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G8MNY  > TECH     13.05.22 20:36z 102 Lines 3370 Bytes #28 (0) @ WW
BID : 21596_GB7CIP
Subj: 3 Rs Physics Problem (Answers)  (Updated Dec 19)
Path: SP7YDD<SR1BSZ<IW0QNL<IZ3LSV<IQ2LB<IR2UFV<SK0BO<SM0YOS<GB7CIP
Sent: 220513/2031Z @:GB7CIP.#32.GBR.EURO #:21596 [Caterham Surrey GBR]
From: G8MNY@GB7CIP.#32.GBR.EURO
To  : TECH@WW

(8 Bit ASCII graphics use code page 437 or 850, Terminal Font)
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
------------------------------------------------------------------

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.
102030  P30=5W, V30=5*30 = 12.25V which is 50% of all R.
            So across all Rs, Voltage = 2x12.25 = 25.5V &
            Total Power = 5+5 = 10W

2/ All in parallel..
Ŀ   As the voltage are all the same, lowest R has 5W.
 10 20 30   P10=5W, V10=5x10 = 7.07V, as other Rs are mutiples
   their power is proportionally less.
            Total Power = 5+(5/2)+(5/3) = 9.166W

3/ Parallel + Series A)..
Ŀ      Here the 2 series Rs = the parallel R, so that is 5W
 10        P30=5W. V=12.25V as in 1/.
   30      Total Power = 5+5 = 10W
 20  


4/ Parallel + Series B)..
Ŀ      Here the 2 series Rs = 4x the parallel R, so that is 5W
 20        P10=5W. V=7.07 as in 2/. Series Rs have 1/5 the power, so
   10      Total Power = 5+(5/5) = 6.00W
 30  


5/ Parallel + Series C)..
Ŀ      Here the 2 series Rs = 2x the parallel R, so that has 5W
 30        P20=5W. V=20*5 = 10V. Series Rs have 1/2 the power, so
   20      Total Power = 5+2.5 = 7.5W
 10  


6/ Series + Parallel a)..
     It is not obvious which R will have the max power!
            Assume 10R has 5W, then I = (5/10) = 0.707A,
10Ŀ   I in 20R = 0.707*(30/50) = 0.424A,
    20 30   So P20 =0.424*0.424*20= 3.6W,
   I in 30R = 0.707*(20/50) = 0.283A,
            So P30 =0.283*0.283*30= 2.4W,
            Total Power = 5+3.6+2.4 = 11W

7/ Series + Parallel b)..
            Assume 20R has 5W, then I =(5/20)= 0.5A
20Ŀ   I in 10R = 0.5*(30/40) = 0.375A,
    30 10   So P20 =0.375*0.375*10= 1.4W,
   I in 30R = 0.5*(10/40) = 0.125A,
            So P30 =0.125*0.125*30= 0.47W,
            Total Power = 5+1.4+0.47 = 6.87W

8/ Series + Parallel c)..
            Assume 30R has 5W, then I=(5/30)= 0.408A
30Ŀ   I in 10R = 0.408*(20/30) = 0.272A,
    10 20   So P10 =0.272*0.272*10= 0.74W
   I in 20R = 0.408*(10/30) = 0.136A,
            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.

-------------------------------------------------------------------------------

Here is another 3 Rs Question that you can do in yor head.

What does this measure?
Ŀ
  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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