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

### Right Answer is:

Zero

#### 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

W_{1} = V_{L}I_{L}cos(30 + φ)

W_{2} = V_{L}I_{L}cos(30 − φ)

**(i) When PF is unity ( φ = 0°)**

W_{1} = V_{L}I_{L}cos30°

W_{2} = V_{L}I_{L}cos30°

Both wattmeters read equal and positive reading i.e upscale reading

**(ii) When PF is 0.5** (**φ = 60°)**

W_{1} = V_{L}I_{L}cos90° = 0

W_{2} = V_{L}I_{L}cos30°

Hence total power is measured by wattmeter W_{2} 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°)**

W_{1} = Negative

W_{2} = positive (since cos(−φ) = cosφ)

The wattmeter W_{2} 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 W_{1}, 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

W_{1} = V_{L}I_{L}cos(30 + 90°) = – V_{L}I_{L}sin30°

W_{2} = V_{L}I_{L}cos(30 − 90°) = V_{L}I_{L}sin30°

In this condition, the two wattmeter reads equal and opposite i.e W_{1} + W_{2} = 0