage across pressure coil,

$V_{BN}=\frac{415}{\sqrt{3}}\angle -{240}^{\circ }A$

$Phaseanglebetween{V}_{BN}and{I}_{c{c}^{{}^{\prime }}}$

$\varphi =-{240}^{\circ }-(-{6.87}^{\circ })={233.13}^{\circ }$

$Wattmeterreading|V_{BN}|\times |I_L|cos\varphi$

$=\frac{415}{\sqrt{3}}\times 4.15\cos 233.13\approx -597W$