Message 21441 in TECH.WW
From: G8MNY
Date: Mon, 27 Apr 26 06:45:00 Z
Newsgroups: TECH.WW
Subject: 3 Rs Physics Problem (Answers) (Updated Dec 19)
Message-ID: <58228_GB7CIP>
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From: G8MNY@GB7CIP.#32.GBR.EURO
To : TECH@WW
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Helmut DK2ZA @ DB0FOR.#BAY.DEU.EU posed this question..
Here is one problem from a German physics contest for 16 year olds:
You have three resistors R1 = 10 Ohm, R2 = 20 Ohm & R3 = 30 Ohm.
Each resistor can absorb at most 5 Watts. You have only one source of
electrical power, the voltage of which can be adjusted to any necessary value.
How do the resistors have to be connected so that when voltage is applied the
total absorbed power is a maximum? How many watts will that be?
Have fun & vy 73 de Helmut, DK2ZA
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My answer..
There are 8 ways to configure the 3 Rs.. (2^3)
1/ All in series..
As currents all the same, highest R dissapates 5W.
So across all Rs, Voltage = 2x12.25 = 25.5V &
Total Power = 5+5 = 10W
2/ All in parallel..
Total Power = 5+(5/2)+(5/3) = 9.166W
3/ Parallel + Series A)..
4/ Parallel + Series B)..
5/ Parallel + Series C)..
6/ Series + Parallel a)..
It is not obvious which R will have the max power!
20 30 So P20 =0.424*0.424*20= 3.6W,
So P30 =0.283*0.283*30= 2.4W,
Total Power = 5+3.6+2.4 = 11W
7/ Series + Parallel b)..
30 10 So P20 =0.375*0.375*10= 1.4W,
So P30 =0.125*0.125*30= 0.47W,
Total Power = 5+1.4+0.47 = 6.87W
8/ Series + Parallel c)..
10 20 So P10 =0.272*0.272*10= 0.74W
So P20 =0.136*0.136*20= 0.37W
Total Power = 5+0.74+0.37 = 6.11W
So the answer is configuration 6/ to give 11 Watts.
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Here is another 3 Rs Question that you can do in your head.
What does this measure?
25 50 75
A simple way to solve it, is to look for a common mutiple. I saw "150".
e.g. 75 is 2x 150s in parallel, 50 is 3 in parallel, & 25 is 6 in parallel,
making a total of 11 150s in parallel = 150/11 = 13.636 QED.
Why don't U send an interesting bul?
73 de John G8MNY @ GB7CIP
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