Current limited voltage source for driving very low resistances.

[engrMunna]

Very basic question: V = I*R

But consider a voltage source which can only provide 1 Amp of current. It is connected across a 1 ohm resistor. If we increase the voltage of the sourse above 1 Volt...what will be the potential across the terminals of the Resistor?

My understanding is that if we connect a voltmeter across the resistor, It will show 1 volt even if the voltage source is providing 2 V. The missing 1 volt drops across the internal output resistance of the Voltage source?

 

[Genovator]

As I know, the potential difference between the 1 ohm resistor will be about zero, no matter the current limit, until there is a load connected at the output of the 1 ohm resistor. The load resistance will determine the current flow and potential difference.

 

[engrMunna]

As I know, the potential difference between the 1 ohm resistor will be about zero, no matter the current limit, until there is a load connected at the output of the 1 ohm resistor. The load resistance will determine the current flow and potential difference.

I think you might be confusing what I said...for clarity lets take another set of values: You have a voltage source, which can provide a maximum current of 1 mA. It is connected across a 1 Kilo-ohm resistor. At 1 Volt the source will provide its maximum current when connected across the resistor (1v / 1Kohm) = 1 mA. But now I increase the voltage to 2 V. What will happen?

I think even when the source is configured to 2 V, The drop across the 1 kilo ohm load resistor will be 1V, while the remaining 1 volt drop internally.

Or perhaps I didn't understand what you said?

 

[BradtheRad]

The missing 1 volt drops across the internal output resistance of the Voltage source?

Yes, this is what happens.

An aging battery often shows this behavior. It develops high internal resistance.

A voltmeter may show it is good when there is no load attached. But then the volt reading falls on its face when you attach even a small load.

There are aging batteries that give a low reading with or without a load.

In my younger days I was used to such batteries. But then on the other hand I found these puzzling batteries that measured good one moment, and low the next. It took me a while to figure out the cause.

 

[engrMunna]

And also with a brand new battery, the explanation is the same right?

 

[FvM]

The behaviour will be different for a voltage source with true current limiting (e.g. a lab power supply) or a source with internal resistance. You didn't clearly say which model you're assuming.

A new battery has most likely a lower internal resistance and delivers a higher current. In so far it's not clear what you mean with the new battery example.

 

[crutschow]

A voltage source with a current limit does as you described. The output voltage remains at the set value until the current limit is reached. Then it reduces the output voltage to keep the current at the limit value.

 

[jeffrey samuel]

only when the source resistance is equal to the load resistance max power is transferred that is the 1 A current you are talking about in all other cases the power dissipated will always be less than that of the Pmax value

This applies only for linear ckts

 

[devilwiss]

Hi
voltage source gives always a constant voltage to the load (in normal conditions), so if you say that the voltage changed from 1 to 2 volts, i think it's an error unless you're talking about another voltage source, in this case the current will change according the relation V=R*I

i hope you understand me

SH

 

[jeffrey samuel]

The constant sources can not vary their responses and So max current can't be more than 1 amp at any time

 

[engrMunna]

What I have in mind as a model for the voltage source is that it has some internal resistance, and also it can not provide infinite current.

So with battery having internal resistance and limited current output, if you connect a 1Kohm resistor with a 2V battery with maximum current that it can provide is 1 mA, then the voltage across the resistor is 1V, the "missing" 1 volt drops internally.

I guess this summary is correct?

 

[FvM]

I guess this summary is correct?

Depends on your definition of maximum battery current. In my view a 2 V battery with 1 mA maximum output current (in short circuit) would have an internal resistance of 2 kOhm. So if you connect a 1 kOhm resistor load, the battery voltage drops to 0.67 V rather than 1 V.

It would be rather arbitrary to say, the maximum output current of 1 mA has to be calculated for 1 V loaded output voltage, isn't it?

 

[Marvel_tronix]

Yup, correct.