Measurement of Power and Power Factor of a Three Phase Load MCQ Quiz - Objective Question with Answer for Measurement of Power and Power Factor of a Three Phase Load - Download Free PDF

Explanation:

Apparent Power Calculation in a Three-Phase System:

Definition: Apparent power is the total power flowing in an electrical circuit and is a combination of both active power (real power) and reactive power. It is measured in volt-amperes (VA) or kilovolt-amperes (kVA). Apparent power represents the vector sum of active power and reactive power in a circuit.

Formula: The relationship between apparent power (S), active power (P), and reactive power (Q) in an electrical circuit is given by the equation:

S = √(P² + Q²)

Where:

• S = Apparent Power (in kVA)
• P = Active Power (in kW)
• Q = Reactive Power (in kVAR)

Given Data:

• Reactive Power (Q) = 12 kVAR
• Active Power (P) = 16 kW

Step-by-Step Solution:

1. Substitute the values of active power (P) and reactive power (Q) into the formula for apparent power:

S = √(P² + Q²)

2. Calculate the squares of active power and reactive power:

P² = (16 kW)² = 256 kW²

Q² = (12 kVAR)² = 144 kVAR²

3. Add the squared values:

P² + Q² = 256 + 144 = 400

4. Find the square root of the sum to calculate the apparent power:

S = √400 = 20 kVA

Correct Answer:

The apparent power of the three-phase system is 20 kVA. Hence, the correct option is:

Option 1: 20 kVA

Additional Information

Analysis of Other Options:

To understand why the other options are incorrect, let’s analyze them:

Option 2: 28 kVA

This value is incorrect because it does not match the calculation based on the formula for apparent power. If we substitute the given values of active and reactive power into the formula, the result is clearly 20 kVA, not 28 kVA. This option might arise from a misunderstanding of the formula or incorrect calculations.

Option 3: 4 kVA

This value is far below the calculated apparent power and does not correspond to the given active and reactive power values. It could be derived from an erroneous assumption or misinterpretation of the relationship between active power, reactive power, and apparent power.

Option 4: 10 kVA

While this value is closer to the actual apparent power, it is still incorrect. The apparent power must be calculated using the formula √(P² + Q²), which yields 20 kVA for the given values of active and reactive power. This option might result from halving or incorrectly approximating the calculated apparent power.

Conclusion:

The correct calculation of apparent power, based on the formula and the given values of active and reactive power, leads to a result of 20 kVA. Understanding the relationship between active power, reactive power, and apparent power is essential for accurate calculations in electrical systems. This ensures that the correct option is chosen, and incorrect options are identified as arising from calculation errors or misunderstandings of the formula.

Concept

For delta connected system:

VL = VP 

For star-connected system:

IL = IP

where, VL and IL are line voltage and line current respectively

VP and IP are phase voltage and phase current respectively

Calculation

Given, VL = V_P = 300 V

IL = 

Power in an AC circuit

The power in an AC circuit is the combination of average and reactive power.

The phasor sum of average and reactive power gives apparent power.

S = P + jQ

where, S = Apparent power

P = Average power

Q = Reactive power

Power Triangle

The apparent power is given by:

S = Vrms × Irms

The average or active power is given by:

P = S cos ϕ

P = Vrms × Irms cos θ

The reactive power is given by:

Q = S sin ϕ

Q = Vrms × Irms sin θ

Concept

In 3ϕ delta connected system, the apparent power is given by:

where, V_p = Phase Voltage

I_p = Phase Current

Calculation

Given, Vp = 415 V

I_L = 70 A

S = 50.32 kVA

Explanation:

Taking the circuit to be star connected.

The active power is given as √3 VLILcosϕ.

In the star connected phasor, it is quite clear from the phasor diagram that the  phase difference  ϕ is the angle between the phase voltage and phase current.

Instantaneous power :

• The product of the instantaneous voltage and the instantaneous current for a circuit.
• The power at any instant of time.
• In Ac circuit the instantaneous electric power is given by P = V I
• For a balanced three-phase system the instantaneous power at any instant of voltage or current will be same or the total power Pa + Pb + Pc = P = 3V_phI_ph cos ϕ will be same.

Instantaneous Power of three-phase balanced load

When phase A is at its peak value

P  = 2000 W 

So power 30° later of phase A 

P = 2000 W = 2 kW

Concept

In 3ϕ delta connected system, the apparent power is given by:

where, V_p = Phase Voltage

I_p = Phase Current

Calculation

Given, Vp = 415 V

I_L = 70 A

S = 50.32 kVA

Concept:

Power consumed in a three-phase system is,

In a star-connected three-phase system,

VL = √3 × Vph

And IL = Iph

In a delta connected three phase system,

VL = Vph

IL = √3 × Iph

Where,

VL is line voltage

Vph is phase voltage

IL is line current

Iph is phase current

Calculation:

Given that,

P = 6.5 kW

V_L = 220 volts

I_L = 20 A

From above concept,

or, 

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

Z = Impedance

S = P + jQ

ϕ is the phase difference between the voltage and current.

Calculation:

Given Vrms  = 200 V, Z = 10∠60, Irms = V_rms/Z = 20∠-60° , 

cos ϕ = cos(60°) = 0.5

Active power PL = 200 × 20 × 0.5  = 2000 W

The total active power supplied by the source to the load remains same with and without capacitor.

Additional Information

Total apparent power after adding capacitor also reduces and can be calculated by

Power factor also improved and it can be calculated as

Power factor 

Power factor can be defined as the ratio of real power to apparent power.

Concept:

Power delivered to a resistance is given as

P = I²R

I is the current passing through the Resistor

R is the resistance value in ohm

Calculation:

Given data:

Generator voltage = 200 V

Internal resistance of generator = 100 Ω

Load resistance = 300 Ω

From the above data, we can assume the circuit as

So, 

Power delivered to a load resistance of 300 Ω is given as

P = (0.5)² × 300

= 75 W

The generator delivers a power of 75 W to load Resistance.

Power Factor: 

• The overall power factor is defined as the cosine of the angle between the voltage and current. 
• In AC circuits, the power factor is also defined as the ratio of the real power flowing to the load to the apparent power in the circuit. Hence power factor can also be defined as the ratio of watts to volt-amperes.
• It is also defined as the ratio of resistance to the impedance of the circuit.
• The capacitor is used for power factor improvement.
   

Power Triangle:

The power factor easily analyzed by the power triangle of the AC circuit.

If each side of the current triangle is multiplied by voltage V, then we get the power triangle as shown.

Power factor = cos ϕ
OA = VI cos φ and represents the active power in watts or kW
AB = VI sin φ and represents the reactive power in VAR or kVAR
OB = VI and represents the apparent power in VA or kVA

OB2 = OA2 + AB2

(kVA)2 = (kW)2 + (kVAR)2

Calculation:

Given, V_L  = 460 V

I_L = 100 A

P = 62 kW

We know that,

P(3ϕ) = √3 V_LI_L cos φ

Hence, 

Concept:

The Power consumed by the coil is given as

Where,

V = voltage across the coil,

I = current through the coil,

cos ϕ = power factor angle

R = resistance offered by the choke coil

Calculation:

Power consumed by the coil,

P = I² R = 10² × 4

= 400 W

Using the Power formula

P = V × I × cos ϕ

⇒ 400 = 200 × 10 × cos ϕ

⇒ cos ϕ = 0.2

power factor = cos ϕ = 0.2 lag

Concept:

When all three resistors are connected in star,

Power absorbed by one resistor is

Calculation:

Given R = 10 Ω, V_L-L = 400 V

V_ph = V_L-L / √3 = 400/√3

V_ph = 230.94 V

The power rating of each resistor

Power in an AC circuit

The power in an AC circuit is the combination of average and reactive power.

The phasor sum of average and reactive power gives apparent power.

S = P + jQ

where, S = Apparent power

P = Average power

Q = Reactive power

Power Triangle

The apparent power is given by:

S = Vrms × Irms

The average or active power is given by:

P = S cos ϕ

P = Vrms × Irms cos θ

The reactive power is given by:

Q = S sin ϕ

Q = Vrms × Irms sin θ

Concept

For delta connected system:

VL = VP 

For star-connected system:

IL = IP

where, VL and IL are line voltage and line current respectively

VP and IP are phase voltage and phase current respectively

Calculation

Given, VL = V_P = 300 V

IL =