5v circuit, now stick a 100ohm resistor in...

the voltage drop is.... 5v?

5/100

i = 0.050 (50ma)

if this was true, there would be 50 (edit) ma current flowing through the circuit, how's that a 5v drop across the resistor?

5v circuit, now stick a 100ohm resistor in...

the voltage drop is.... 5v?

5/100

i = 0.050 (50ma)

if this was true, there would be 50 (edit) ma current flowing through the circuit, how's that a 5v drop across the resistor?

0.05 is 50mA

Rob

This is what I was (am?) struggling with to understand.

- The net voltage in any closed path circuit must be 0v .. this is a constant then.

so let's take a 1mohm resistor and a 1v power supply, the voltage drop has to be 1v regardless of resistor used then?

so let's take a 1mohm resistor and a 1v power supply, the voltage drop has to be 1v regardless of resistor used then?

Yes.

Regardless of what value of resistor you use as well, the battery keeps pumping current into it until it is 1V.

Of course in practice the battery has an internal resistance that can be thought of as a series resistor inside the battery. That is why the voltage output of the battery will drop for some values of load resistor.

interesting... thanks mike

Have a look at this

http://www.thebox.myzen.co.uk/Raspberry/Understanding_Outputs.html

It talks about internal impedance of batteries and any power source in general.

cjdelphi:

50 (edit) ma current flowing through the circuit, how's that a 5v drop across the resistor?

Because of the laws of the universe (and in particular Ohm's law), that's how.

fungus:

cjdelphi:

50 (edit) ma current flowing through the circuit, how's that a 5v drop across the resistor?Because of the laws of the universe (and in particular Ohm's law), that's how.

Ohms law may not one day may not "apply" to all.

cjdelphi:

Ohms law may not one day may not "apply" to all.

Ohms law was correct all the time but the world was not perfect, voltage source had internal impedance, even the connection wire that you used had resistance.

You have to remember real life isn't the same as the 'perfect' circuit.

As Mike says, a power source may have an internal impedance.

If measuring using an instrument then that will have an impedance too.

Both of these can affect what you may measure in a real-life circuit vs what you calculate.

You have to be aware of these factors to understand why initial calculation and measurements may not agree. Once you understand the extra factors then you can add those into your calculations and find agreement, within certain inherent tolerances.

I wish I have 18 multimeters in my hamshack handy..

Yeah. I have to make do with one DMM, and a 2 channel o'scope. I could do with a couple more bits of measurement kit really. And a dedicated shelf to keep them all on

pito:

I wish I have 18 multimeters in my hamshack handy..

I think there is something slightly suspect about some one with that many multimeters.

Grumpy_Mike:

pito:

I wish I have 18 multimeters in my hamshack handy..I think there is something slightly suspect about some one with that many multimeters.

Two is often useful - see volts and amps at the same time. Even cheap $5 meters will do for watching voltages.

Two is often useful

Yes but 18!!

You have to have multimeters to measure the multimeters, which in turn have to me measured. Gotta account for every single one of those pesky ohms...

Confucius say...Man with two multimeters never know real voltage.

Rob