Supposed you have a dc source of 1 ampere and you short it... from I=V/R, R=0, what then is the current produced??
Also suppose the voltage is 10 volts, and I want the maximum current, from I=V/R. What is the least ohms value can I use, 10 ohms? 1 ohm? Suppose the maximum current of the source is 1 ampere. What value of current would I get when I use a 1 ohm resistor in the 10 volt circuit? The I=v/r gives out I=10/1ohm= 10 Ampere which doesn't make sense. What is the rule of units used? Thanks.
Lorraine
P.S. May I know what are other electronic newsgroups with lots of participants, tnx.
What you probably want, is maximum effect over the connected load. Then, you have to know the internal resistance in the source. Most power will go into the load, if the internal and load resistance is the same. However, you'll then make as much effect in the source as the load...
R will never become 0, but for instance 1*10^-15. With 10 volts, I=10/(1*10^-15)=1*10^16. However, if your dc source is current-limited at 1 amp., drawn current will be 1 amp while the voltage is lowered to
I see. So there is a lower limit to the resistor used to make constant volage. For example. Voltage is constant at 10 volts if the resistor varies from
10 ohm to 1 Gigaohm in a 1A source. Right? And if the resistor gets lower than 10 ohm, the voltage gets down. Now are there applications in which they intentional make the resistance go down the critical value to make the voltage lower??
Also why is that in larger load, the current is lesser in value. How come the electrons don't just get stuck up at the entry point to the load? How did the battery knows how much current to send to the given load? What's the feedback mechanism?
Not exactly. Normally, you would use a power-source that can provide enough current for your application. Not the other way around. The lowerage of the voltage that I mentioned, is only used as a possible security measure in current-limited power-sources, just like a fuse in a regular supply.
It depends what you mean with "larger load". The more _resistance_ your application has, the lower the current.
The battery doesn't "send" electrons, so electrons can't "get stuck" somewhere. The electrons float (through your application) from one pole to another, because the charge of the 2 poles differs.
Almost but forget the 1A source bit. Your statement should read..
"Voltage is constant at 10 volts if the resistor varies from 10 ohm to 1 Gigaohm.
In a perfect world a "constant voltage source" would provide a constant voltage no matter how much current the load "sucked" out of it. In practice voltage sources are not perfect they have some internal resistance of their own. This means that the voltage they provide isn't quite constant. It falls slightly as the load takes more current. The power supply (in a computer for example) might be rated at 5V +/- 2% for currents upto 20A. This means that when the load takes taking 20A the voltage might fall to 5V-2%= 4.9V.
You could look uo "potential divider circuit" but that's complicated stuff at your level.
Ah I see your confusion. Try and think of it slightly differently... It's not the battery that controls the current it's the load. With no load connected the resistance of the load is infinitly high. This means that the current is zero because I=V/R and R is very big. Now work out what happens as R is reduced. The current starts to increase.
They don't for the same reason that water in a tap doesn't "get stuck up" just before a tap when it's switched off. The pressure in the pipe stops the water (eg current) flowing so no more water accululates at the tap.
Current flowing in the resistor causes a voltage drop accross the resistor. That voltage drop opposes the voltage of the battery reducing the current.
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