How Does Frequency Variation Affect Phasor Current in a Series Circuit?

Click For Summary

Homework Help Overview

The discussion revolves around the effects of frequency variation on phasor current in a series RL circuit. Participants explore how changes in frequency influence impedance and current values, particularly focusing on the maximum possible current that can be achieved under varying conditions.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the relationship between frequency, impedance, and current in an RL circuit. There are attempts to calculate new impedance values and current magnitudes based on frequency changes. Questions arise regarding the assumptions made about phase relationships and how they affect calculations.

Discussion Status

Some participants have provided guidance on recalculating impedance and considering the phase changes that occur with frequency variation. There is ongoing exploration of how to determine the maximum current value, with various interpretations of impedance behavior being discussed.

Contextual Notes

Participants are working under the constraints of a homework problem, which may limit the information available for calculations. There is an emphasis on understanding the implications of frequency changes on circuit behavior without arriving at definitive solutions.

viciado123
Messages
53
Reaction score
0
Impedance -- Current phasor

In a series circuit of two elements, the voltage and current, for w = 2000 rad/s, are V = 150|-45° and I = 4,74|-116,6°.That variation in the frequency of the source would result in a phasor current of 6 amps? Assuming an unlimited variation in the frequency, what is the maximum possible value for the current?


The answer is 23.6% reduction in f
And the maximum 15 amps


I not find the answer
My attempt:

Z = V/I = 31,6|71,6° = 9,97 + 30i
Xl = 30
wL = 30
L = 0,015 and f = 318Hz

For I = 6 amps I assumed I' = 6|-116,6
Z' = V/I' = 25|71,6° = 7,89 + 23,72i
Xl = 23,72
2pif'L = 23,72
f' = 252 Hz

I find f' reduction 20,76% is correct ? How I find the maximum value for the current ?
 
Physics news on Phys.org


viciado123 said:
In a series circuit of two elements, the voltage and current, for w = 2000 rad/s, are V = 150|-45° and I = 4,74|-116,6°.That variation in the frequency of the source would result in a phasor current of 6 amps? Assuming an unlimited variation in the frequency, what is the maximum possible value for the current?


The answer is 23.6% reduction in f
And the maximum 15 amps


I not find the answer
My attempt:

Z = V/I = 31,6|71,6° = 9,97 + 30i
Xl = 30
wL = 30
L = 0,015 and f = 318Hz

For I = 6 amps I assumed I' = 6|-116,6
Z' = V/I' = 25|71,6° = 7,89 + 23,72i
Xl = 23,72
2pif'L = 23,72
f' = 252 Hz

I find f' reduction 20,76% is correct ? How I find the maximum value for the current ?
Your mistake is in assuming the phase of the 6-amp current is the same as the previous phase. As the frequency changes, the phase of the impedance changes, so the phase relationship of the voltage and current will change.

You have an RL circuit, so the impedance is given by Z=R+iωL. At the new frequency ω', you should have Z'=R+iω'L. Note that the real part of the impedance doesn't change between the two cases; only the imaginary part does.
 


vela said:
Your mistake is in assuming the phase of the 6-amp current is the same as the previous phase. As the frequency changes, the phase of the impedance changes, so the phase relationship of the voltage and current will change.

You have an RL circuit, so the impedance is given by Z=R+iωL. At the new frequency ω', you should have Z'=R+iω'L. Note that the real part of the impedance doesn't change between the two cases; only the imaginary part does.

To find w' I need z'. How calculate z' ?
 


Try solving for the magnitude of Z'. Since you know the real part, you can find the imaginary part and solve for ω.
 


vela said:
Try solving for the magnitude of Z'. Since you know the real part, you can find the imaginary part and solve for ω.

Xl = sqrt(25^2 - 9,97^2)
Xl = 22,93
w' = 1528,67 and f' = 243,3 (23,6% reduction)
Thanks

How I find maximum possible value for the current ?
 


viciado123 said:
How I find maximum possible value for the current ?

Think about how the magnitude of impedance changes in time. Does it stay constant? Does it have a minimum or a maximum? What should it be to yield the maximum value for the current?
 


cartonn30gel said:
Think about how the magnitude of impedance changes in time. Does it stay constant? Does it have a minimum or a maximum? What should it be to yield the maximum value for the current?

The maximum value is when the imaginary part is 0.
Z = V / I = 150 / 9,97(real part) = 15A
 

Similar threads

Replies
4
Views
2K
Replies
2
Views
2K
  • · Replies 20 ·
Replies
20
Views
5K
Replies
14
Views
5K
  • · Replies 26 ·
Replies
26
Views
3K
  • · Replies 4 ·
Replies
4
Views
7K
  • · Replies 4 ·
Replies
4
Views
6K
  • · Replies 17 ·
Replies
17
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
6K