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HelicopterMain Discussion › can i use v=ir
10-20-2004 04:56 PM  13 years agoPost 21
washout

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london, england

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is it V=IR
V/R=I
SO 15/15.5OHM= .96amps so nearly 1amp
now resistor on other line is 10ohms so how does this effect the overall current? is it in parallel what was 1/v=1/r1+1/r2 if in parallel???
remember the stated output is 70ma
ive resoldered it back so it will live

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10-20-2004 05:16 PM  13 years agoPost 22
ifixairplanes

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Haverhill, MA. USA

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if the resistor on the pther side is connected to the LED then it dosnt do anything for the circuit you are playing with, just keeping the voltage to an acceptable level for the LED.

sean

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10-20-2004 05:18 PM  13 years agoPost 23
Peter Wales

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Orlando Fl

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That tells me that the transformer is a 12v nominal voltage one. To increase the charge current wire a 1/2 watt 22 ohm resisor across the 15ohm one. That will take it to 110mA.

The other resistor and LED are across the existing resistor sensing that it is dropping voltage. You may need to change the 10 ohm to 4.7 ohms

However, I must say, I agree with the others who say that 11.1v is a fine terminal voltage for a 9.6v battery. Now you know how to do it, dont bother and leave it all alone

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10-20-2004 05:20 PM  13 years agoPost 24
washout

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london, england

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but it must as why else is it putting out 70ma (by the way i pm'd u a second ago i was wrong)
if the other resistor didnt effect it then it would put out 1000ma or 1amp

speaking of which how do computer chargers work does the cpu vary a variable resistor and how thru relays(as otherwise the cpu would fry) its very interesting if u think about it.

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10-20-2004 05:22 PM  13 years agoPost 25
washout

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london, england

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peter. how did u figure that out (show ur calculations as if u were in school again)
thanks and yes i wil leave it alone AFTER i find out how

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10-20-2004 07:12 PM  13 years agoPost 26
Peter Wales

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Orlando Fl

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You are assuming the charge voltage is 12 or 15v. Its not. In this case its 12v - 9.6v which is the battery voltage ie 2.4v

The makers of the charger have assumed that the fully charged voltage of the battery will be 10.9v ie the voltage difference will be 1.1v. When the voltage difference is 1.1v they want to pump 70mA into the battery and from V=IR 1.1/.07 = 15.9 or so ohms.

This means that when the battery is very low, it will get a lot more current pumped into it and when it is high, less current, but as it gets less current, so the terminal volatage of the transformer will rise etc etc

The charge current is going to vary enormously depending on the battery voltage, so, simply assume that it is 70mA as you said, which is going to be the final trickle charge rate

If the resistor is 15ohms, then in order to increase the current, you need to reduce the resistance. Rather than back calculate it, I proportioned it from the 15 ohm value. ie if I added another 15 ohms in parallel, I would double the current. To get 110mA gave me a value of 10 ohms, roughly. The nearest empirical value would be 22 ohms in parallel with 15 ohms to give me 10 ohms

from 1/15 + 1/X=1/10 where X is the new resistor.

I hope it makes sense.

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10-20-2004 07:27 PM  13 years agoPost 27
washout

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london, england

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VERY CLEVER peter u figured it out thanks il leave it alone after all that

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