Power in A.C. Circuits MCQ Quiz - Objective Question with Answer for Power in A.C. Circuits - Download Free PDF

Power Triangle

In the given figure, the given terms represent the type of power:

P = Active Power

Q = Reactive Power 

S = Apparent Power

Explanation

The relationship between P, Q, and S is:

$S=\sqrt{P^2+Q^2}$

S2 = P2 + Q2

Power in an AC circuit

The complex power in an AC circuit is defined as the product of phasor voltage and conjugate of phasor current.

S = V × I* 

where, S = Complex power

V = Voltage

I* = Conjugate of current

Calculation

Given, V = 150∠60° 

I = 50∠30°

S = (150∠60°) (50∠30°)^*

S = (150∠60°) (50∠-30°)

S = 7.5∠30° kVA

Concept:

For a series RLC circuit, the net impedance is given by:

Z = R + j (X_L - X_C)

X_L = Inductive Reactance given by:

X_L = ωL

X_C = Capacitive Reactance given by:

X_C = 1/ωC

Impedance = Resistance + j Reactance

The magnitude of the impedance is given by:

$|Z|=\sqrt{R^2+(X_L-X_C{)}^2}$

Application:

Given:

$V_C=250V$, f = 50 Hz

Z = R + j (X_L - X_C)

$X_C=\frac{V_C}{I}=\frac{250}{0.5}=500$ Ω 

$X_C=\frac{1}{2\pi \times f\times C}$

$500=\frac{1}{2\pi \times 50\times C}$

C = 6.37 μF

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 = Vm­sin ωt

I = Im sin (ωt – ϕ)

P = V I = VmIm sin ωt sin (ωt – ϕ)

$=\frac{V_m{I}_m}{2}\cos \varphi -\frac{V_m{I}_m}{2}\cos \left(2\omega +t-\varphi \right)$

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

Concept

The reactive power is given by:

Q = (I)² X_c

The current in a circuit is given by:

$I=\frac{V_s}{Z}$

where, Q = Reactive power

V = Voltage

I = Current

Z = Net impedance

Xc = Capacitive Reactance

Calculation

Given, V_s = 250 V

Z = 19.6 Ω 

$I=\frac{250}{19.6}=12.755A$

$X_C=\frac{1}{2\pi fC}$

$X_C=\frac{1}{2\pi \times 50\times 178\times {10}^{-6}}=17.891\mathrm{\Omega }$

Q = (12.755)² × 17.891

Q = 2912.5 VAR

Concept

The reactive power is given by:

Q = (I)² X_c

The current in a circuit is given by:

$I=\frac{V_s}{Z}$

where, Q = Reactive power

V = Voltage

I = Current

Z = Net impedance

Xc = Capacitive Reactance

Calculation

Given, V_s = 250 V

Z = 19.6 Ω 

$I=\frac{250}{19.6}=12.755A$

$X_C=\frac{1}{2\pi fC}$

$X_C=\frac{1}{2\pi \times 50\times 178\times {10}^{-6}}=17.891\mathrm{\Omega }$

Q = (12.755)² × 17.891

Q = 2912.5 VAR

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

$i_1=\frac{V_1}{Z}\sin \left(\omega t-\varphi \right)=i_m\sin \left(\omega t-\varphi \right)$

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)]

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

$i\left(t\right)=\frac{1}{12}\sin 3tA$

Power supplied will be:

$p\left(t\right)=4\cos 3t\times \frac{1}{12}\sin 3t$

$=\frac{1}{3}\times \sin 3t\cos 3t$

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

$p(t)=\frac{1}{6}\times 2\sin 3t\cos 3t$

Using trig. property:

2 sin3t cos3t = sin 6t

∴ The power supplied will be:

$p(t)=\frac{1}{6}\sin 6tW$

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.

Concept:

Complex Power of an AC circuit is given as:

$P=V\times I^{\ast }$

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:

The power triangle is shown below.

P = Active power (or) Real power in W = VrIr cos ϕ = (Ir)2R = S cos ϕ

Q = Reactive power in VAR = VrIr sin ϕ

S = Apparent power in VA = VrIr = (Ir)2Z

Where,

Vr and Ir is the RMS value of voltage and current.

R is resistance Ω.

Z is impedance in Ω.

S = P + jQ

$S=\sqrt{P^2+Q^2}$

ϕ is the phase difference between the voltage and current

Power factor $\cos \varphi =\frac{P}{S}=\frac{R}{Z}$

Calculation:

Given, Z = (4 - j3) Ω

∴ $|Z|=\sqrt{4^2+3^2}=5\mathrm{\Omega }$

Hence, Power factor will be,

$cos\varphi ={\displaystyle \frac{R}{Z}}={\displaystyle \frac{4}{5}}=0.8$

Given, i = 5cos (100πt + 100) A

Where, I_m = 5 A

∴ RMS Value of current, (I_r) = 5/√2 A

From above concept,

P = S cos ϕ = (Ir)2Z cos ϕ

∴ $P=({\displaystyle \frac{5}{\sqrt{2}}}{)}^2\times 5\times 0.8=50W$

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 $\cos \varphi =\frac{P}{S}$

Calculation:

V = 100 V

Impedance, Z = 3 + j4

Power factor, $\cos \varphi =\frac{R}{Z}=\frac{3}{\sqrt{3^2+4^2}}=0.6$

Current (I) = V/Z = 100/5 = 20 A

Real power (P) = 100 × 20 × 0.6 = 1200 W

Reactive power (Q) = 100 × 20 × 0.8 = 1600 W

Apparent power (S) = 100 × 20 = 2000 W

The corresponding power triangle is as shown below.

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 $\cos \varphi =\frac{P}{S}$

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

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 $\cos \varphi =\frac{P}{S}$

Calculation:

Given that, V = 50 sin ωt

I = 25 sin (ωt – 53°)

Apparent power, $P=\frac{50}{\sqrt{2}}\times \frac{25}{\sqrt{2}}=625VA$

Concept:

The power triangle is as shown below.

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

Q = Reactive power in VAR = Vrms Irms sin ϕ

S = Apparent power in VA = Vrms Irms

If V(t) = A₁ + B₁ cos ω₁t + C₁ cos ω₂ t ....

I(t) = A2 + B2 cos (ω1t + ϕ₁)  + C2 cos (ω2t + ϕ₂)  ....

Average power P = A₁A₂ + B₁B₂ cos ϕ1 + C₁C₂ cos ϕ₂ ...

Calculation:

Given,

V(t) = 5 – 10 cos (ωt + 60°) V

i(t) = 5 + X cos (ωt) A

ω = 100 π radian/sec

P_avg = V_rms I_rms cosϕ = 0

$\Rightarrow \left(5\right)\left(5\right)-\frac{10}{\sqrt{2}}\times \frac{X}{\sqrt{2}}\times \cos {60}^{\circ }=0$

$\Rightarrow 25=\frac{10X}{2}\times \frac{1}{2}$

⇒ X = 10 A