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Solutions
Two wattmeter method:
The connection diagram using wattmeters is shown below.

\({W_1} = {I_R}{V_{RB}}\cos \left( {{I_R} \ ^\wedge{V_{RB}}} \right)\)
\({W_2} = {I_Y}{V_{YB}}\cos \left( {{I_Y}^\wedge{V_{YB}}} \right)\)

From the phasor diagram
\({I_R}^\wedge{V_{RB}} = 30 - ϕ \)
\({I_Y}^\wedge{V_{YB}} = 30 + ϕ \)
\({W_1} = {I_R}{V_{RB}}\cos \left( {30 - ϕ } \right)\)
\(\Rightarrow {W_1} = {V_L}{I_L}\cos \left( {30 - ϕ } \right)\)
\({W_2} = {I_Y}{V_{YB}}\cos \left( {30 + ϕ } \right)\)
\(\Rightarrow {W_2} = {V_L}{I_L}\cos \left( {30 + ϕ } \right)\)
\({W_1} + {W_2} = {V_L}{I_L}\left[ {\cos \left( {30 - ϕ } \right) + \cos \left( {30 + ϕ } \right)} \right]\)
\(= \sqrt 3 {V_L}{I_L}\cos ϕ\)
⇒ Total three-phase power \( = {\rm{\;}}{{\rm{W}}_1} + {{\rm{W}}_2} = \sqrt 3 {V_L}{I_L}\cos ϕ \)
Total three-phase power is the sum of two wattmeters.
\({W_1} = {V_L}{I_L}\cos \left( {30 - ϕ } \right)\)
\({W_2} = {V_L}{I_L}\cos \left( {30 + ϕ } \right)\)
\({W_1} - {W_2} = \sqrt 3 {V_{ph}}{I_{ph}}\sin ϕ\)
\(\sqrt 3 \left( {{W_1} - {W_2}} \right) = 3{V_{ph}}{I_{ph}}\sin ϕ \)
Reactive power \(= \surd 3\;\left( {{W_1}-{W_2}} \right)\)
Reactive power is equal to √3 times the difference between the readings of the two wattmeters.
\({W_1} + {W_2} = 3{V_{ph}}{I_{ph}}\cos ϕ\)
\(\sqrt 3 \left( {{W_1} - {W_2}} \right) = 3{V_{ph}}{I_{ph}}\cos ϕ \)
\(\Rightarrow ϕ = {\tan ^{ - 1}}\left( {\frac{{\sqrt 3 \left( {{W_1} - {W_2}} \right)}}{{{W_1} + {W_2}}}} \right)\)
Power factor \(= cos\;ϕ\)
\(cosϕ = \cos \left[ {{{\tan }^{ - 1}}\left( {\frac{{\sqrt 3 \left( {{W_1} - {W_2}} \right)}}{{{W_1} + {W_2}}}} \right)} \right]\)
Points to remember:
- If the power factor is between zero and 0.5 or the power factor angle is between 60° and 90° then one of the wattmeters shows a negative value.
- If ϕ = 60° or power factor is 0.5 then one of the wattmeter shows zero reading