Let´s do an example of Kirchhoff´s law and see what we get here.
For example, if we had a 9 volt battery connected to three resistors
with resistances 2, 4, and 6 and we have the 6 ohm resistor
in parallel with the 2 ohm and 4 ohm resistors, and those two are in series with each other.
How much power would be dissipated in the 2 ohm resistor?
We´ve already introduced an expression for how much power is dissipated
in a resistor if we know the current that´s flowing through it
and we can use our resistance additional laws as well as the voltage in our circuit,
and Ohm´s law to find out what that current is.
So go ahead and use those laws and see if you can give this a shot
and find the power that´s being dissipated in that particular resistor
and then we´ll try it here as well.
Hopefully if you did this it looked something like what we´ll have here.
This problem is describing a battery which has some voltage,
and this battery goes through one or two paths either a 6 ohm resistor
or a 2 and a 4 ohm resistor. So, we have 2 and 4 and then a 6 ohm resistor here.
The question we´re being asked is in this resistor here,
'How much power is being dissipated?'
So the way we would do this is remember our equation for power which is I squared times R.
So this is the current flowing through your resistor times the resistance of that resistor
and we already know the resistance is 2 ohms, so the question is can we find this current?
So, that´s what we´ll do just using Ohm´s law.
So the first thing we can do here is simplify this just a little bit.
So we have our battery and we´re going to add the 2 and the 4 ohm resistor together
to be a 6 ohm resister. Because these are in series,
we simply add them normally and so we get 6 ohms.
We already know, given in this problem the voltage of this battery is 9 volts
and now we can apply one of our voltage laws that we discussed.
We said that the voltage as it drops going through either one of these paths
has to be the same and the reason for this is if I was using Kirchhoff´s law,
tracing my way around the path. I could´ve picked the left path here
and gone around and I know I would need to dissipate my entire 9 volts
but I could also have picked the right path here
and I know I would also have to dissipate the same 9 volts.
So, the important thing is that this 9 volts has dissipated on either path.
I could´ve picked either one. So we already know exactly what the voltage drop here has to be.
So now, all we need to do is use Ohm´s law. We know that V equals I times R.
The voltage drop equals the current times the resistance.
So the current in this path then, in this upper path that we´re trying to find,
will be equal to the voltage divided by the resistance in that path
and both of these are things we already know.
We know that the voltage is 9 volts from our battery.
We know the resistance is 6 ohms because we´ve already added these 2 and 4 ohm resistors together.
So this is 1.5 amps flowing through this wire.
Now what we´ll do is the opposite and we?ll go back and look at this whole system
with the 2 and the 4 ohm resistor and we know that because these are part of the same wire
just by splitting them back up, I haven?t changed the paths.
They´re part of the same wire so this 1.5 amps must be going through both of these resistors.
So this 1.5 amps is exactly what is the current through the 2 ohm resistor.
So now we can solve for the power will be 1.5 amps squared
times the resistance and we´re looking for the power dissipated in the 2 ohm resistor, times 2 ohms.
Now here´s a quick math trick if we don?t want to try to think about 1.5 squared,
seems like a messy number. We can think about 1.5 as 3/2 that might be a little simpler.
Little tricks like this can help us to square this value.
3 squared is 9, 2 squared is 4, 2 and then our units, amps times ohms.
We don´t have to think too hard about this because we know we´re using all of our standard units
and we´re looking for a unit of power so this would be watts.
Let´s simplify our numbers here.
We see that our answer is 9/2 and again we have watts for our units.
So this is simply 4 and a half watts of power being dissipated in our 2 ohm resistor.
So this was an example of how you could use both Ohm's law
and Kirchhoff´s law by looking at the entire path
and recognizing that the voltage drop must be the same
between two parallel paths to find the power that is dissipated in a particular resistor
of a more complicated circuit.