# Circuit problem

Discussion in 'Homework Help' started by tracccccccccy, Jul 16, 2010.

Jul 16, 2010
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2. ### Georacer Moderator

Nov 25, 2009
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I will try a purely mathematical approach. If someone has a more intuitive one, please let us know your line of thought.

Here is the answer for question 2:
The power consumed on R is
$P_R=V_R-I \\
=I^2 \cdot R\\
=(frac{6}{20+5+R})^2 \cdot R\\
=frac{36 \cdot R}{(25 +R)^2}\\
=f(R)\\
\text{wich is max when its derivative reaches zero:}\\
f'(R)=36 \cdot frac{(25+R)^2-R \cdot 2 \cdot (25+R)}{(25+R)^4}=0 \Leftrightarrow\\
R^2=625 \Leftrightarrow \\
R=25$

and thus the statement is false.

By the same method, questions 1 and 3 are true.

The general idea is that when you want max power transfer on a load attached to a circuit, you want the resistance of the load to be equal to the output resistance of that circuit. However, I can't see how that applies to this circuit.

I might have done a logical mistake here, so somebody confirm or reject my result.

3. ### The Electrician AAC Fanatic!

Oct 9, 2007
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326
Before anything else, I should mention that in the real world the light bulb would not have a constant resistance of 20Ω as R is varied. But, for this problem, it's assumed to be just another resistor.

1) Since the 6V source by itself is assumed to have no internal resistance (the internal resistance is shown explicitly in this problem, as a 5Ω resistor), the power delivered by the cell (I think this must a battery of cells; I don't know of any readily available cells that can deliver 6V, but no mind) will be 6*I, so the greater current delivered by the cell, the greater power delivered by the cell.

2) Allow the 20Ω resistance of the bulb to become part of the internal resistance of the cell, which now becomes 25Ω. Then the maximum power theorem tells us that maximum power will be dissipated in R when its value is 25Ω, so 2) is false. And, of course, the same result can be obtained by just solving the circuit.

3) Trying the same trick of allowing the resistance of R to become part of the internal resistance of the cell won't work because the maximum power theorem only applies when it's the load which is varying. In this case, since it's R that varies, allowing it to become part of the internal resistance of the battery won't allow use of the maximum power theorem. Solving the circuit will show that maximum power is consumed by the bulb when R=0. Part 3) is false.

4. ### Georacer Moderator

Nov 25, 2009
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1,266
Isn't it more accurate to ask for the power that the voltage source ALONG with its internal resistance are offering? And thus calculating the power as P=(V-5*I)*I?

5. ### tracccccccccy Thread Starter New Member

Jul 16, 2010
2
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why does the 20 resistance become part of the internal resistance?

6. ### The Electrician AAC Fanatic!

Oct 9, 2007
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It depends on how you interpret this phrase "...P given out by the cell..." from the problem statement.

The problem creator already made an error by referring to the 6 volt source as a "cell"; there are no 6 volt chemical "cells" outside of research labs; it's a battery (of cells).

I thought it was reasonable to interpret "cell" as the electrochemical part in the dotted box, since the problem didn't say to consider the dotted box as a single physical "cell". It seemed to me that the 5Ω resistor was shown as an explicit, separate item for pedagogical purposes.

At any rate, it doesn't change the final result; maximum power is supplied by the energy source when R is zero, whether or not you consider the power dissipated by the 5Ω resistor.

7. ### The Electrician AAC Fanatic!

Oct 9, 2007
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It doesn't have to do so, but allowing the 20Ω resistor to be absorbed by the 5Ω resistor, you can treat the source as though it had a 25Ω internal resistance and then use the maximum power theorem.