Ohm's law

[Opara]

Hi, using Ohm's law of V=IR in a circuit of say a 10v battery in series with a 0.5 ohm resistor, then wouldn't the voltage across it be 20v?

Doesn't this then contradict Kirchoff's voltage law stating the sum of voltages around a loop are equal to zero?

Thanks

[Ultuous]

No... All Ohm's law is saying (in your example) is that the Current(R)=20Amps

[motco]

Ultuous said:

No... All Ohm's law is saying (in your example) is that the Current(R)=20Amps

It's a race between the battery going flat and the resistor burning out!

[Opara]

Sorry moment of madness, this electronics stuff is frying my brain

[Flibble]

motco said:

It's a race between the battery going flat and the resistor burning out!

Or the battery catching fire!

[Ganglandboss]

Opara said:

Hi, using Ohm's law of V=IR in a circuit of say a 10v battery in series with a 0.5 ohm resistor, then wouldn't the voltage across it be 20v?

Doesn't this then contradict Kirchoff's voltage law stating the sum of voltages around a loop are equal to zero?

Thanks

Ohm's law tells us voltage, current or resistance when we know the value of any two.

V=IR

I=V/R

R=V/I

Current is always dependent on the voltage and the resistance of the connected load, so we calculate current as:

I=V/R

I=10/0.5

I=20 amperes

Now say we have the above circuit but nothing is labelled. We have a voltmeter and an ammeter so we can determine V=10v and I=20A. To find the resistance of the load as follows:

R=V/I

R=10/20

R=0.5

Now we have an ammeter and an Ohmeter. We know the current is 20A and the resistance of the load is 0.5 ohm. To determine voltage:

V=IR

V=0.5 x 20

V=10v

Edited by Ganglandboss on Tuesday 12th November 12:33

[Dr John]

Don't forget the internal resistance of the battery. It may (or may not) be significant.