Question

The voltage-flux adjustment of a certain 1-phase 220 V induction watt-hour meter is altered so that the phase angle between the applied voltage and the flux due to it is ${85}^{\circ }$ (instead of ${90}^{\circ }$). The errors introduced in the reading of this meter when the current is 5 A at power factors of unity and 0.5 lagging are respectively

• A. $3.8mW,77.4mW$
• B. $-3.8mW,-77.4mW$
• C. $-4.2W,-85.1W$
• D. $4.2W,85.1W$

Answer 1

The correct option is C $-4.2W,-85.1W$

$\text{Measured value = VI sin}(\Delta -\varphi )$

Where,
$\Delta \text{= Phase angle between voltage and flux}$

$\text{cos}\varphi \text{= Power factor}$

$\text{True value =VI cos}\varphi$

$\text{Error=Measured value-True value}$

Case-I:
$\Delta ={85}^{\circ },pf=cos\varphi =1,\varphi =0^{\circ }$

$V=220V,I=5A$

$Error=VIsin(\Delta -\varphi )-VIcos\varphi$

$=220\times 5sin(85-0^{\circ })-220\times 5\times 1$

$\approx -4.2W$

Case-II:
$\Delta ={85}^{\circ },ph=cos\varphi =0.5\Rightarrow \varphi ={60}^{\circ }$

$V=220V,I=5A$

$Error=VIsin(\Delta -phi)-VIcos\varphi$

$=220\times 5\times sin(85-{60}^{\circ })-220\times 5\times 0.5$

$=-85.1W$