In three-phase power measurement in a delta-connected load using 2 watt meters, when the power factor of the load is zero, readings of 2 watt meters will be __________.

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MPPGCL JE Electrical 28 April 2023 Shift 3 Official Paper

\(\sqrt{3} \times \mathrm{V}_{\mathrm{L}} \times \mathrm{I}_{\mathrm{L}} \) and \(-\sqrt{3} \times \mathrm{V}_{\mathrm{L}} \times \mathrm{I}_{\mathrm{L}} \)
\(\frac{\sqrt{3}}{2} \times \mathrm{V}_{\mathrm{L}} \times \mathrm{I}_{\mathrm{L}} \) and \(-\frac{\sqrt{3}}{2} \times \mathrm{V}_{\mathrm{L}} \times \mathrm{I}_{\mathrm{L}} \)
\(\sqrt{3} \times \mathrm{V}_{\mathrm{L}} \times \mathrm{I}_{\mathrm{L}}\) and \(\sqrt{3} \times \mathrm{V}_{\mathrm{L}} \times \mathrm{I}_{\mathrm{L}} \)
\(\frac{\sqrt{3}}{2} \times \mathrm{V}_{\mathrm{L}} \times \mathrm{I}_{\mathrm{L}} \) and \(\frac{\sqrt{3}}{2} \times \mathrm{V}_{\mathrm{L}} \times \mathrm{I}_{\mathrm{L}}\)

Answer 1

Two Wattmeter Method of Power Measurement

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The wattmeter meter readings are:

\(W_1=V_LI_Lcos(30-ϕ)\)

\(W_2=V_LI_Lcos(30+ϕ)\)

where, \(ϕ=tan^{-1}({√{3}(W_1-W_2)\over W_1+W_2})\)

The power factor is given by cosϕ.

Calculation:

Given, cosϕ = 0 ⇒ ϕ = 90°

\(W_1=V_LI_Lcos(30-ϕ)\)

\(W_1=V_LI_Lcos(30-90)\)

\(W_1={1\over 2}\times V_L\times I_L\)

\(W_2=V_LI_Lcos(30+ϕ)\)

\(W_2=V_LI_Lcos(30+90)\)

\(W_2=-{1\over 2}\times V_L\times I_L\)

If we multiply both W₁ and W₂ by √3, then:

\(W_1={\sqrt{3}\over 2}\times V_L\times I_L\) and \(W_2=-{\sqrt{3}\over 2}\times V_L\times I_L\)