EE MCQ

SSC JE Electrical 2019 with solution SET-2
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In the two-wattmeter method of three-phase power measurement of a balanced load, if the reading of one meter is −200W, then the power factor of the load is

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°

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

(ii) When PF is 0.5 (φ = 60°)

W1 = VLILcos90° = 0

W2 = VLILcos30°

Hence total power is measured by wattmeter W2 alone

(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 negative (i.e. downscale) reading.

SSC JE Electrical 2019 with solution SET-2
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Observe the given figure. Find Thevenin’s resistance as seen from open-circuited terminals?

To find the Thevenin’s resistance the current source is open-circuited and the voltage source is short-circuited.