Hi,

By process of elimination the 50 ohm resistor is the load resistor. They want to know what values you would choose for R1 and R2 given that the values can range from 0 to 100 ohms each to get the maximum possible power dissipation in the 50 ohm resistor.

This isnt your typical maximum power transfer question either it's actually simpler than that.

If we look at the circuit, since R2 is in parallel with the load when R2 is low it 'steals' power from the load, and so when R2 is at its maximum it 'steals' the minimum possible power from the load.
If we look at R1, when R1 is lowest is supplies the load with the maximum possible power, and when R1 is at max it supplies the most power to the load.

Taking this two arguments together i think you should be able to figure out what the best values are. If not, maybe you should look for another job opening

 

Hi,

By process of elimination the 50 ohm resistor is the load resistor. They want to know what values you would choose for R1 and R2 given that the values can range from 0 to 100 ohms each to get the maximum possible power dissipation in the 50 ohm resistor.

This isnt your typical maximum power transfer question either it's actually simpler than that.

If we look at the circuit, since R2 is in parallel with the load when R2 is low it 'steals' power from the load, and so when R2 is at its maximum it 'steals' the minimum possible power from the load.
If we look at R1, when R1 is lowest is supplies the load with the maximum possible power, and when R1 is at max it supplies the most power to the load.

Taking this two arguments together i think you should be able to figure out what the best values are. If not, maybe you should look for another job opening

KINDLY GO THROUGH THE ATTACHMENT WHERE I TRIED TO SOLVE AND GOT STUCK IN THE MID.

 

KINDLY GO THROUGH THE ATTACHMENT WHERE I TRIED TO SOLVE AND GOT STUCK IN THE MID.

Hi again,

As i stated in the second paragraph of post #21 this is not the typical power transfer question. We are not trying to find the the best load for the source or the best impedance for the source in general, we are just trying to find out how we can get the max power into the load when we can only vary the two resistors within a certain set range.

If you want to do this more analytically, then we can calculate the power in the load from the circuit and we get:
PL=(Vin^2*R2^2*RL)/(R2*RL+R1*RL+R1*R2)^2

and we also know that the maximum power that you can EVER get into a resistive load with a DC voltage source when the other elements are all passive too is:
Pm=Vin^2/RL

Now when we calculate the power from PL above with R1=0 and R2>0 we get:
PL=Vin^2/RL

which is the maximum we can ever get, so we know R1 must be zero.

Next, because R2 is in parallel with the load it does not matter that much because any value with R1=0 will still produce the mex power in the load of Vin^2/RL except for one value and that is zero ohms.

So strictly speaking R1=0 and R2 can be anything except zero, but it makes more sense to set it at the highest value so it does not consume power. The more exact result however is:
R1=0 and 0<R2<=100.

You may also want to note that for two parallel resistors R1 and R2 that equal a certain value (like 50 ohms in your pdf) then we get solution for R1 as:
R1=(50*R2)/(R2-50)

and you can note that R1 can be a lot of values depending on what you make R2. So there is no one solution for that problem. With a DC source with zero resistance available however you would not want to do it that way anyway.

 

Hi again,

As i stated in the second paragraph of post #21 this is not the typical power transfer question. We are not trying to find the the best load for the source or the best impedance for the source in general, we are just trying to find out how we can get the max power into the load when we can only vary the two resistors within a certain set range.

If you want to do this more analytically, then we can calculate the power in the load from the circuit and we get:
PL=(Vin^2*R2^2*RL)/(R2*RL+R1*RL+R1*R2)^2

and we also know that the maximum power that you can EVER get into a resistive load with a DC voltage source when the other elements are all passive too is:
Pm=Vin^2/RL

Now when we calculate the power from PL above with R1=0 and R2>0 we get:
PL=Vin^2/RL

which is the maximum we can ever get, so we know R1 must be zero.

Next, because R2 is in parallel with the load it does not matter that much because any value with R1=0 will still produce the mex power in the load of Vin^2/RL except for one value and that is zero ohms.

So strictly speaking R1=0 and R2 can be anything except zero, but it makes more sense to set it at the highest value so it does not consume power. The more exact result however is:
R1=0 and 0<R2<=100.

You may also want to note that for two parallel resistors R1 and R2 that equal a certain value (like 50 ohms in your pdf) then we get solution for R1 as:
R1=(50*R2)/(R2-50)

and you can note that R1 can be a lot of values depending on what you make R2. So there is no one solution for that problem. With a DC source with zero resistance available however you would not want to do it that way anyway.

 

I mean according to your conclusion literally finding the values R1 & R2 is not possible. Am I right? please reply me.

 

1. The maximum power transfer theorem is applicable only when the source resistance is known and it is required to determine the value of the load resistance at which maximum power is transferred.

2. This case is vice versa i.e. the load resistance is known and it is required to determine the value of the source resistance at which maximum power is transferred. The obvious answer is 'zero' since full voltage is available to the load only when the source resistance is zero. Hence R1=0.

3. With R1 being zero, the value of R2 has no bearing on the voltage applied to the load but contributes to a fixed loss within the source which can be kept low at R2=100.

4. Maximum power at load resistance = (load resistance) x (load current)² = 50 x (100/50)² = 200W.

Regards,

Nandu.

I agree with you that the way how you said R1=0.But I failed to understand how you fixed the value of R2=100 ohms.Could please clarify me?

 

is it not 100 ohm for both resistors?
it is maximum power transfer isn't it. I have been doing that recently at college. maximum power transfer when the output resistance is the same as the load, so r2 is 50 ohm and load is 50 ohm = maximum power transfer.
I thought that two same resistor values in parallel would equal half the resistance of one, R1||R2 = 100||100 = 50 ohms. its like a voltage divider R2 over R1 + R2 times voltage?
then again, im not very good at this stuff just a beginner really!

 

is it not 100 ohm for both resistors?
it is maximum power transfer isn't it. I have been doing that recently at college. maximum power transfer when the output resistance is the same as the load, so r2 is 50 ohm and load is 50 ohm = maximum power transfer.

Let's test your hypothesis and, thus, your understanding of the maximum power transfer theorem. You have a 12 V source that has a 2 Ω source resistance. It is driving and a 2 Ω load, which you claim results in the maximum possible power transfer to the load. What power is being delivered to the load? Now you are able to lower the source resistance to 1 Ω. How much power is being delivered to the load now? What does that tell you about the maximum power transfer theorem and/or your understanding of it?

 

Then SHOW us your attempt to solve it that way. Go back and try using the approach I suggested in Post #5. You've been given plenty of hints that should lead you to see the solution by inspection, but the fact that you aren't seeing how to exploit them is a strong indicator that you will be best served by slugging through it. At some point you will have an, "Ah hah!" moment and see the light.

You asked me to show my attempt. I posted it(my attempt) in the form of post kindly have a glance and reply me

 

I agree with you that the way how you said R1=0.But I failed to understand how you fixed the value of R2=100 ohms.Could please clarify me?

how come load current=100/50=2Amps. Could please justify it? Because it is important for me.please reply

 

Don't mistake this question with a question on impedance matching.

When the impedance of the source is non-zero, maximum power is delivered to the load when the load impedance matches that of the source (not the other way around, i.e. you don't match the source to the load).

In this question, maximum power is delivered when R1 = 0.

 

You asked me to show my attempt. I posted it(my attempt) in the form of post kindly have a glance and reply me

Your problem is EXACTLY what the last several posts have been talking about (which is good since it means that the thread hasn't been hijacked too badly). Since you won't do what I asked back in Posts #5 and #20, at least do what I asked ninjaman to do in Post #29. The problem is in your understanding of what the maximum power transfer theorem does and does not say.

 

Hello again,

Yes, as others have said many times now at least from 0 to 100 times is that this problem is not about the maximum power transfer it's just about finding the max power, so your attempt is entirely incorrect because it is not applicable. You did try and that was good, but now since we've seen it we can tell you what is wrong with your approach. It's not a matter of using that approach and finding R1, it's an entirely different process we must go through to get the right values.

And as you have been told, R2 is not fixed but can vary. It makes sense to set it at 100 ohms, but to satisfy the problem more exactly it can be almost anything (except zero).

What you are doing is trying to match the impedance of the load to the impedance of the source, which you DO NOT have to do here, because this is not an impedance matching problem. So you do not calculate the Thevenin resistance or anything like that. That is NOT the right approach for this problem.

Again, the power in the load for this circuit is:
PL=(Vin^2*R2^2*RL)/(R2*RL+R1*RL+R1*R2)^2

Dont be afraid to plug in some values for Vin, then play around with the values of R1 and R2 and see what the highest power you can get is.
You'll find that when you plug in R1=0 and R2= anything above zero, you'll get:
P=Vin^2/RL

which is the maximum power you can ever get in the load, period.

There's no impedance matching going on here.

It may take a while for this to sink in because you seem to be so determined to use impedance matching. You must abandon that now and move on.

 

Oh, it's about finding the maximum power transfer, it's just that the maximum power transfer theorem does not apply because part of the maximum power transfer theorem is a caveat that people conveniently forget: "For a fixed source impedance...."

 

You asked me to show my attempt. I posted it(my attempt) in the form of post kindly have a glance and reply me

What is your math background -- specifically, do you have any calculus at all behind you?

 

In other words power at 50 Ohm resistor is: P=VI=V(V/R). So when R1=0, all the voltage appears across the load, and P=100(100/50)=200 Watts.

It is effing algebra. No need for Thevenin "stuff" or any other theorems. Just do the effing math dude!

 

When R1=0, R2 is just another resistance parallel to the 50 ohms load. Its value has no effect on the load current. Since the specified range is 0-100 ohms (both values inclusive) 100 ohms is the maximum value which can be chosen to keep the power loss due to R2 low.

Regards,

Nandu.

Though, in all fairness, the problem places no spec on minimizing power loss. But it is a reasonable and justifiable way to choose R2, without further information, particularly if it is noted that the specific value of R2 (other than the mathematical problem associated with R2=0Ω) is immaterial but that maximizing it is a reasonable thing to do.

 

With R1 equal to zero, R2 can be anything from 1e-99 ohms to 1e99 ohms. It just doesn't matter. The maximum power on the load is when the total source voltage is applied to the load. R2 has no bearing on the load power. In the world of perfect sources, who cares about the value of R2.

Of course, when dealing with real voltage sources, you would want R2 as high as possible.

Remember the question only asked about load power.

 

It's fascinating that this question was used in an interview. Given the informative speculation & comment in this thread, I can envisage some interesting / robust discussion arising during some of the interviews.

 

Hi,

Yes, and the next question on that page was, "On the lines provided below, describe your TOE (Theory Of Everything)"
Or maybe, "If John has 9 marbles and Mary has 3 marbles and John gives 2 marbles to Mary, how many marbles do you have?"

Seriously though the OP is going to have to give the resistors/power problem some deep thought.