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When an electric current coil of a wattmeter is connected to the electric current path R and the voltage coil is connected between the electric current R path and the neutral side of the balanced three phase supply, it shows 10 kW. The line voltage along the phase sequence RYB is 400 V. The current drawn by the current path is 54 A. What is the new reading of the wattmeter if the voltage coil is connected between the lines B and the Y paths, even if the electric current coil is continuously present in the line R?

When an electric current coil of a wattmeter is connected to the electric current path R and the voltage coil is connected between the electric current R path and the neutral side of the balanced three phase supply, it shows 10 kW. The line voltage along the phase sequence RYB is 400 V. The current drawn by the current path is 54 A. What is the new reading of the wattmeter if the voltage coil is connected between the lines B and the Y paths, even if the electric current coil is continuously present in the line R? Correct Answer -13 kW

Concept:

If current coil (CC) is connected in B phase and potential coil (PC) is connected between R and Y phase, then

Wattmeter reading, W = VLL IL cos ϕ 

Where, ϕ is angle between VLL and IL

VL-L = line to line voltage like VBY

IL = line current like IR

Explanation:

Given, PC is connected between R and N (neutral) and CC in B phase.

Phasor diagram:

Additional Information

For star connection, 

VL-L = √3 Vph  and  IL = Iph

Phase voltage lags line voltage by 30° in RYB phase or +ve phase sequence.

For delta connection, 

VL-L = Vph and IL = √3 Iph

Phase current leads line current by 30° in RYB phase or +ve phase sequence.

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