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Kirchoff's Laws : Electricity equations
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Kirchoff's Laws : Electricity equations
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andy2405
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Kirchoff's Laws : Electricity equations
«
on:
04/03/2004 20:38:23 »
the field of a serias motor has a restince of 0.6 ohms and is connect to the diverter resitor(parallel) which has three steps 5 ohms 4 ohms and 2 ohms, assuming that the current is fixed at 28 amps what is the current at each step of the diverter
The answers should be a)25
b)24.4
c)21.6
how do i get the answers, what are the equations used, hepthis has been driving me nuts for days.[?]
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NakedScientist
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Re: Kirchoff's Laws : Electricity equations
«
Reply #1 on:
05/03/2004 04:36:01 »
Dear Andy
I'm not sure exactly what a 'diverter' is, but is Kirchoff's Law any use for resistors in parallel ?
1/ R
eq
= 1 / R
1
+ 1 / R
2
where R1 and R2 are the parallel resistors. You can extend the law to add further terms for more resistors.
Chris
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andy2405
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Re: Kirchoff's Laws : Electricity equations
«
Reply #2 on:
05/03/2004 07:48:32 »
to work out the parrell side using x-1 on the sceintific calculator.
rt= r1/1 + r2/1 + r3/1=?/1=
5/1 + 4/1 + 2/1 = 0.95/1 = 1.05 ohms
plus the serias equation gives me 1.65263 ohms
and v=i x r
i=v/r
r=v/i
so my voltage is 28a x 1.652ohm = 46.256v
therefor i1=v/r 46.256v / 5 = 9.2512a
i2=v/r " " / 4 = 11.564a
i3=v/r " " / 2 = 23.128a
i4=v/r " " / 0.6 = 77.09a
but i'm supposed to get the amperage passing through each resitor in the parrel circuit (5,4,2 ohm restior) to the answers a)25a b)24.4a c)21.6 and whatever do it does'nt work out.where am i going wrong. ?
is it a typo in my question book, help ?[xx(],
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Quantumcat
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Re: Kirchoff's Laws : Electricity equations
«
Reply #3 on:
05/03/2004 08:33:38 »
A diverter resistor is a variable resistor I think (you should be able to tell by the fact it has settings with different resistances chris!). I remember doing this last year !!! I wish we were doing this here instead of stupid Work (mutters) but anyway, I don't really remember much since it was like a whole year ago (lol) I would try and answer but I don't ever remember a question with the electricity producer (is that what a series motor is?) having resistance ... anyway, maybe you could work out the voltage with V = IR in the motor since you know its ampage there, then with the known voltage and known resistances you can work out the current. so, v = 28 x 0.6 = 16.8 = I x 5 I = 3.36 well that's not even near 25 so I guess that's not the way to do it, shouldn't have tried, lol.
Am I dead? Am I alive? I'm both!
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Re: Kirchoff's Laws : Electricity equations
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Reply #3 on:
05/03/2004 08:33:38 »
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