Hello,
there is somthing i dont understand about ohm's law. let say i have a circuit with 9 volts battery and a 1 ohm resistor in searis. the battery can suplly current of 0.1 ampere. what is going to be the voltage on the resistor? 9v volts or 0.1 volts?
Thanks.

 

http://farside.ph.utexas.edu/teaching/302l/lectures/node57.html
http://physicsnet.co.uk/a-level-phy.../electromotive-force-and-internal-resistance/

 

Put a meter on it and measure it. Get two meters. Measure current and voltage at same time.

 

Apply Kirchhoffs Voltage Law (KVL). The voltage on the resistor will be 9V minus the voltage drop on the internal resistor of the battery. Get a meter, measure voltage on the resistance and you won't probably get exactly 9V.

 

Thanks! i understand it much better now

 

Your battery may be able to supply a PEAK voltage of 9V, but if it cannot support the current required by the load then the voltage will drop, and the voltage across your resistive load will be V = I*R, or 0.1V

 

Hello,
there is somthing i dont understand about ohm's law. let say i have a circuit with 9 volts battery and a 1 ohm resistor in searis. the battery can suplly current of 0.1 ampere. what is going to be the voltage on the resistor? 9v volts or 0.1 volts?
Thanks.

Ohm's Law is

I = V / R

I = 9V / 1Ω = 9A

Your battery cannot supply 9A. Hence it will have a short painful life. It will die very quickly.

Measure the voltage across the resistor. The battery is no longer supplying 9V. Ohm's Law still applies.
I = V / R

The voltage is no longer 9V and the current is not 9A.
Realistically, the total resistance in the circuit is not 1Ω. You have introduced some additional resistance. Where is it?
It must be in the battery, assuming that your connecting wires make up less than 1Ω.
Batteries have internal resistance too. You have to take into account the internal resistance of the battery.
Now see if you can apply Ohm's Law knowing that there is some hidden resistance in the battery.