Kirchoffs law

[sunny1982]

Hi Folks
I desperately needed help on this question, I have to calculate the value of the RL load, I'm guessing I have to find the resistance of the whole circuit and what ever resitance is that is the resistance of RL, I have to do this using kirchoffs, where have I gone wrong? I need to calculate the value of RL then calculate the maximum power transfer. Any help please?

 

[MrAl]

Hello,

Short out the voltage supply and calculate the resistance across E and F.

 

[WTP Pepper]

Agree with MrAl. Alternatively try Thevenins equivalent theorm. Max Power is when RL=Rsource (resistance looking back into the network with an ideal voltage source of zero ohms or a short circuit) which I reckon without a calculator to be 8ish ohms range.

 

[panic mode]

if you are looking for shortcomings, then look at the circuit, where are all the currents and what are their respective directions? you don't have to use Thevenin, but some extra legwork may be required, depending on approach (there are many ways to skin the cat).

your teacher may appreciate if you choose less obvious method.

for example use voltage divider rule (there is a cascade), then you can express power of RL as V^2/RL.
once you have expression for power of the RL, then solve dP/dRL=0

 

[sunny1982]

I've tried this task again, I know that the resitance in RL is correct, but I think my presentation of my work is suffering as I'm not sure if my diagram is correct and if the formulas are correct according to the thevenins equivalent circuit I was wondering if someone could help me out here I would be very grateful.

 

[panic mode]

Vb=E5*R8/(R7+R8)
Ve=Vb*R10/(R9+R10)

therefore

Ve = E5*R8*R10/[(R7+R8) (R10+R9)]

But if we have a load, then
we must use

R = Rload || R10

instead of R10

R=Rload*R10/(Rload+R10)

Ve = E5*R8*Rload*R10/[(R7+R8)(Rload+R10)(Rload*R10/(Rload+R10)+R9)]

Now that we have voltage across load, it is easy to compute power

Pload= Ve^2 / Rload

and to find max power transfer we need to solve

dPload/Rload=0

good luck with differentiating.
tip, try to separate the constant terms (anything without R10) and terms that include R10.
you should get something of form

A/B=0

which is true if A=0, and this will give you result that Rload = Rth = Rinternal as seen from the load

 

[Ratchit]

sunny1982,

I desperately needed help on this question, I have to calculate the value of the RL load,

Under what conditions? RL can be anything.

I'm guessing I have to find the resistance of the whole circuit and what ever resitance is that is the resistance of RL,

What do you define as the "whole circuit"? Everything including RL?

I have to do this using kirchoffs,

You mean Kirchoff's law? Then you can't use Thevenin's theorem, can you? That would be against your teachers instructions, wouldn't it?

where have I gone wrong?

You are implementing K's law by loop equations, which is good. But, your application is faulty. Look at the way I have written them out, and see how it compares with your work.

I need to calculate the value of RL then calculate the maximum power transfer. Any help please?

Why do you even need to know the maximum power transferred? You want to find the value of RL which give you the maximum power transfer.

OK, let's get started. Look at the BIG PICTURE. You want to find the value of RL so that the most power is transferred to it. You can do this without differentiating. First the loop equations according to K's law, assuming all currents exist in a clockwise direction.

For the west loop: 15+9*I1+(I1-I2)*25=0
For the middle loop: (I2-I1)*25+(I2-I3)*24+I2*7=0
For the east loop: (I3-I2)*24+I3*RL=0

This is easily solved for;
I1 = -(120*(96+7*RL))/(11112+1279*RL)
I3 = -9000/(11112+1279*RL)

We don't care about I2. According to the max power transfer theorem, RL will can only dissipate a maximum of one-half the power supplied by the voltage source. So:

(I1*15)/2 =((I3)^2)*RL ======>-9000*RL/(11112+1279*RL) = -(900*(96+7*RL))/(11112+1279*RL)

which is easily solved for RL = 32

Ratch

 

[JoeJester]

I got a Rth of 8.688 Ω so the RL should not be 32 ohms for maximum power transfer. The Max Power pdf could not open so I have no idea what the OP has for RL.

 

[MikeMl]

Joe is right; Ratch=Bzzzzt.

Look at the blue trace which is power in RL as a function of the value of RL as RL is varied from 7Ω to 10Ω (the independent variable of the simulation, dependent values are voltage at node E and the current through RL).

To the OP. Simulation is not what you should be doing right now, but you should take away that it can easily solve a system of equations in several unknowns based solely on an input of a circuit, and that the max power point occurs when the derivative of power goes to zero...

 

[sunny1982]

Sorry about that the PDF keeps going corrupt for some reason but I've posted it again so if anyone can be kind enough to look and advise me I would be grateful.

 

[Ratchit]

JoeJester,

I got a Rth of 8.688 Ω so the RL should not be 32 ohms for maximum power transfer. The Max Power pdf could not open so I have no idea what the OP has for RL.

Yes, you are correct. That is the Thevenin value, but the instructions were not to use Thevenin's theorem. I was wrong in assuming that the power dissipated in a network is the same as the the power of the Thevenin equivalent. For the example given, the total power supplied by the 15 volt source is 12.70 watts, whereas the power dissipated by the equivalent load is only 2.85 watts. Don't use equivalent resistances for power calculations. What I should have done was calculate the open circuit voltage with RL = ∞, and the short circuit current with RL = 0 using loop equations. Then I could get the circuit impedance looking back from RL.

So, from my previous post:
I1 = -(120*(96+7*RL))/(11112+1279*RL)
I2 = -(375*(24+RL))/(11112+1279*RL)
I3 = -9000/(11112+1279*RL)

Setting RL to ∞ in I2 and multiplying by R10=24 gives the open circuit voltage of 7.04 volts.
Setting RL to 0 in I3 gives the short circuit current of 0.81 amps.

Dividing 7.04 by 0.81 gives an impedance of 8.69 ohms.

Ratch

 

[Ratchit]

MikeMl,

Joe is right; Ratch=Bzzzzt.

The value is correct, the method is not. The problem should not be worked with Thevenin's theorem or a simulator. It would have been more enlightening if you explained where I went wrong is my solution. My corrected solution is using K's method the way the OP wanted.

Ratch

 

[MikeMl]

MikeMl, ...The problem should not be worked with ... a simulator. ...

I just knew you were going to say that, which is why I said "Simulation is not what you should be doing right now, but you should take away...

At least I didn't just work the entire problem for the student; where is the learning in that?

 

[sunny1982]

JoeJester,



Yes, you are correct. That is the Thevenin value, but the instructions were not to use Thevenin's theorem. I was wrong in assuming that the power dissipated in a network is the same as the the power of the Thevenin equivalent. For the example given, the total power supplied by the 15 volt source is 12.70 watts, whereas the power dissipated by the equivalent load is only 2.85 watts. Don't use equivalent resistances for power calculations. What I should have done was calculate the open circuit voltage with RL = ∞, and the short circuit current with RL = 0 using loop equations. Then I could get the circuit impedance looking back from RL.

So, from my previous post:
I1 = -(120*(96+7*RL))/(11112+1279*RL)
I2 = -(375*(24+RL))/(11112+1279*RL)
I3 = -9000/(11112+1279*RL)

Setting RL to ∞ in I2 and multiplying by R10=24 gives the open circuit voltage of 7.04 volts.
Setting RL to 0 in I3 gives the short circuit current of 0.81 amps.

Dividing 7.04 by 0.81 gives an impedance of 8.69 ohms.

Ratch

This is briliant stuff so you did this using kirchoffs so this is how I need to do it as this task has to be done solely using kirchoffs law. But I would be very grateful if you can clarify a few things for me please.

What do you mean by set RL to ∞ and set RL to 0?

 

[KeepItSimpleStupid]

This equation 0v = 7ΩI₂ + 25Ω (I₂ - I₁) + 24ΩI₂ is wrong. try 0v = 7ΩI₂ + 25Ω (I₂ - I₁) + 24Ω(I₂-I3)

It helps to actually draw the loop currents.

 

[MikeMl]

...What do you mean by set RL to ∞ and set RL to 0?

He is trying to find the current that the network will put out if RL=0 (i.e. replaced with a short) and the voltage across RL if RL=∞ (i.e. the network is open circuited). That can be done using only KCL. However, you must have studied Thevenin to know that the network can be replaced with a new voltage source of the same voltage as the open circuit voltage, in series with a resistor of Reff=Voc/Ish.

Having those two values, you have to know apriori that the max power transfer occurs when RL is set to Reff. If you dont know that apriori, you have to set up a voltage divider consisting of Reff and RL, calculate the power in RL as a function of RL, differentiate that equation and find where it crosses zero...

All of the things that Spice knows how to do intrinsically...

 

[sunny1982]

He is trying to find the current that the network will put out if RL=0 (i.e. replaced with a short) and the voltage across RL if RL=∞ (i.e. the network is open circuited). That can be done using only KCL. However, you must have studied Thevenin to know that the network can be replaced with a new voltage source of the same voltage as the open circuit voltage, in series with a resistor of Reff=Voc/Ish.

Having those two values, you have to know apriori that the max power transfer occurs when RL is set to Reff. If you dont know that apriori, you have to set up a voltage divider consisting of Reff and RL, calculate the power in RL as a function of RL, differentiate that equation and find where it crosses zero...

All of the things that Spice knows how to do intrinsically...

So are these are all the equations I should add to complete this task?
I1 = -(120*(96+7*RL))/(11112+1279*RL)
I2 = -(375*(24+RL))/(11112+1279*RL)
I3 = -9000/(11112+1279*RL)

 

[Ratchit]

MikeMl,

I just knew you were going to say that, which is why I said "Simulation is not what you should be doing right now, but you should take away...

At least I didn't just work the entire problem for the student; where is the learning in that?

This thread has been burning since 3/35/2013. The OP did learn something. Specifically, he saw his mistake in the loop equations.

All of the things that Spice knows how to do intrinsically...

Where in Spice is there a feature concerning the max power transfer theorem?

Ratch

 

[Ratchit]

Kiss,

This equation 0v = 7ΩI₂ + 25Ω (I₂ - I₁) + 24ΩI₂ is wrong. try 0v = 7ΩI₂ + 25Ω (I₂ - I₁) + 24Ω(I₂-I3)

It helps to actually draw the loop currents.

Which post # shows the wrong equation? Doesn't post #7 show the correct equation? It helps to specify the post # where the error is.

Ratch

 

[sunny1982]

Kiss,



Which post # shows the wrong equation? Doesn't post #7 show the correct equation? It helps to specify the post # where the error is.

Ratch

I think he means its wrong in my pdf, which I have corrected now and I am just going to add all the equations on that you have helped me with.