Test: Apparent Power and Complex Power - Electrical Engineering (EE) MCQ

Concept:

The instantaneous power delivered to any device is given by the product of the instantaneous voltage across the device and the instantaneous current through it.

Let the voltage and current equation be

Calculation:

Given that

V_an = 220 sin (100 πt) V

i_a = 10 sin (100 πt) A

V_bn = 220 cos (100 πt) V

i_b = 10 cos (100 πt) A

Instantaneous power,

P = V_ani_a + V_bni_b

= 220 sin (100 πt) 10 sin (100 πt) + 220 cos (100 πt) 10 cos (100 πt)

= 2200 [sin² (100 πt) + cos² (100 πt)]

= 2200 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

ϕ is the phase difference between the voltage and current

Power factor 

Calculation:

Given that, power factor (cos ϕ) = 0.8

⇒ sin ϕ = 0.6

Reactive power (Q) = VI sin ϕ = 300 VAR

⇒ VI = 500 VA

⇒ Active power (P) = VI cos ϕ = 500 × 0.8 = 400 W

Active Power or True power:

• The actual amount of power being dissipate or used, in a circuit is called a true power
• It is measured in watts and symbolized by the capital letter P

Reactive Power:

• It is measured in Volt-Amps-Reactive (kVAR).
• The mathematical symbol for reactive power is the capital letter Q.

Apparent Power:

• The combination of reactive power and true power is called apparent power.
• Apparent power is measured in the unit of Volt-Amps (kVA) and is symbolized by the capital letter S.

Let impedance per phase of the 3-ϕ connected R-L load is Z_ph = Z∠ϕ

Let the three-phase supply voltage be

V₁ = V_m sin ωt

V₂ = V_m sin (ωt + 120°)

V₃ = V_m sin (ωt + 240°)

Now, the three-phase currents will be

i₂ = i_m sin (ωt + 120° - ϕ)

i₃ = i_m sin (ωt + 240° - ϕ)

Instantaneous power, P = v₁i₁ + v₂i₂ + v₃i₃

= V_mI_m [ sin ωt sin (ωt – ϕ) + sin (ωt + 120°) sin (ωt + 120° - ϕ) + sin (ωt + 240°) sin (ωt + 240° - ϕ)]

= 3V_mI_m cos ϕ = constant.

Given that

L = 1 mH

X_L = inductive reactance of the circuit.

XL = 2 × π × ω × L

XL = 2 ×  π × 50 ×10^-3 Ω 

XL = 0.31 Ω 

Current through the inductive branch will be:

IL = 9.54 ∠-17.43^∘ 

∴ Complex power absorbed by the inductive circuit:

S = V× (I_L)^*

S = (10∠0)(9.54 ∠17.43∘ )

S = 95.40∠17.43^∘ 

S = (91.01 + j28.57) VA

It is given that 'C' takes 10 VAR

Xc = (V_c)²/Q_c

Where

Xc = Capacitive reactance of the circuit.

Vc = voltage across capacitor

Qc = Reactive power supplied by capacitor

Xc = (20)²/10 = 40 Ω 

C = 1/(2 × π × f × Xc)

C = 79.57 μF

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 + jQ

ϕ is the phase difference between the voltage and current.

Calculation:

Given Vrms  = 230 V, Irms = 10 A, cos ϕ = 0.7

Active power P_L = 230 × 10 × 0.7 = 1610 W

Reactive power Q_L = 230 × 10 × 0.714 = 1642.52 VAR

When the capacitor is added in parallel to the load it supplies the reactive power (Q_C) to the load.

So the total reactive power supplied by the source (Q) to load will decreases, it can be shown as

Q = Q_L - Q_C

But the total active power supplied by the source to the load remains same.

Concept:

Single-phase power:

The value of instantaneous power is obtained by multiplying instantaneous voltage with instantaneous current. If at any instant of time t, the voltage, and current values are represented by sine functions as

V = V_m­sin ωt

I = I_m­ sin (ωt – ϕ)

P = V I = V_m­I_m­ sin ωt sin (ωt – ϕ)

The instantaneous power in a single-phase circuit varies sinusoidally with double the supply frequency.

Three-phase power:

In the balanced three-phase case, each phase's instantaneous power is pulsating, but the three pulsating power waves are 120 degrees displaced from each other.

At any instant in time, the total of these three instantaneous power waves is a constant, and it is given by

P = 3 |V| | I | cos φ

So, the total power consumed in a three-phase balanced system is not pulsating.

Concept

The reactive power is given by:

Q = (I)² X_c

The current in a circuit is given by:

where, Q = Reactive power

V = Voltage

I = Current

Z = Net impedance

Xc = Capacitive Reactance

Calculation

Given, V_s = 250 V

Z = 19.6 Ω 

Q = (12.755)² × 17.891

Q = 2912.5 VAR

Concept:

Complex Power of an AC circuit is given as:

P = complex power

V = Voltage

I* = Conjugate of current

For lagging loads, the current lags the voltage by an angle ϕ.

Voltage = V∠0°

Current = I∠-ϕ°

Complex power = VI* = (V∠0°) (I∠ϕ°)

= VI cos ϕ + VI sin ϕ = P + jQ

For leading loads, current leads the voltage by an angle ϕ.

Voltage = V∠0°

Current = I∠ϕ°

Complex power = VI* = (V∠0°) (I∠-ϕ°)

= VI cos ϕ – VI sin ϕ = P – jQ

Calculation:

Load 1: Capacitive load, 10 kW and 40 kVAR

P₁ = (10 – j40) kVA

Load 2: Inductive load, 35 kW and 120 kVAR

P₂ = (35 + j120) kVA

Load 3: Resistive load of 15 kW

P₃ = 15 kW

Total power (P) = P₁ + P₂ + P₃

= 10 – j40 + 35 + j120 + 15

= (60 + j80) kVA

Concept:

Power supplied by a source is given by:

p(t) = v(t) × i(t)

v(t) is the voltage

i(t) is the current

Trigonometric Property:

sin 2θ = 2 sinθ cosθ

Calculation:

With v(t) = 4 cos 3t V and

Multiplying both the numerator and denominator by 2, we get:

Using trig. property:

2 sin3t cos3t = sin 6t

∴ The power supplied will be: