Maximizing Power Transfer in Circuits

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Discussion Overview

The discussion revolves around solving circuit analysis problems using mesh analysis and Thevenin's theorem. Participants are attempting to find the voltage across a resistor and the power absorbed by it, as well as the Thevenin equivalent circuit and conditions for maximum power transfer.

Discussion Character

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

Main Points Raised

  • One participant expresses frustration with dependent sources and seeks help with mesh analysis for finding voltage and power in the circuit.
  • Another participant suggests that the Thevenin equivalent resistance can be calculated using the open-circuit voltage and short-circuit current.
  • Several participants share their calculations for voltages and currents, with varying results, indicating uncertainty about their correctness.
  • There are discussions about the values of voltage sources and resistances, with some participants questioning the assumptions made in their calculations.
  • Participants propose different methods for calculating the Thevenin equivalent and express confusion about the relationships between various circuit parameters.
  • Some participants ask for clarification on specific equations and methods, indicating a lack of consensus on the approach to take.
  • There is mention of using supermesh analysis and the need to account for dependent sources in the calculations.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correct values for voltages and resistances, and multiple competing views on the methods for analysis remain. The discussion reflects a range of interpretations and calculations, with some participants expressing confusion and seeking further clarification.

Contextual Notes

Participants exhibit uncertainty regarding the correct application of mesh analysis and Thevenin's theorem, with some calculations leading to negative values for current and resistance, which raises questions about their validity. There are also unresolved assumptions about the circuit configuration and the impact of dependent sources on the analysis.

Who May Find This Useful

This discussion may be useful for students learning circuit analysis, particularly those struggling with mesh analysis, Thevenin's theorem, and the handling of dependent sources in circuits.

anthony0118
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Help!New to the circuit

Homework Statement


Question 1
Use the Mesh analysis to find v
0 and power absorbed by the 20 resistor in the circuit
Question 2
a) Find the Thevenin equivalent circuit external to the resistor R
b) Find the condition under which maximum power is transferred to R
c) Determine the maximum power transfer to R

Does anyone know how to solve them I tried them but it didnt seen right and the dependent source is so annoying is there any way to get rid of them? thanks

p.s untitiled is question 1 and 2 for question 2 thankz
 

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i got Vo + 24v don't know if that's right and quesrion 2 I am bit confused i got Rth 10ohms sound right?
Need great help thankz
 
There is no way to cancel dependent sources. Show what you have done in question 1 in order to be helped.
For question 2. Va is the difference between the 10V source and the open circuit voltage Vab. Develop your analysis and you will have one equation with one unknown: Vab.
In order to calculate the equivalent resistance, the best approach with dependent sources is to use Rth = Voc/Isc. You already have Voc. Short circuit terminals a and b and calculate Isc.
 
i get number one now i have I1 as -2.585A ,I2=-3.732A ,I3=-5A
AND V0= (I1-I2)R= 23V AND POWER = 26.45W
But i still don't know question2 where is the 10v u talking about here is and is Vab = Vth?
 
anthony0118 said:
i get number one now i have I1 as -2.585A ,I2=-3.732A ,I3=-5A
AND V0= (I1-I2)R= 23V AND POWER = 26.45W
But i still don't know question2 where is the 10v u talking about here is and is Vab = Vth?

Sorry! I meant 12V, not 10V. And Vab = Vth.
 
so is Rth = 4+2+4=10?
and Vth =
(Vth-2Va)/4 +Vth -12/4 -4=0
5Vth-4Va=84

what's Va?
12V/4=3A
3+4=7A
7*4=28v=Va?
Sorry I am trying my best to understand it
 
anthony0118 said:
so is Rth = 4+2+4=10?
and Vth =
(Vth-2Va)/4 +Vth -12/4 -4=0
5Vth-4Va=84

what's Va?
12V/4=3A
3+4=7A
7*4=28v=Va?
Sorry I am trying my best to understand it

Write the equation for the outside mesh, using Va as unknown. Remember that the current through the resistors 2 ohm and 4 ohm after the controlled source is the sum of the current through the firsr 4 ohm resistor with 4A of the current source. From this I found Va = -24V and Vab = Vth = 36V.
Next, short circuit the points a and b and calculate the current through them, Isc.

Rth = Vth/Isc.
 
CEL can you write out the equation coz i can't get anywhere close to 36V for Vth!
I did node a (Vth-12)/4 -4+ 2Va+Vth/6=0
Vth equals 14 sth and not even -24 or 36 I am really really lost and i spend any hours on this still can't finish part A how do u do part B and C
Max power is = Rth as well as Rn but where do i go from there

damn i feel stupid with this circuits stuff!
 
anthony0118 said:
CEL can you write out the equation coz i can't get anywhere close to 36V for Vth!
I did node a (Vth-12)/4 -4+ 2Va+Vth/6=0
Vth equals 14 sth and not even -24 or 36 I am really really lost and i spend any hours on this still can't finish part A how do u do part B and C
Max power is = Rth as well as Rn but where do i go from there

damn i feel stupid with this circuits stuff!

Calling Ia the current trough the 4ohm resistor:
Ia = Va/4
The outermost mesh:
-12 + Va - 2Va +(2+4)(Ia + 4) = 0
or
-12 -Va + 6Va/4 + 24 = 0
12 + Va/2 = 0
Va = -24V.

Vth = 12 - Va = 12 + 24 = 36V.
 
  • #10
Wow are u a teacher? That was quick so u did a supper mesh and find out Va and Vth is an open circuit ya? and Now how do i find R's resistor Rth is a short circuits right? only way i know is Rth=Vth/Isc any other way to solve other then the one i menotion any hint to start with finding Isc?

sorry my head is just not working atm and been on this question like 9 or 10 hrs already and havnt sleep for longlong time


I alomst finishing my self study! YOU should teach me I'm kind of doing it as self study so hard without a teacher
 
  • #11
anthony0118 said:
Wow are u a teacher? That was quick so u did a supper mesh and find out Va and Vth is an open circuit ya? and Now how do i find R's resistor Rth is a short circuits right? only way i know is Rth=Vth/Isc any other way to solve other then the one i menotion any hint to start with finding Isc?

sorry my head is just not working atm and been on this question like 9 or 10 hrs already and havnt sleep for longlong time


I alomst finishing my self study! YOU should teach me I'm kind of doing it as self study so hard without a teacher

When you short circuit points a and b you get two independent meshes: one with the two independent sources an one resistor and the other with the dependent source ant two resistors. Work from there.
 
  • #12
k i have loop 1 as -12 +4I1+16=0 and I1 =-1A

loop2 as -48+6I2=0 then I2=8A

is this what u mean in the last post and if so Isc= (I1-I2) = -9A
is Rth become -4?

max power would be?
 
  • #13
anthony0118 said:
k i have loop 1 as -12 +4I1+16=0 and I1 =-1A

loop2 as -48+6I2=0 then I2=8A

is this what u mean in the last post and if so Isc= (I1-I2) = -9A
is Rth become -4?

max power would be?

Where did you get the value 16 in the first equation? Loop 1 comprises the 12V source, the 4 ohm resistor and the short circuit ab.
 
  • #14
so u mean loop1 as -12+4I1=0 SINCE SHORT CIRCUIT AB IS THERE?
I1=3 BUT IS LOOP 2 RIGHT?

THEN ISC = -5A, Rth = -36/5

somehow i hva a feeling of that's wrong
 
  • #15
anthony0118 said:
so u mean loop1 as -12+4I1=0 SINCE SHORT CIRCUIT AB IS THERE?
I1=3 BUT IS LOOP 2 RIGHT?

THEN ISC = -5A, Rth = -36/5

somehow i hva a feeling of that's wrong
It is wrong. Don't forget that the current source also flows through the short circuit and that Va has now a new value that must be substituted in the dependent source.
 
  • #16
so what do u do with the current source then


ya VA was sub and loop2 = 2*-24+2i2+4i2=0
-48+6i2=0
i2=8A isn't that it?

and is ISC=(I1-I2)? IM KIND OF CONFUSED
 
  • #17
anthony0118 said:
so what do u do with the current source then


ya VA was sub and loop2 = 2*-24+2i2+4i2=0
-48+6i2=0
i2=8A isn't that it?

and is ISC=(I1-I2)? IM KIND OF CONFUSED

Where did you get the -24 value? Va = 4i1!
You have 3 currents through the short circuit: i1, i2 and 4A from the current source.
 
  • #18
im really confused
so how many loop is it

is it one on the lefthand side and second on the right of short circuit
 
  • #19
-12+4i1+16=0 loop 1
what do i do with the 4A in this loop do u do source converction? and change it to voltage? then that's 16V 4A*4OHMS


coz VA =-24V AND 2VA=-48V for the dependent source
2 and 4 ohms are added up
-48+6i2=0 and this is loop2


and one thing is Isc=(i1- i2)? if so how that make sense i have a negative number for current
that is how i arrive at that point and I am sorry i can't get it any further
 
  • #20
are u saying 4i1+6i2=0 for a second equation?
 
  • #21
demonstrate ur methods please master

i really need a proper teacher to teach me all this

thankz for ur time really
i helps me to think a bit faster :)
 
  • #22
anthony0118 said:
im really confused
so how many loop is it

is it one on the lefthand side and second on the right of short circuit

There are 3 loops:
1. The independent voltage source plus the 4 ohm resistor.
2. The current source.
3. The dependent voltage source and two resistors in series,
 
  • #23
anthony0118 said:
-12+4i1+16=0 loop 1
Where did you get the 16 V value?
what do i do with the 4A in this loop do u do source converction? and change it to voltage? then that's 16V 4A*4OHMS

The source and the short circuit form a loop.

coz VA =-24V AND 2VA=-48V for the dependent source
2 and 4 ohms are added up
-48+6i2=0 and this is loop2
No, Va was -24V when there was an open circuit between points a and b. Now there is a short circuit there, so you must write new equations.
and one thing is Isc=(i1- i2)? if so how that make sense i have a negative number for current
that is how i arrive at that point and I am sorry i can't get it any further
No, there are 3 currents through the short circuit.
 
  • #24
anthony0118 said:
are u saying 4i1+6i2=0 for a second equation?

No, i1 and i2 belong to independent loops.
 
  • #25
anthony0118 said:
demonstrate ur methods please master

i really need a proper teacher to teach me all this

thankz for ur time really
i helps me to think a bit faster :)

According to the rules of the forum, you must show some work before being helped.
 

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