In a 3-phase power measurement by two wattmeter method, both the wattmeters had identical readings. The power factor of the load was

Question

Question

This question was previously asked in

WBPSC JE Electrical 2018 (Held on 18th Feb 2018) Official Paper

Answer 1

Concept:

A two-watt-meter method is used to measure three-phase power. Readings of both watt meters are

\({W_1} = \sqrt 3 {V_p}{I_p}\cos \left( {30 - \phi } \right)\)

\({W_2} = \sqrt 3 {V_p}{I_p}\cos \left( {30 + \phi } \right)\)

In a two-watt-meter method, the Power factor can be calculated as

\(\cos \phi = \cos \left[ {{{\tan }^{ - 1}}\sqrt 3 \left( {\frac{{{W_1} - {W_2}}}{{{W_1} + {W_2}}}} \right)} \right]\)

In a two-watt-meter method, the total power can be calculated as

P = W1 + W2

Where Vp = Voltage across pressure coil

Ip = Current through the current coil

ϕ = Phase angle

Case 1st

For ϕ = 0°,
cos ϕ = p.f = 1
\({W_1} = \sqrt 3 {V_p}{I_p}\cos \left( {30} \right) = \frac{3}{2}{V_p}{I_p}\)
\({W_2} = \sqrt 3 {V_p}{Z_p}\cos \left( {30} \right) = \frac{3}{2}{V_p}{I_p}\)
∴ W1 = W2
Hence two watt-meters are used to measure three-phase power reads equal reading at unity power factor.

Case 2nd

For ϕ = 30°,
cos ϕ = p.f. = 0.867
\({W_1} = \sqrt 3 {V_p}{I_p}\cos \left( 0 \right) = \sqrt 3 {V_p}{I_p}\)
\({W_2} = \sqrt 3 {V_p}{I_p}\cos \left( {60^\circ } \right) = \frac{{\sqrt 3 }}{2}{V_p}{I_p}\)
∴ W1 = 2 W2
Reading of one watt-meter is half of other

Case 3rd

For ϕ = 60°,
cos ϕ = p.f. = 0.5
\({W_1} = \sqrt 3 {V_p}{I_p}\cos \left( { - 30^\circ } \right) = \frac{{\sqrt 3 }}{2}{V_p}{I_p}\)
\({W_2} = \sqrt 3 {V_p}{I_p}\cos \left( {90^\circ } \right) = 0\)
One of the wattmeters has zero reading and the other reads complete power

Note:

For an angle, more than 60° or cosϕ < 0.5, the wattmeter will give a negative reading.
To take negative readings either the potential coil or current coil connection should be reversed.

Case 4th

For ϕ = 90°,
cos ϕ = p.f. = 0
\({W_1} = \sqrt 3 {V_p}{I_p}\cos \left( {30^\circ } \right) = \frac{{\sqrt 3 }}{2}{V_p}{I_p}\)
\({W_2} = \sqrt 3 {V_p}{I_p}\cos \left( {120^\circ } \right) = \frac{{ - \sqrt 3 }}{2}{V_p}{I_p}\)
∴ W1 = -W2 (zero p.f.)
One of the wattmeters has negative reading