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ElectricalBasic Electrical
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Resistance is 25 ohm and impedance is 50 ohm, what is value of power factor?

A

0

B

0.5

C

1

D

2

Correct Answer

Direct FormulaElectricalBasic Electrical
Option A

0

Quick Summary: Given: Resistance R = 25 Ohm, Impedance Z = 50 Ohm

ЁЯУЛ Given

Resistance R = 25 Ohm, Impedance Z = 50 Ohm

ЁЯФв Formula Used

cosтБб(╧Х)=RZ\cos(\phi) = \frac{R}{Z}cos(╧Х)=ZRтАЛ

ЁЯУК Diagram / Illustration
R=25╬йR = 25 \OmegaR=25╬й
X=Z2тИТR2X = \sqrt{Z^2 - R^2}X=Z2тИТR2тАЛ
Z=50╬йZ = 50 \OmegaZ=50╬й
╧Х\phi╧Х
ЁЯФв Step-by-Step Solution
1

Identify given parameters

We are provided with the resistance (RRR) and the impedance (ZZZ) of the AC circuit.

R=25╬й,Z=50╬йR = 25 \Omega, Z = 50 \OmegaR=25╬й,Z=50╬й

2

Recall the power factor formula

The power factor in an AC circuit is defined as the ratio of resistance to impedance.

cosтБб(╧Х)=RZ\cos(\phi) = \frac{R}{Z}cos(╧Х)=ZRтАЛ

3

Calculate the power factor

Substitute the given values into the formula to find the power factor.

cosтБб(╧Х)=2550=0.5\cos(\phi) = \frac{25}{50} = 0.5cos(╧Х)=5025тАЛ=0.5

тЬЕ

B is correct because the power factor is the ratio of resistance to impedance, which yields 0.5 for the given values.

Core Concepts Used
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AC Circuits Impedance Triangle Power Factor
ЁЯТб EXAM TIP

In AC power analysis, remember that the power factor is also equal to cosтБб(╧Х)\cos(\phi)cos(╧Х), where ╧Х\phi╧Х is the phase angle between voltage and current; this is fundamental for calculating real power P=VIcosтБб(╧Х)P = VI \cos(\phi)P=VIcos(╧Х).

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