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Kirchhoff's Law: Example

by Jared Rovny
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    00:01 Let´s do an example of Kirchhoff´s law and see what we get here.

    00:04 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.

    00:16 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.

    00:34 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.

    00:43 Hopefully if you did this it looked something like what we´ll have here.

    00:46 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.

    01:06 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.

    01:23 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.

    01:34 So the first thing we can do here is simplify this just a little bit.

    01:37 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.

    01:52 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.

    02:02 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.

    02:25 So, the important thing is that this 9 volts has dissipated on either path.

    02:28 I could´ve picked either one. So we already know exactly what the voltage drop here has to be.

    02:33 So now, all we need to do is use Ohm´s law. We know that V equals I times R.

    02:39 The voltage drop equals the current times the resistance.

    02:42 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.

    02:52 We know that the voltage is 9 volts from our battery.

    02:54 We know the resistance is 6 ohms because we´ve already added these 2 and 4 ohm resistors together.

    03:00 So this is 1.5 amps flowing through this wire.

    03:06 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.

    03:18 They´re part of the same wire so this 1.5 amps must be going through both of these resistors.

    03:24 So this 1.5 amps is exactly what is the current through the 2 ohm resistor.

    03:30 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.

    03:42 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.

    03:51 Little tricks like this can help us to square this value.

    03:56 3 squared is 9, 2 squared is 4, 2 and then our units, amps times ohms.

    04:04 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.

    04:11 Let´s simplify our numbers here.

    04:13 We see that our answer is 9/2 and again we have watts for our units.

    04:19 So this is simply 4 and a half watts of power being dissipated in our 2 ohm resistor.

    04:26 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.


    About the Lecture

    The lecture Kirchhoff's Law: Example by Jared Rovny is from the course Circuit Elements.


    Author of lecture Kirchhoff's Law: Example

     Jared Rovny

    Jared Rovny


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