Problems with currents and phase

  • Thread starter Thread starter MissP.25_5
  • Start date Start date
  • Tags Tags
    Currents Phase
Click For Summary

Discussion Overview

The discussion revolves around understanding current phasors, impedance calculations, and phase differences in an electrical circuit context. Participants explore the relationships between various currents and their respective phasors, as well as the implications of these relationships for calculating impedance and phase angles.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant seeks clarification on the sum of currents, specifically questioning the role of I4 in the equation I1 = I2 + I3.
  • Another participant suggests that I2 splits into I4 and I5, indicating a connection to a parallel RL pair.
  • Multiple participants confirm the impedance calculation, with one noting the importance of including units in the final answer.
  • There is a discussion about the phase difference between I3 and I4, with one participant questioning whether arg(I3/I4) is equivalent to arg(I3) - arg(I4), leading to different results in their calculations.
  • Some participants agree that in polar form, dividing one phasor by another involves subtracting angles, while others express uncertainty about calculating specific angle differences without a calculator.

Areas of Agreement / Disagreement

Participants generally agree on the method for calculating impedance and the relationship between phase angles in polar form. However, there is uncertainty regarding the calculation of phase differences and whether different methods yield the same results.

Contextual Notes

Participants express limitations in their calculations, particularly regarding the use of calculators for arctan values and the implications of different approaches to calculating phase differences.

MissP.25_5
Messages
329
Reaction score
0
I am still trying to solve this but I first, I need someone to explain to me about the sum of the currents, which I don't understand.

The phasor of current I1 is 2+3j[A], and the phasor of current I2 is 1[A].
1) Find the value of impedance Z.
2) Let the phase difference (arg(I3/I4)) between I3 and I4 be θ. Find the value of tan θ.

Ok, firstly, I am told that I1 = I2 + I3. My question is, what happened to I4?
 

Attachments

  • 2b.jpg
    2b.jpg
    61.7 KB · Views: 454
Physics news on Phys.org
I2 will split into I4 and whatever current continues on to the inductor (Maybe call it "I5", so that I2 = I4 + I5). Think of the segment labeled with I2 as a wire that connects to the parallel RL pair.
 
gneill said:
I2 will split into I4 and whatever current continues on to the inductor (Maybe call it "I5", so that I2 = I4 + I5). Think of the segment labeled with I2 as a wire that connects to the parallel RL pair.

Can you check if I got the impedance right, please?
 

Attachments

  • IMG_4377.jpg
    IMG_4377.jpg
    23.8 KB · Views: 415
Looks good! Be sure to include the units with the final value.
 
gneill said:
Looks good! Be sure to include the units with the final value.

Is this considered as a complete answer?

|Z| = (√50)/25 [Ω] , argZ = arctan(-1/7)
 
MissP.25_5 said:
Is this considered as a complete answer?

|Z| = (√50)/25 [Ω] , argZ = arctan(-1/7)

The magnitude could be further reduced to √2 / 5. There's not much you can do with the arctan unless you resort to using a calculator.
 
gneill said:
The magnitude could be further reduced to √2 / 5. There's not much you can do with the arctan unless you resort to using a calculator.

So I guess, I could just leave the answer in its complex number form and put the unit Ω.
By the way, in the question it says that phase difference is arg(I3/I4), is it the same as arg(I3)-arg(I4)? I calculated both ways and got different answers, so it means they are different?
 
Last edited:
MissP.25_5 said:
So I guess, I could just leave the answer in its complex number form and put the unit Ω.
By the way, in the question it says that phase difference is arg(I3/I4), is it the same as arg(I3)-arg(I4)? I calculated both ways and got different answers, so it means they are different?
The result should be the same. In polar form, when dividing one value by another you subtract the denominator's angle from the numerator's angle (and when you multiply, you add the angles).
 
Here is my solution, can you check please?
 

Attachments

  • IMG_4378.jpg
    IMG_4378.jpg
    35.1 KB · Views: 423
  • #10
gneill said:
The result should be the same. In polar form, when dividing one value by another you subtract the denominator's angle from the numerator's angle (and when you multiply, you add the angles).

Oh, so it is like logarithm?
 
  • #11
In your final line, if you take the arctan of both sides of the line above you should find that tan(θ) = 1/7. Otherwise, looks fine.
 
  • #12
gneill said:
In your final line, if you take the arctan of both sides of the line above you should find that tan(θ) = 1/7. Otherwise, looks fine.

Oh yes, I got it:) Ok, so I tried to calculate argI3-argI4 and got tan^-1(3)-tan^-1(2). Is there a way to calculate that without calculator?
 
  • #13
MissP.25_5 said:
Oh, so it is like logarithm?
I suppose that it's similar, at least in a procedural way.
 
  • #14
gneill said:
I suppose that it's similar, at least in a procedural way.

Yes, I meant the procedure. So question 2 is solved, then! Thanks again. I just posted a new question, hope you could help me out.
 
  • #15
MissP.25_5 said:
Oh yes, I got it:) Ok, so I tried to calculate argI3-argI4 and got tan^-1(3)-tan^-1(2). Is there a way to calculate that without calculator?

Offhand, I'm not sure; there may be a way via trig identities and/or geometry, but I'd have to give it some thought. Being essentially lazy, I tend to reach for a calculator :smile:
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 26 ·
Replies
26
Views
3K
Replies
4
Views
1K
  • · Replies 187 ·
7
Replies
187
Views
59K
  • · Replies 25 ·
Replies
25
Views
4K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 9 ·
Replies
9
Views
4K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K