# In two wattmeter method if the power factor is 0.5, then one of the wattmeter will read

In two wattmeter method if the power factor is 0.5, then one of the wattmeter will read

#### SOLUTION

In two wattmeter method if the power factor is 0.5, then one of the wattmeters will read Zero.

The reading of two wattmeters can be expressed as

W1 =  VLILcos(30 + φ)
W2 =  VLILcos(30 − φ)

(i) When PF is unity ( φ = 0°)

W1 =  VLILcos30°
W2 =  VLILcos30°

(ii) When PF is 0.5 (φ = 60°)

W1 =  VLILcos90° = 0
W2 =  VLILcos30°

Hence total power is measured by wattmeter W2 alone. Hence one of the wattmeters will read Zero.

(iii) When PF is less than 0.5 but greater than 0 i.e ( 90° > φ > 60°)

W1 = Negative
W2 = positive (since cos(−φ) = cosφ)

The wattmeter W2 reads positive (i.e.upscale) because for the given conditions (i.e. ( 90° > φ > 60°), the phase angle between voltage and current will be less than 90°. However, in wattmeter W1, the phase angle between voltage and current shall be more than 90° and hence the wattmeter gives a negative (i.e. downscale) reading.

Wattmeter cannot show negative reading as it has only a positive scale. An indication of negative reading is that pointer tries to deflect in a negative direction i.e. to the left of zero. In such a case, reading can be converted to positive by interchanging either pressure coil connections or by interchanging current coil connections. Remember that interchanging connections of both the coils will have no effect on wattmeter reading.

(iv) When P.F reads zero (φ = 90°)

Such a case occurs when the load consist of pure inductance or pure capacitance

W1 =  VLILcos(30 + 90°) = – VLILsin30°
W2 =  VLILcos(30 − 90°) = VLILsin30°

In this condition, the two wattmeter reads equal and opposite i.e W1 + W2 = 0

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