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

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

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

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

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

For a delta-connected load being measured for power by the two-wattmet

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

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

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