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

Concept:
Impedance triangle for the series RC circuit


Impedance triangle for series RL circuit

Calculation:

Power factor=  \(cos(\theta )\)

Where power factor angle

By the Impedance triangle of the series RC circuit, we can write

\(tan(\theta)=\frac{Opposite}{Adjucent}=\frac{X_c}{R}\)

\(Therefore \ Power\ factor \ angle \ \theta =tan^{-1}(\frac{1}{\omega RC})\)

The correct answer is option 3):(0.707)

Concept:

The load impedance is given by: Z = R + j X (for lagging load)

Z = R - j X (for leading load)

\(cosϕ = {R \over Z}\)

Calculation:

Given

A circuit load impedance is (25 - j25) Ω 

\(cosϕ = {R \over \sqrt{R^2+X^2}}\)

\(cosϕ = {25 \over \sqrt{(25)^2+(25)^2}}\)

= \(25 \over 35.35\)

= 0.707

Concept:

The load power factor is given by:

\(cos ϕ _L = {P_L \over S_L}\)

P_L = V_1(RMS) × I_1(RMS) ×  cosϕ₁

S_L = V(RMS) × I(RMS) 

Calculation:

\(P_L = {100 \over \sqrt {2}}\times{10 \over \sqrt {2}}\times\cos60\)

\(P_L = 250\space W\)

\(S_L = {100 \over \sqrt{2}}\times {\sqrt{({10\over\sqrt2})^2+({2\over\sqrt2})^2+({5\over\sqrt2}})^2}\)

\(S_L = 567.89 \space VA\)

\(cos ϕ _L = {250 \over 567.89}\)

\(cos\phi_L= 0.44\)

Power factor:

• The power factor is defined as the ratio of the active power to the apparent power.
• It is denoted by cos ϕ.
   

Formula:

Power factor = (Active power /Apparent power)

In series circuit,

Power factor = (Resistance / Impedance)cos⁡ϕ=R|Z|" tabindex="0">cosϕ=R|Z|

 \((cos\phi\ =\frac{R}{Z})\)

Total Equivalent impedance is given by,

Z = R ±  j X

Where,

R = Resistance of Circuit

+ j X = Inductive Reactance of Circuit

- j X = Capacitive Reactance of Circuit

The magnitude of Equivalent Impedance is given by

\((\left| Z \right| = \sqrt {{R^2} + {X^2}})\)

Calculation:

Equivalent Impedance Z_eq = j2 + ( - j2 ∥ j4 )

\( =j2+\dfrac{-j2\times j4}{-j2+j4}\)

|Z|=R2+X2" tabindex="0">|Z| = R₂ + X₂

= j2 - j4 Ω

= - j2 Ω

= - j X Ω

Negative Imaginary Term without Real Part Indicates a Pure Capacitive Circuit.

Therefore, a capacitive Reactance of 2 Ω with zero power factor leading.

Concept:

• Low power factor actually leads to higher energy losses. Power factor is a measure of how effectively electrical power is being utilized in a system. A low power factor indicates that the reactive power component is high compared to the active power component. Reactive power does not perform useful work but circulates between the source and the load, causing additional current flow. This increased current results in higher losses in the transmission and distribution systems, leading to inefficiencies and wastage of energy.
• Power factor and voltage regulation are closely related. When the power factor is low, the reactive power component is high, which causes increased voltage drops and poorer voltage regulation. The voltage drops occur due to the additional current required to supply the reactive power, resulting in a decrease in the voltage at the load. This can have negative impacts on the performance and operation of electrical equipment.
• Improving power factor to unity (a power factor of 1) is generally advisable. Unity power factor means that the reactive power component is minimized, and the power being consumed is effectively utilized for useful work. By improving power factor, the system can reduce energy losses, enhance voltage regulation, and improve overall efficiency. This can lead to cost savings, improved equipment performance, and a more stable and reliable electrical system.
• In conclusion, the statement "1) Low power factor leads to less energy loss" is not true about power factor. Low power factor actually leads to higher energy losses.

In the AC circuit, the load power factor is given by:

Load power factor (cos φ) \(= \frac{{Total\;Active\;power\;consumed\;by\;load\left( P \right)}}{{Total\;Apparent\;Power\left( S \right)}}\)

Total apparent power is given by

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

Where,

P = Total active power consumed

Q = Total reactive power consumed

Load power factor (cos φ) = \(\frac{P}{{\sqrt {{P^2} + {Q^2}} }}=\frac{P}{S}\)

Application:

Given that, P = 5.6 kW = 5600 W

S = 7000 VA

From above concept,

(cos φ) = \(\frac{P}{{\sqrt {{P^2} + {Q^2}} }}=\frac{P}{S}\) = \(\frac{5600}{7000}=0.8\)

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 (VAR).
• 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 (VA) and is symbolized by the capital letter S.

In the AC circuit, the load power factor is given by:

Load power factor (cos φ) \(= \frac{{Total\;Active\;power\;consumed\;by\;load\left( P \right)}}{{Total\;Apparent\;Power\left( S \right)}}\)

Total apparent power is given by

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

Where,

P = Total active power consumed

Q = Total reactive power consumed

Load power factor (cos φ) = \(\frac{P}{{\sqrt {{P^2} + {Q^2}} }}=\frac{P}{S}\)

Application:

Given that, P = 5.6 kW = 5600 W

S = 7000 VA

From above concept,

Q = \(\sqrt{7000^2-5600^2}=4200\ VAr\)

Hence, Q = 4.2 kVAr

The form factor is defined as the ratio of the RMS value to the average value of an alternating quantity.

F.F. (Form factor) = \(\frac{{R.M.S\;Value}}{{Average\;Value}}\)

Crest Factor ‘or’ Peak Factor is defined as the ratio of the maximum value to the R.M.S value of an alternating quantity.

C.F. ‘or’ P.F. = \(\frac{{Maximum\;Value}}{{R.M.S\;Value}}\)

The form and crest factor for different waveforms is as shown: 

Concept:

In AC circuits, the power factor is defined as the ratio of the real power flowing to the load to the apparent power in the circuit.

Power factor = cos ϕ

Where ϕ is the angle between voltage and current.

If the current lags the voltage, the power factor will be lagging.

If the current leads the voltage, the power factor will be leading.

Power factor = cos ϕ = R/Z

Z = R ± j X 

\(Z=\sqrt{{{R}^{2}}+{{X}^{2}}} \;\angle (\pm\;tan ^{-1}({\dfrac{X}{R}}))\)

Where,

R = Resistance

Z = Impedance

X = Reactance (X = X_l - X_c)

X_l = Inductive Reactance

X_c = Capacitive Reactance

Positive angle = Inductive circuit, power factor lagging in nature

Negative angle = Capacitive circuit, power factor leading in nature

The power factor is unity when the circuit is purely resistive.

• The maximum value of the power factor is unity (1)
• The power factor for a purely resistive load is unity (1)
• The power factor for RL load is less than unity (1) and is lagging in nature
• The power factor for RC load is also less than unity (1) but leading in nature.

Calculation:

The impedance seen from the input

\(Z = 6 + \frac{{\left( { - j2} \right)\left( {j5 - j2} \right)}}{{ - j2 + j5 - j2}}\) 

= \(6 + \frac{{10 - 4}}{j}\)

= 6 – j6

= \(\sqrt {{6^2} + {6^2}} {\tan ^{ - 1}}\left( {\frac{{ - 6}}{6}} \right)\) 

= \(6\sqrt 2 \;\angle - 45^\circ \)

∴ \(pf = \cos 45^\circ = \frac{1}{{\sqrt 2 }}\)

As the impedance is in the form of R - j X

So the nature of the circuit is leading.