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G8MNY > TECH 17.10.06 23:35l 86 Lines 2871 Bytes #999 (0) @ WW
BID : 51322_GB7CIP
Read: GUEST DL1LCA DL8FBH
Subj: Physics Problem (Answers)
Path: DB0FHN<DB0MRW<OK0PPL<DB0RES<ON0AR<GB7CIP
Sent: 061017/1716Z @:GB7CIP.#32.GBR.EU #:51322 [Caterham] $:51322_GB7CIP
From: G8MNY@GB7CIP.#32.GBR.EU
To : TECH@WW
Helut 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 and 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 and vy 73 de Helmut, DK2ZA
------------------------------------------------------------------
My answer..
There are 8 ways to configure the 3 Rs..
1/ All in series..
As currents all the same, highest R dissapates 5W.
Ä10Ä20Ä30Ä P30=5W, V30=û5*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..
ÄÄÂÄÄÂÄÄ¿ As the voltage are all the same, lowest R has 5W.
10 20 30 P10=5W, V10=û5x10 = 7.07V, as other Rs are mutiples
ÄÄÁÄÄÁÄÄÙ their power is proportionally less.
Total Power = 5+(5/2)+(5/3) = 9.166W
3/ Parallel + Series A)..
ÄÄÂÄÄ¿ Here the 2 series Rs = the parralel R, so that is 5W
10 ³ P30=5W. V=12.5V as above.
³ 30 Total Power = 5+5 = 10W
20 ³
ÄÄÁÄÄÙ
4/ Parallel + Series B)..
ÄÄÂÄÄ¿ Here the 2 series Rs = 4x the parralel R, so that is 5W
20 ³ P10=5W. V=7.07 as above. Series Rs have 1/4 the power, so
³ 10 Total Power = 5+(5/4) = 6.25W
30 ³
ÄÄÁÄÄÙ
5/ Parallel + Series C)..
ÄÄÂÄÄ¿ Here the 2 series Rs = 2x the parralel R, so that has 5W
30 ³ P20=5W. V=û20*5 = 10V. Series Rs habe 1/2 the power, so
³ 20 Total Power = 5+2.5 = 7.5W
10 ³
ÄÄÁÄÄÙ
6/ Series + Parallel a)..
It is not obvious which R will have the max power!
Assume 10R has 5W, then I = û(5/10) = 0.707A
ÄÄ10ÄÂÄÄ¿ I in 20R = 0.707*(30/50) = 0.424A,
20 30 So P20 =0.424*0.424*20= 3.6W
ÄÄÄÄÄÁÄÄÙ 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 =û(5/20)= 0.5A
ÄÄ20ÄÂÄÄ¿ I in 10R = 0.5*(30/40) = 0.375A,
30 10 So P20 =0.375*0.375*10= 1.4W
ÄÄÄÄÄÁÄÄÙ 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=û(5/30)= 0.408A
ÄÄ30ÄÂÄÄ¿ I in 10R = 0.408*(20/30) = 0.272A,
10 20 So P10 =0.272*0.272*10= 0.74W
ÄÄÄÄÄÁÄÄÙ 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 cofiguration 6/ to give 11 Watts.
Y don't U send an interesting bul?
73 de John G8MNY @ GB7CIP
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