For a star-connected load being measured for power by the two-wattmeter method, if W1 = VLIL cos (30 - ϕ) and W2 = VLIL cos (30 + ϕ), then the total 3-phase power will be:

\(\sqrt 3 {V_{ph}}{I_{ph}}\sin \left( {30 + \phi } \right)\)

\(3{V_{ph}}{I_{ph}}\sin \left( \phi \right)\)

\(\sqrt 3 {V_L}{I_L}\cos \left( \phi \right)\)

\(\sqrt 3 {V_L}{I_L}\cos \left( {30 + \phi } \right)\)

For a star-connected load being measured for power by the two-wattmete

1.	For a star-connected load being measured for power by the two-wattmeter method, if W1 = VLIL cos (30 - ϕ) and W2 = VLIL cos (30 + ϕ), then the total 3-phase power will be:
A.	\(3{V_{ph}}{I_{ph}}\sin \left( \phi \right)\)
B.	\(\sqrt 3 {V_L}{I_L}\cos \left( \phi \right)\)
C.	\(\sqrt 3 {V_{ph}}{I_{ph}}\sin \left( {30 + \phi } \right)\)
D.	\(\sqrt 3 {V_L}{I_L}\cos \left( {30 + \phi } \right)\)
Answer» C. \(\sqrt 3 {V_{ph}}{I_{ph}}\sin \left( {30 + \phi } \right)\)

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For a star-connected load being measured for power by the two-wattmeter method, if W1 = VLIL cos (30 - ϕ) and W2 = VLIL cos (30 + ϕ), then the total 3-phase power will be:

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