Measurement of AC Power MCQ Quiz - Objective Question with Answer for Measurement of AC Power - Download Free PDF

Construction of wattmeter

• By using a current and potential transformer in conjunction with a wattmeter, high power measurements can be done.
• The current and potential transformers are connected between the line and current and voltage coils of a wattmeter respectively.
• CT is connected in series with the wattmeter current coil, and PT is connected in parallel with the voltage coil.
• Both transformers step down the high voltage and current values to a value measurable by the voltage and current coils of the wattmeter.
• In order to extend the range of the wattmeter, many current transformers and potential transformers of varied ratios are to be used along with the current and voltage coils of the wattmeter respectively. Thereby, a single wattmeter can cover a wide range of power measurements.

Two Wattmeter Method of Power Measurement

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

Two Wattmeter Method of Power Measurement

The wattmeter meter readings are:

\(W_1=V_LI_Lcos(30+ϕ)\)

\(W_2=V_LI_Lcos(30-ϕ)\)

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

The power factor is given by cosϕ.

Calculation:

Given, W1 = 200 Watts and W2 = -200 Watts

\(ϕ=tan^{-1}({\sqrt{3}(200-(-200))\over 200-200})=tan\space (\infty)\)

ϕ = 90° 

Power factor = cos 90° = 0

Explanation:

Two Wattmeter Method in a Three-Phase System

Definition: The two wattmeter method is a technique used to measure power in a three-phase system. It involves connecting two wattmeters to the system in such a way that they measure the power consumed in the three-phase load. This method is particularly useful for both balanced and unbalanced loads.

Working Principle: In the two wattmeter method, the wattmeters are connected in such a way that each wattmeter measures the power in one phase of the three-phase system. The total power consumed by the load is the algebraic sum of the readings of the two wattmeters. The readings of the wattmeters depend on the power factor of the load and the phase angle between the line voltages and the currents.

Measurement and Analysis:

• Power Measured by Wattmeters: Let the readings of the two wattmeters be W1 and W2.
• In a balanced three-phase system, the total power (P) consumed by the load is given by:
  P = W1 + W2
• The power factor (pf) of the load can be determined using the readings of the two wattmeters:
  pf = cos(θ) = (W1 - W2) / (W1 + W2)
• Given: The readings of the wattmeters are +1000 W and -1000 W.

Calculations:

• Total Power:
  P = W1 + W2 = 1000 W + (-1000 W) = 0 W
• Power Factor:
  pf = cos(θ) = (W1 - W2) / (W1 + W2) = (1000 W - (-1000 W)) / (1000 W + (-1000 W)) = 2000 W / 0 W
  Since the denominator is zero, the power factor is undefined in a real-world sense, but it indicates that the phase angle is 90 degrees, leading to a power factor of zero.

Conclusion:

Option 1: Power factor of load is zero and total power is zero.

This option is correct because the total power calculated is zero, and the power factor is zero, which corresponds to a purely reactive load where the voltage and current are 90 degrees out of phase.

Additional Information

To further understand the analysis, let’s evaluate the other options:

Option 2: Power factor of load is one and total power is zero.

This is incorrect because a power factor of one (unity) implies that the load is purely resistive, and in such a case, the wattmeter readings would not be equal and opposite. Instead, both wattmeters would show positive readings, and their sum would not be zero.

Option 3: Power factor of load is zero and total power is 2000.

This is incorrect because while the power factor is indeed zero, the total power calculated from the given wattmeter readings is zero, not 2000 W.

Option 4: Power factor of load is one and total power is 2000.

This is incorrect because a power factor of one implies a purely resistive load, which would result in both wattmeters showing positive readings. Additionally, the total power calculated from the given readings is zero, not 2000 W.

Conclusion:

The two wattmeter method provides valuable insights into the power consumption and power factor of a three-phase system. By analyzing the readings of the wattmeters, one can determine the total power and the nature of the load. In this case, the readings indicate a purely reactive load with a power factor of zero and a total power of zero, making option 1 the correct choice.

Two Wattmeter Method of Power Measurement

The wattmeter meter readings are:

\(W_1=V_LI_Lcos(30+ϕ)\)

\(W_2=V_LI_Lcos(30-ϕ)\)

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

The power factor is given by cosϕ.

Calculation:

Given, W₁ = 200 Watts and W₂ = -200 Watts

\(ϕ=tan^{-1}({\sqrt{3}(200-(-200))\over 200-200})=tan\space (\infty)\)

ϕ = 90° 

Power factor = cos 90° = 0

Wattmeter:

A wattmeter is an electrical instrument that is used to measure the electric power of any electrical circuit.

The internal construction of a wattmeter is such that it consists of two coils. One of the coils is in series and the other is connected in parallel.

The coil that is connected in series with the circuit is known as the current coil and the one that is connected in parallel with the circuit is known as the voltage coil

Wattmeter measures the average power.

The reading of wattmeter will be,

Average power = Vpc Icc cos ϕ

Where,

Vpc is the voltage across pressure coil

Icc is current flows through the current coil

ϕ is the phase angle between Vpc and Icc

So, wattmeter is a device capable of detecting voltage, current and the angle between the voltage and the current to provide power readings.

Calculation:

Given,

Vpc = V = 200 ∠0° V

Z = Impedance = 4 +j3 = 5∠36.87° 

I = Current from supply = \({V\over Z}={200\angle0 \over 5\angle 36.87}=40\angle -36.87\)

Icc = 40 × (5/50) = 4 A

ϕ = 36.87°

From concept

Average power = 200 ×  4 ×  cos 36.87° = 640 W 

Concept:

The power triangle is as 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 + jQ

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

ϕ is the phase difference between the voltage and current

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

Calculation:

i = 10 sin (ωt + 30)

v = 10 sin (ωt – 30)

ϕ = - 30 – (30) = - 60°

Real power consumed, \(P = \frac{{10}}{{\sqrt 2 }} \times \frac{{10}}{{\sqrt 2 }} \times \cos 60^\circ = 25\;W\)

The circuit configuration can be drawn as,

Given,

R_c = 0.03 Ω

R_p = 6000 Ω

Where R_c and R_p are current coil resistance and potential coil resistance respectively.

I_L = 20 A

V_L = 220 V

cos ϕ = 0.6

Actual Power measured by wattmeter (P_a) is given as,

P_a = V_L I_L cos ϕ = 220 × 20 × 0.6 = 2640 W

Therefore error is due to the current coil is given as, 

\(e=I_L^2R_c=20^2\times0.03=12 \ W\)

Wattmeter reading, Pm = Pa + e = 2640 W + 12 W = 2652 W

 \(\%e = \frac{{{P_m} - {P_a}}}{{{P_a}}} × 100 = \frac{{12}}{{2640}} × 100\)

%e = 0.45

Measurement of power

According to Blondel's Theorem, for the measurement of power in n-phase, n-wire without neutral, (n - 1) wattmeters are required.

For measurement of power in 3-phase, (3 - 1 = 2) wattmeters are required.

The readings of both the wattmeters are given by:

\(W_1=V_LI_Lcos(30+ϕ)\)

\(W_2=V_LI_Lcos(30-ϕ)\)

The power factor is given by:

Power factor = cosϕ 

At zero power, cosϕ = 0

At, ϕ = 90°, the readings of both the wattmeters are:

\(W_1=V_LI_Lcos(30+90)={-0.5V_LI_L}\)

\(W_2=V_LI_Lcos(30-90)={+0.5V_LI_L}\)

Thus, W1 = -ve, W2  = +Ve 

Thus at zero power factor, the reading of both wattmeters are equal and opposite.

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

Wattmeter:

A wattmeter is used to measure the power in the circuits.

The wattmeter reading can be expressed as

W = V I cos ϕ Watts

V = RMS value of pressure coil voltage

I = RMS value of the current in the current coil 

cos ϕ = Power factor

Calculation:

Given that,

Current through current coil (I) = ammeter reading = 12 A.

Power factor = cos ϕ = 1

Resistance of current coil = R_cc = 0.2 Ω

Ammeter resistance = R_A = 0.1 Ω

The pressure coil voltage = The voltage drop across the current coil and ammeter + voltmeter reading.

V = I (R_cc + R_A) + 150

V = 12 (0.2 + 0.1) + 150

∴ V= 153. 6 V

Watt meter reading is 

W = VI cos ϕ

= (153.6) (12) (1)

Measurement of power

According to Blondel's Theorem, for the measurement of power in n-phase, n-wire without neutral, (n-1) wattmeters are required.

Therefore, for a 6-phase, 6-wire without a neutral system, a 5-wattmeter is required.

For measurement of power in n-phase, (n+1) wire with neutral, n wattmeters are required.

Additional Information For the measurement of power in the 3ϕ system, 2 wattmeters are required.

The readings of both the wattmeters are given by:

\(W_1=V_LI_Lcos(30+ϕ)\)

\(W_2=V_LI_Lcos(30-ϕ)\)

The total power is given by:

\(W=W_1+W_2=\sqrt{3}V_LI_L\space cosϕ\)

where VL and IL are line voltage and line current respectively and cosϕ is the power factor.

Concept:

Real Power transmitted  to the load can be given as,

P = Vrms × Irms × cos ϕ 

Where,

Vrms = load terminal voltage in volt

Irms = load current in amper

cos ϕ  is the power factor of the load

ϕ is the power factor angle

Calculation:

v = 100√2  sin (500t) V

I = 10√2 sin (500t + π/3) A

Power factor angle \( \phi=\dfrac{\pi }{3} - 0= \dfrac{\pi }{3}\)

Active power consumed by load = V_rms× I_rms × cos ϕ

= \(\dfrac{{100\sqrt2}}{{√ 2 }} × \dfrac{{10\sqrt2}}{{√ 2 }} × \cos \dfrac{\pi }{3} = 100 × 10× \dfrac{1}{2} \)

= 500 W

We have,

Error = 1.2 % of true value

Wattmeter has full scale range of 2500 W

True Power Measured = 1250 W

True Power Measured = 1250

⇒ Error = ±1.2 % of 1250 = 0.012 × 1250 = ±15 W

Hence, range of Wattmeter reading will be,

From (1250 - 15) W to (1250 + 15) W

∴ Reading of wattmeter = 1235 W to 1265 W

Measurement of power

According to Blondel's Theorem, for the measurement of power in n-phase, n-wire without neutral, (n-1) wattmeters are required.

For measurement of power in 3-phase, (3-1 = 2) wattmeters are required.

The readings of both the wattmeters are given by:

\(W_1=V_LI_Lcos(30+ϕ)\)

\(W_2=V_LI_Lcos(30-ϕ)\)

At ϕ = 0°, the readings of both the wattmeters are the same and are equal to \(W_1=W_2={\sqrt{3}V_LI_L\over 2}\)

The power factor at ϕ = 0° is given by:

Power factor = cosϕ 

Power factor = cos 0° = Unity