Maximum power transfer and kirchoffs

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Discussion Overview

The discussion revolves around the application of Kirchhoff's laws and the Maximum Power Transfer theorem in the context of calculating the load resistance (RL) in a circuit. Participants are exploring how to find RL using Kirchhoff's laws and how to apply the Maximum Power Transfer theorem to maximize power transfer in the circuit.

Discussion Character

  • Homework-related
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant expresses confusion about whether to calculate the resistance of the load (RL) or the current in the load (RL) for the maximum power transfer calculation.
  • Another participant points out inconsistencies in the calculations presented, specifically regarding the units of current and voltage.
  • Some participants suggest using Thevenin's theorem to simplify the circuit before applying the Maximum Power Transfer theorem.
  • There is a repeated assertion that finding the Thevenin equivalent is necessary before determining the optimal load resistance for maximum power transfer.
  • Several participants express uncertainty about the relationship between Kirchhoff's laws and the Maximum Power Transfer theorem, questioning if they can directly use Kirchhoff's laws to find RL.
  • One participant mentions salvaging previous calculations and seeks clarification on how to calculate the voltage output without a load.
  • There are discussions about measuring voltage across specific components and determining the open circuit voltage (Vth) for the Thevenin equivalent.
  • Participants engage in calculations to find the short circuit current (Isc) and the Thevenin resistance based on their circuit analysis.

Areas of Agreement / Disagreement

Participants generally agree that the Thevenin equivalent is necessary for applying the Maximum Power Transfer theorem, but there is disagreement on the initial approach to finding the load resistance (RL) and the role of Kirchhoff's laws in this process. The discussion remains unresolved regarding the best method to calculate RL.

Contextual Notes

Some participants express confusion about the application of Kirchhoff's laws versus Thevenin's theorem, indicating a potential misunderstanding of circuit analysis techniques. There are also unresolved mathematical steps in the calculations presented.

Andrew187
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Hi Folks
I need a bit of help, My weakness is kirchhoffs law and algebra, I know but I still give it a try. For this task I am required to calculate the value of the load (RL) using kirchhoffs and then I have to find the maximum power transfer for the circuit. I have attempted to calculate the load (RL) using kirchhoff now the bit that I am confused on is am I looking for the resistance of the Load (RL) or the current in the load (RL). Which figure is required to calculate the maximum power transfer? is it the current or resistance of the load (RL)? If it is the current is my calculations correct? and if its the resistance I need to calculate, how do I figure that out because I have no clue where to start:(
 

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The value that you found for I2 is okay. Things go off the rails when you get to the section "Sub I2 into equation (1). You write:
$$I_1 = 15v + \frac{0.293a}{15v} = 15.02$$
which doesn't make sense. The units are not consistent; current is not the same as voltage. Later you have RL = 15.03a, but RL is a resistance, not a current.

Can you state the "Max Power theorem"? Have you looked at Thevenin's theorem (and hence Thevenin equivalent circuits) yet?
 
This is my weakness this assignment is long overdue, I covered this last year now when I am going back through my notes it doesn't make sense:( But come to think of it yes you are correct I need to find the resistance of RL and then use that to find the maximum power transfer. I have covered thevenins and nortons but this task has to be calculated using kirchhoffs to find the resistance of RL, I am lost.
 
Use KVL and KCL to find the Thevenin equivalent. Then apply the Max Power theorem to find RL.
 
I still don't get it by thevanin equivalent and why I should use thevanin equivlent for this? I thought I had to use max power theorum once I determine the resistance for RL?
 
Andrew187 said:
I still don't get it by thevanin equivalent and why I should use thevanin equivlent for this? I thought I had to use max power theorum once I determine the resistance for RL?

That's exactly backwards :smile: Find the Thevenin equivalent first, than add the load RL and use the Max Power theorem to find the value of RL that maximizes the power in RL.
 
I don't get it can't I find the resistance of the load (RL) by using Kirchoffs? and then find the maximum power transfer using the maximum power theorum?
 
Andrew187 said:
I don't get it can't I find the resistance of the load (RL) by using Kirchoffs? and then find the maximum power transfer using the maximum power theorum?

Nope. Any value of RL will form a valid circuit that can be analyzed with KVL/KCL. There's no way to pick a particular value of RL from that.

The problem wants you to find the value of RL that maximizes the power transfer. To do that you need to apply the maximum power theorem. To apply the maximum power theorem you need to simplify the circuit to a form where it can be applied. The Thevenin equivalent is such a circuit.

So. First use Thevenin to reduce the network that will drive the load to its Thevenin equivalent. Then apply the max power theorem to the result to determine RL.
 
I'm lost now, it isn't making sense to me at all.
 
  • #10
Andrew187 said:
I'm lost now, it isn't making sense to me at all.

Perhaps you should review the Max Power Transfer theorem. Take a look at the circuit used to prove it; The Max Power theorem allows you to find the load resistance RL that yields the maximum power transfer from a given source voltage with a series resistance.

This is the same form you should reduce your circuit to in order to apply the theorem to find your value of RL.
 
  • #11
So from what point do I change what I have done in my work? or do I just start all over again?
 
  • #12
Andrew187 said:
So from what point do I change what I have done in my work? or do I just start all over again?

You'll have to decide how you want to go about finding the Thevenin equivalent for the network. You may be able to salvage some of the work you've done. For example, your value of I2 will let you find the open circuit voltage at the network terminals.
 
  • #13
OK I am going to salvage the first part which for I2 I got 0.293a I believe the second bit is nonsense so how do I calculate the voltage output (Vo=V_EF) without load (Rl=∞)?
 
  • #14
Andrew187 said:
OK I am going to salvage the first part which for I2 I got 0.293a I believe the second bit is nonsense so how do I calculate the voltage output (Vo=V_EF) without load (Rl=∞)?

What component is the Vo across?
 
  • #15
gneill said:
What component is the Vo across?

Is it across all the resistors, R7,R8,R9,R10?
 
  • #16
Andrew187 said:
Is it across all the resistors, R7,R8,R9,R10?

Pick one... which resistor would you place the leads of a meter across in order to measure Vo?
 
  • #17
gneill said:
Pick one... which resistor would you place the leads of a meter across in order to measure Vo?

Between R8 and R10 ?
 
  • #18
Andrew187 said:
Between R8 and R10 ?

No. The load (RL) is not connected between R8 and R10. What component is RL in parallel with?
 
  • #19
R10 so would I measure across E and F?
 
  • #20
Andrew187 said:
R10 so would I measure across E and F?

Right.

attachment.php?attachmentid=57884&stc=1&d=1366035456.gif


Now, where does your I2 flow...
 

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  • #21
I2 flows in R8 R9 and R10?
 
  • #22
Andrew187 said:
I2 flows in R8 R9 and R10?

Yes, so what's Vo?
 
  • #23
gneill said:
Yes, so what's Vo?

R8+R9+R10/ 0.293a= 25Ω+7Ω+24Ω/ 0.293a= 5.23v is that correct?
 
  • #24
Andrew187 said:
R8+R9+R10/ 0.293a= 25Ω+7Ω+24Ω/ 0.293a= 5.23v is that correct?

No. Vo is across R10 only! You know the current through R10. Vo is equal to the potential drop across R10. So use Ohm's law on R10.
 
  • #25
gneill said:
No. Vo is across R10 only! You know the current through R10. Vo is equal to the potential drop across R10. So use Ohm's law on R10.

V =IR= V= 0.293a*24Ω=7.032v
 
  • #26
Andrew187 said:
V =IR= V= 0.293a*24Ω=7.032v

Yes. (If you keep a few more decimal places for intermediate values you'd find 7.037 V).

So that's the open circuit potential for the circuit, and thus the Thevenin voltage Vth.
What's the next step for finding the Thevenin equivalent?
 
  • #27
I think its Isc? but for that I need to find I1 so how would I go about that?
 
  • #28
You can either find Isc or directly determine the equivalent resistance of the network as seen from Vo.

To find Isc, redraw the circuit with the short in place and determine the current. To find Req, suppress the source and determine the equivalent resistance via simplification of the network.
 
  • #29
Is the drawing correct?
 

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  • #30
Looks good.
 

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