Calculating Resistance & Reactance of Load with PF 0.866 and 200V, 5A

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Discussion Overview

The discussion revolves around calculating the equivalent series resistance and reactance of a load with a power factor of 0.866, given a voltage of 200V and a current of 5A. The context is primarily homework-related, focusing on the application of electrical concepts and equations.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant states the power factor is 0.866 lagging, with voltage and current provided, and seeks help in calculating resistance and reactance.
  • Another participant corrects the initial equation regarding voltage and impedance, stating that voltage equals current times impedance.
  • A participant calculates impedance as 40 ohms but questions the teacher's answer of 20 ohms, expressing uncertainty about the definitions of equivalent series reactance and impedance.
  • One participant clarifies the distinction between impedance as a complex number and reactance as its imaginary part, providing the formula for power factor.
  • Another participant suggests that the reactance can be calculated using the known impedance and power factor, indicating a method to find both resistance and reactance.
  • There is a suggestion to use sine to calculate the reactance based on the triangle formed by resistance and impedance.

Areas of Agreement / Disagreement

Participants express uncertainty regarding the correct calculation of resistance and reactance, with some confusion about the teacher's answer. There is no consensus on the correct values or methods to arrive at them.

Contextual Notes

Participants have not resolved the discrepancies between their calculations and the teacher's answer. There are also unresolved assumptions regarding the definitions and relationships between impedance, resistance, and reactance.

tofushop193
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Homework Statement


the power factor of a load is 0.866 lagging.the voltage is 200v and the current is 5A.caculate the equivalent series reactance and resistance of the load.



Homework Equations



cos.theta = power factor = resistance / impedence= 0.866
volt = ampere / impendence

The Attempt at a Solution



trying but still stuck at this question..can someone help me
btw i am new here if i did anything wrong please guide me...thanks guys
 
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tofushop193 said:

Homework Statement


the power factor of a load is 0.866 lagging.the voltage is 200v and the current is 5A.caculate the equivalent series reactance and resistance of the load.



Homework Equations



cos.theta = power factor = resistance / impedence= 0.866
volt = ampere / impendence

The Attempt at a Solution



trying but still stuck at this question..can someone help me
btw i am new here if i did anything wrong please guide me...thanks guys

First of all: Voltage = Current * Impedance (not /)
You know the voltage and the current. What is the total impedance?
The resistance and the inductive impedance form the sides of a right triangle, whose hypotenuse is the total impedance.
 
sorry about the wrong equiation given but i tried
200v=5A * impendence
so impendence = 200/5
= 40 ohm
but the answer given by my teacher is 20 ohm
is he wrong?
does equivalent series reactance means impendence??
 
[tex]\tilde{}[/tex]I think you are mixing terms.

Impedance is the name given to the whole complex number, defined by [tex]Z=\frac{\tilde{V}}{\tilde{I}}=R+jX[/tex] (tilde denotes phasors, hence both amplitudes and phases)

While reactance is defined as only the imaginary part of Z: X.

Also the power factor will be [tex]cos(\theta}=\frac{R}{|Z|}=\frac{R}{\sqrt{R^{2}+X^{2}}[/tex]
(You got it almost right actually, I just want you to distinguish the impedance, which encodes amplitude and phase, from only the amplitude of the impedance)
 
hmm so how do i get the 20 ohm that my teacher did?
 
tofushop193 said:
hmm so how do i get the 20 ohm that my teacher did?
The reactance is one of the sides of the triangle. You already know the hypotenuse (40 ohm) and the cosine of the angle between the hypotenuse and the side representing the resistance (0.866).
You can calculate both sides.
 
so i take sine30*40 to get the 20 ohms?
 
tofushop193 said:
so i take sine30*40 to get the 20 ohms?
Exactly!
 

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