Average power dissipated in resistor

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Discussion Overview

The discussion revolves around calculating the average power dissipated in a 30 Ω resistor within a circuit involving a current source and mesh analysis. Participants explore various approaches to set up mesh equations and analyze the circuit behavior, focusing on the relationship between mesh currents and the power calculation.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant attempts to use mesh analysis but encounters difficulties with their equations and assumptions about the currents in the circuit.
  • Another participant suggests finding the RMS current through the resistor to calculate average power, noting the importance of distinguishing between peak and RMS values.
  • There is a discussion about the correct formulation of mesh equations, with participants questioning the contributions of different currents and components in the circuit.
  • Some participants express confusion over the roles of mesh currents and the dependent source, leading to multiple iterations of equations and substitutions.
  • One participant calculates a peak current and attempts to convert it to RMS, while others challenge the accuracy of their approach and the assumptions made about current flow through components.
  • Participants discuss the implications of using different variables and the need to ensure all components are accounted for in the mesh equations.
  • There is a back-and-forth regarding the correct application of mesh analysis and the treatment of the dependent source in relation to the mesh currents.
  • Eventually, a participant arrives at a power calculation, but there is still uncertainty about the correctness of their approach and the values used.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the best approach to set up the mesh equations or the correct interpretation of the currents in the circuit. There are multiple competing views on how to handle the dependent source and the relationships between the mesh currents.

Contextual Notes

Participants express uncertainty regarding the correct application of mesh analysis, the treatment of dependent sources, and the conversion between peak and RMS values. There are unresolved issues related to the number of equations versus variables in the analysis.

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Homework Statement


Find the average power dissipated in the 30 Ω resistor in the circuit seen in the figure if ig=7cos20,000tA.

Homework Equations


KVL/KCL equations
P=VI* (possibly) or I^2(R)

The Attempt at a Solution


I tried to do a mesh, but got I∆(60-j130) = 0, which isn't right . . . I'm assuming I find the current through the resistor and just do I^2(R)?
 

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You'll want to find the RMS current through the resistor if you're looking for average power. The source specification would appear to be a time-domain peak current...

Show your work!
 
I = I∆

30I(-j40) + 30I(30) - I(j10) = 0

Doesn't quite add up . . .
 
I don't see the controlled voltage source in there, and I don't see a contribution from the independent current source ig from the first mesh. What are your mesh equations?
 
That was my one mesh equation . . .
 
Bluestribute said:
That was my one mesh equation . . .
But there are two loops...
 
But wouldn't the other one just be Ig? Or is it 5mH(I-Ig) = 0? Converting mH to ohms of course
 
Bluestribute said:
But wouldn't the other one just be Ig? Or is it 5mH(I-Ig) = 0? Converting mH to ohms of course
Well, you can call it ig if you wish, but it's still a separate mesh current.

Fig1.gif


In the figure, i1 = ig and iΔ is the net current through the inductor.
 
Right right. So across the inductor, it'd be (I1 - I∆)10j = 0? And then for the other mesh, 30I(-j40) + 30I(30) - I(j10) = 0?
 
  • #10
Bluestribute said:
Right right. So across the inductor, it'd be (I1 - I∆)10j = 0?
No, I∆ is the net current through the inductor. The first mesh is already "solved" for its current since it's identical to that of the fixed source. Thus ##i_\Delta = i_g - i_2##.
And then for the other mesh, 30I(-j40) + 30I(30) - I(j10) = 0?
Remember that there are two mesh currents flowing through the inductor. And the source is dependent on ##I_\Delta##, not I.
 
  • #11
Wait, you are very confusing right now.

So I∆ = Ig - I2? Ok, I can dig that.

So for my other mesh, do I replace I∆ with I2? And solve for I2?
 
  • #12
Bluestribute said:
Wait, you are very confusing right now.

So I∆ = Ig - I2? Ok, I can dig that.

So for my other mesh, do I replace I∆ with I2? And solve for I2?
You just wrote an equation for I∆ above! Use that.
 
  • #13
Wait, you lost me again. What two equations am I using? Am I solving I∆? Do I solve I∆ then use I^2(R) to find power?
 
  • #14
What current is flowing through the resistor? It's ##i_2##, right? So that's what you need to find. That's the mesh current for the second loop. ##i_\Delta## is just the net current through the inductor, and happens to be the "sense current" controlling the dependent source. You don't necessarily have to solve for it explicitly, but you do need to replace it with its equivalent in terms of mesh currents otherwise you'll have another variable on your hands.The first mesh current is effectively already solved for since it's identical to ##i_g##. You don't need to write a mesh equation for it. But you do need to use the auxiliary equation for ##I_\Delta## since that's needed to write the equation for the second loop (for the controlled source).
 
  • #15
30(Ig - I2)(-j40) + 30(Ig-I2)(30) - (Ig-I2)(j10) = 0?

I just replaced I∆ with the equivalent current of Ig - I2

But I still have one less equation than variables . . . So this gives me one equation two variables. Adding in the Ig relationship gives two equations three variables.
 
  • #16
Bluestribute said:
30(Ig - I2)(-j40) + 30(Ig-I2)(30) - (Ig-I2)(j10) = 0?

I just replaced I∆ with the equivalent current of Ig - I2

But I still have one less equation than variables . . . So this gives me one equation two variables. Adding in the Ig relationship gives two equations three variables.
The only variable I see in your equation is ##I_2##.
 
  • #17
Oh yeah, it says what Ig is.

So when I do my calculations, I can use 7 for Ig, right, since its angle is 0?

I just need to make sure my first attempt is correct. Since, you know, Mastering Engineering. I already failed one for putting the magnitude of j instead of the complex form of the impedance (it didn't ask for the complex form, just magnitude).
 
  • #18
Bluestribute said:
Oh yeah, it says what Ig is.

So when I do my calculations, I can use 7 for Ig, right, since its angle is 0?
You can. Just keep in mind that 7 is the peak value of the source current, not the RMS value. So you'll end up calculating the peak value of the current through the resistor.
I just need to make sure my first attempt is correct. Since, you know, Mastering Engineering. I already failed one for putting the magnitude of j instead of the complex form of the impedance (it didn't ask for the complex form, just magnitude).
Yeah, I'm not a big fan of computer based quizzes myself.
 
  • #19
So use 7/root2 as I go through just to be safe . . . got it.

Yep, this is pretty much the only online place where I actually learn something. Which is a shame 'cause that's what Mastering is SUPPOSED to do.
 
  • #20
Bluestribute said:
So use 7/root2 as I go through just to be safe . . . got it.
If you want you can use 7 and find the peak current, then convert peak to RMS at the end. Saves carrying the √2 through all the maths. It's just a scaling factor for a sinusoidal waveform.
 
  • #21
For space, I'm just using I as my variable. So I plugged in 7 for Ig and solved that long equation (30(Ig - I2)(-j40) + 30(Ig-I2)(30) - (Ig-I2)(j10) = 0)

-8400j + I1200j + 6300 - 6300I -70j + i10j = 0
I(6300 + 1210j) = (6300 - 8470j)

Gives me a peak of 1.6455 and an Arms of 1.16354

That gives me a power, using I^2(R), of 40.614 W

Did I screw up anywhere?
 
  • #22
Bluestribute said:
Did I screw up anywhere?
Yup. You've got (Ig - I2) flowing through all the components. It doesn't. I2 is the mesh current that flows through all the components of the loop (with Ig flowing through the inductor too, of course, since it's a component shared by the two loops).
 
  • #23
So only the inductor and dependent source use I∆ and the rest use I2?
 
  • #24
Bluestribute said:
So only the inductor and dependent source use I∆ and the rest use I2?

I∆ is a parameter of the dependent source. I2 flows through it (and through the capacitor, and through the resistor). The inductor has two mesh currents because it borders two meshes.
 
  • #25
So

I2(-j40) + I2(30) + (I2 - (Ig - I2)) = 0?
 
  • #26
Bluestribute said:
So

I2(-j40) + I2(30) + (I2 - (Ig - I2)) = 0?
You've left out the impedance of the inductor, and the current through the inductor totals ##i_\Delta##, which is (Ig - I2).

You keep treating ##i_\Delta## as though it were a mesh current. It is not. It's the sum of two mesh currents. It just happens to be what they've called that sum, and is used as a parameter for the dependent source.

When you write a mesh equation, just use the mesh currents first. Then go back and make appropriate substitutions for defined parameters. That should help to avoid confusion.
 
  • #27
I2(-j40) + I2(30) - (Ig - I2)(j10) = 0?

I∆ is screwing me up.
 
  • #28
Bluestribute said:
I2(-j40) + I2(30) - (Ig - I2)(j10) = 0?

I∆ is screwing me up.
You're only missing the dependent source now...
 
  • #29
I2(-j40) + I2(30) - (Ig - I2)(j10) + 30(Ig - I2) = 0?

Ig = 7

This gives me a peak value after converting to polar.
Divide by root 2 to find Arms.
Arms^2(R) = W
 
  • #30
Looks reasonable. So what's your final result?
 

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