Measurement of Power MCQ Quiz in मल्याळम - Objective Question with Answer for Measurement of Power - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

In two wattmeter method, the wattmeter readings are given by

W1 = VLIL cos (30 – ϕ)

W2 = VLIL cos (30 + ϕ)

Total power = W1 + W2

Reactive power = √3 (W1 – W2)

Power factor = cos ϕ

Where \(\phi = {\tan ^{ - 1}}\left( {\frac{{\sqrt 3 \left( {{W_1} - {W_2}} \right)}}{{\left( {{W_1} + {W_2}} \right)}}} \right)\)

Important Point:

In wattmeter, the arrangement of the current coil is in series with the load and the arrangement of the potential coil is across the load.

Hence the wattmeter measures the angle between the current phasor detected by the current coil and the voltage phasor detected by the voltage coil.

The wattmeter measures the average power and it is given by

\(P = {V_{ph}}{I_{ph}}\cos \left( \phi \right)\)

ϕ is the angle between V_ph and I_ph

Two wattmeter method:

Two wattmeter method is used to measure the power in three-phase circuits for both balanced and unbalanced load.

The wattmeter readings are

\({W_1} = {V_L}{I_L}\cos \left( {30 - ϕ } \right)\)

\( {W_2} = {V_L}{I_L}\cos \left( {30 + ϕ } \right)\)

\({W_1} + {W_2} = {V_L}{I_L}\left[ {\cos \left( {30 - ϕ } \right) + \cos \left( {30 + ϕ } \right)} \right]\)

\(= \sqrt 3 {V_L}{I_L}\cos ϕ\)

∴Total three-phase power is

 \( W= {\rm{\;}}{{\rm{W}}_1} + {{\rm{W}}_2} = \sqrt 3 {V_L}{I_L}\cos ϕ \)

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 \(= cos\;ϕ\)

\(cosϕ = \cos \left[ {{{\tan }^{ - 1}}\left( {\frac{{\sqrt 3 \left( {{W_1} - {W_2}} \right)}}{{{W_1} + {W_2}}}} \right)} \right]\)

Nature of the load based on wattmeter readings:

The nature of the load w.r.t the readings shown in the below table.

P = Active power value in Watts

Points to remember:

The value 0 < cos ϕ < 0.5 or 60° < ϕ < 90° means that one of the wattmeters reads negative value.

Given that,

Phase voltage (V_ph) = 200 V

Phase current (I_ph) = 10 A

Power factor = cos ϕ = 0.5

⇒ ϕ = cos^-1 (0.5) = 60°

In two wattmeter method,

W₁ = V_LI_L cos (30 + ϕ)

And

W₂ = V_LI_L cos (30 – ϕ)

\({W_1} = \sqrt 3 \times 200 \times 10\cos \left( {90^\circ } \right) = 0\;W\)

• In the circuit shown in figure (a), the pressure coil is connected on the supply side of the current coil; The same current passes through the load and current coil of the wattmeter
• But the voltage across the pressure coil is more than that across the load by an amount equal to the voltage drop in the current coil
• This connection is used under light loads (or) where the load impedance is high
• This connection is used for high power factor circuits
• In the circuit shown in figure (b), the pressure coil is connected on the load side of the current coil
• The voltage across the pressure coil is the same as that across the load, but the current in the current coil is more than in the load by an amount equal to the current in the pressure coil
• This connection is used under heavy loads (or) where the load impedance is low
• This connection is used for low power factor circuits

Concept:

In a two-wattmeter method,

The reading of first wattmeter (W1) = VL IL cos (30 – ϕ)

The reading of second wattmeter (W2) = VL IL cos (30 + ϕ)

Total power in the circuit (P) = W1 + W2

Total reactive power in the circuit \(Q = \sqrt 3 \left( {{W_1} - {W_2}} \right)\)

Power factor = cos ϕ

\(ϕ = {\rm{ta}}{{\rm{n}}^{ - 1}}\left( {\frac{{\sqrt 3 \left( {{W_1} - {W_2}} \right)}}{{{W_1} + {W_2}}}} \right)\)

Explanation:

W₁ = -ve, W₂ = +ve

|W₁| ≠ |W₂|, then the powere factor range is given as

The range of ϕ is, 60 < ϕ < 90

Now the range of power factor is, 0.5 > cos ϕ > 0.

Concept:

The power triangle is shown below.

P = Active power (or) Real power in W =  Vrms Irms cos ϕ

Q = Reactive power in VAR =  Vrms Irms sin ϕ

S = Apparent power in VA = Vrms Irms

S = P + j Q

\(S = \sqrt {{P^2} + {Q^2}} \)

ϕ is the phase difference between the voltage and current

Power factor \(\cos \phi = \frac{P}{S}\)

Calculation:

Given that,

\(\begin{array}{l} V = \sqrt 2 sin\;\left( {100\pi t} \right)\\ I = \sqrt 2 cos\;\left( {100\pi t} \right) \end{array}\)

The phase difference between V and I is = 90°

Power = VI cos ϕ

In two wattmeter method, the power factor is given by

\(cos\phi = \cos \left[ {{{\tan }^{ - 1}}\left( {\frac{{\sqrt 3 \left( {{W_1} - {W_2}} \right)}}{{{W_1} + {W_2}}}} \right)} \right]\)

Here, \(\phi = {\tan ^{ - 1}}\left( {\frac{{\sqrt 3 \left( {{W_1} - {W_2}} \right)}}{{{W_1} + {W_2}}}} \right)\)

Given that, one reading is negative.

Hence the term (W1 + W2) is always lesser than (W1 - W2).

Now the minimum value of the term \(\frac{{\sqrt 3 \left( {{W_1} - {W_2}} \right)}}{{{W_1} + {W_2}}}\) is √3 and it occurs at one of the reading is zero. Here no reading is zero. Hence the value should be greater than √3 and hence ϕ will be greater than 60°.

Now the maximum value of the term \(\frac{{\sqrt 3 \left( {{W_1} - {W_2}} \right)}}{{{W_1} + {W_2}}}\) is infinite and it occurs at (W1 = -W2). In this case ϕ will be 90°.

Hence ϕ varies between 60 to 90 degrees.

Now the range of ϕ is, \(60^\circ < \phi < 90^\circ \)

Now the range of power factor is, \(0.5 > \cos \phi > 0\).

The power factor of the system is greater than or equal to 0 but less than 0.5.

Two-wattmeter method:

• If one of the wattmeters shows -ve value then this indicates that the load is reactive.
• Element i.e. an inductor or a capacitor that consumes the reactive power.
• When p.f. angle = 60°, only a single wattmeter measures the entire 3-phase power.
• When p.f. angle = 60°, only one wattmeter shows zero deflection.
• When p.f. angle > 60°, one of the wattmeters shown -ve deflection.
• When p.f. angle < 60°, one of the wattmeters shown +ve deflection.
• When one of the wattmeters shows -ve deflection, the range of p.f. is from 0 to 0.5.
• When two wattmeter show +ve deflection, then a range of the p.f. is from 0 to 1.

Important Points

The correct answer is option 4):

Concept:

Two wattmeter method:

• The two-wattmeter method uses two voltage measurements referenced to the same phase (line) and the two currents flowing into that phase. The assumption is that the three-phase system is balanced, i.e., the summation of all voltages = 0 V and the summation of all currents = 0 A.
• This is true if there is no leakage current from neutral to ground.

Reading of both wattmeters are

W₁ = √3 V_pI_pcos (30 - ϕ )

W₂ = √3 V_pI_pcos (30 + ϕ)

where V_p → Voltage across pressure coil I

p → Current through coil ϕ → Phase angle

\(\mathrm{Power\, Factor} = \cos \phi = \cos\left [ \tan^{-1}\sqrt{3}\left ( \frac{w_{1}-w_{2}}{w_{1}+w_{2}} \right ) \right ]\)

• The result can be summarised as At zero power factor lagging, W1 reads the positive value and W2 reads the negative value.
• For power factors lying between 0 to 0.5 lagging, W1 reads the positive value and W2 reads the negative value.
• At 0.5 power factor lagging, W1 reads the positive value and W2 reads the zero value.Hence option 4 is correct
• For power factors lying between 0.5 to 1 lagging, W1 reads the positive value and W2 reads the positive value.
• At unity power factor lagging, W1 reads the positive value and W2 reads the positive value.