Very simple circuit - and still stuck

Discussion in 'Homework Help' started by pianolife, Dec 30, 2012.

1. pianolife Thread Starter New Member

Nov 15, 2012
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0
Dear members of AllAboutCircuits, I am really sorry for asking such a simple question, but I am really a newbie in electronics.

For the circuit shown in the attachment, what value of RL will result in maximum power transfer? Determine the maximum power dissipated in RL.

What I know is that the max power transfer is when Rs=RL. I started to find the total Rs. But:
- shall I consider the 1Ω resistor in parallel or in series? If I "get rid" of the load, that resistor become in series with 2Ω.
- assuming this is in parallel, is there any specific order to find the total resistance? I mean, if I calculate 3Ω||4Ω then in series with 2Ω I get a value, if I find 4Ω||2Ω then in series with 3Ω I get obviously a different one.
Which is correct?

Please forgive me for this question.
Thank you so much.

All the best

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2. panic mode Senior Member

Oct 10, 2011
1,328
305
correct, max power transfer is when RL matches internal resistance of the source.

there are different ways to do this. one common way is to use Thevenin.
when circuit is open and without RL, you get certain voltage out. this is your Eth.

when you short output, current will not flow through 1Ohm because path of least resistance is the short. you need to calculate that short circuit current.
then you can divide the two results to get Rth which will be same value as RL you are looking for.

Rth=Eth / Isc

Eth= Thevenin voltage
Isc = short circuit current
Rth = thevenin resistance

3. mbohuntr Active Member

Apr 6, 2009
413
32
Draw the flow of current using arrows, and things will get clearer. ALL current must pass thru the left side 2Ω resistor, all other resistors are in parallel to it. The total resistance will be 2Ω + (4 Ω parallel with 3 Ω)

4. pianolife Thread Starter New Member

Nov 15, 2012
28
0
Thank you very very much.
Then, just for double check, I got Total Resistance= 3.7, and the power dissipated in RL is 100/3.7
Is that correct?

5. panic mode Senior Member

Oct 10, 2011
1,328
305
no....

1. decide on method,
2. choose what to calculate
3. calculate
4. repeat until done

and if you need help, you need to provide steps and values of your attempt.

in your case you seem to have decided to find total resistance of the circuit WITHOUT load RL. This produced approximate result of total circuit resistance (without RL). then you calculated power supplied by 10V source to circuit WITHOUT RL. this is NOT power consumed by RL. In fact RL is not even part of the circuit yet.

information like this can be used to find some additional things about circuit but it is a LONG way to find what you are after and to get correct result you need clear approach.

6. panic mode Senior Member

Oct 10, 2011
1,328
305
your "source" is the "circuit on the left" (10V source and 4 resistors).

this source is what would provide power to RL if RL is connected to it.

power dissipated by RL depends on resistance of RL and voltage or current provided by source.

note, voltage of "source" is NOT the 10V, it is voltage across 1 Ohm resistor (because THIS is where RL gets connected).

we REALLY need to ANNOTATE circuit before anything can take place. for example, I suggest to call the elements:
E1= 10 V
R1= 2 Ohm
R2= 4 Ohm
R3= 2 Ohm
R4= 1 Ohm
Eth = voltage across R4 (and accross RL)

note. when RL is not connected Eth will be some value A.
when RL is connected, Eth will be some value B.
and A>B

power consumed by RL is PL=B^2/RL
you need to find value of RL that makes this power PL maximum.

note B^2 is not 100....

note 3.7 Ohm is resistance of the circuit without RL as seen from E1. this is NOT the same as resistance seen by the load.

note, question does NOT ask what is that power. you do not need numeric answer for this. only the resistance or RL for which this is true.

7. pianolife Thread Starter New Member

Nov 15, 2012
28
0
Thank you so much for your help, panic mode. And I'm sorry for wasting your time. I guess I need to review the basic stuff as well, I get stuck even on the easiest things.

Thank you very much, and please forgive me.

8. mbohuntr Active Member

Apr 6, 2009
413
32
Panic mode is absolutely correct, I did not give you the whole method, I should have been more specific in my response not including the load. I apologise for the confusion I caused. I was trying to lead you along to show how the methodology works. You are NOT wasting anyones time, we all learn something from the exchange of knowledge. Keep up the studies and questions.

9. pianolife Thread Starter New Member

Nov 15, 2012
28
0
Thank you so much to all of you!

Another question, sorry: could you please suggest any book/website/pdf/whatever to study on?
I am using Fundamentals of Electric Circuits (Alexander/Sadiku)

Apr 5, 2008
15,799
2,385
Hello,

Just take a look at our eBook:

Bertus

11. pianolife Thread Starter New Member

Nov 15, 2012
28
0
Brilliant, Bertus, thank you so much.

Coming back to my question, can I find the equivalent resistance by doing R1+(R2||R3), then find the voltage drop across Req and R4.
Calculating the current in the circuit, and knowing that the Voltage Drop across R4 must be the same as the drop across RL, could I find the resistance?

And then, by having the voltage drop across RL, the resistor value, could I find the power?

Just guessing...

12. panic mode Senior Member

Oct 10, 2011
1,328
305
hello,

for case1 (without RL, output is open and we are determining Eth) you can use the Req=R1+(R2||(R3+R4)). this can be used to find total current draw from the E1 and thus voltage drop accross R1. then using Kirchoffs Law you can get voltage across R2. then you can find current through R3 and R4 and Eth.

for case2 (without RL, but R4 is shorted) you can use the Req=R1+(R2||(R3+R4)). but because R4 is shorted (0 Ohm), equation becomes Req=R1+(R2||R3). this can be used to find total current draw from the E1 and thus voltage drop accross R1. then using Kirchoffs Law you can get voltage across R2. then you can find current through R3 and R4. This current is Isc (btw. compare this part with "case1").

calculate Rth from known Eth and Isc. RL is same as Rth. done....

the long way is to include RL into circuit, derive equation of power for RL and solve dPL/dRL=0 which is a good exercise. you will use this many times to find min and max of some function (dy/dx=0).

pianolife likes this.