Kirchhoff’s voltage law | Circuit analysis | Electrical engineering | Khan Academy

Kirchhoff’s voltage law | Circuit analysis | Electrical engineering | Khan Academy

– [Voiceover] Now we’re
ready to start hooking up our components into circuits,
and one of the two things that are going to be very useful
to us are Kirchhoff’s laws. In this video we’re gonna talk about Kirchhoff’s voltage law. If we look at this circuit here, this is a voltage source, let’s
just say this is 10 volts. We’ll put a resistor connected to it and let’s say the resistor is 200 ohms. Just for something to talk about. One of the things I can do
here is I can label this with voltages on the different nodes. Here’s one node down here. I’m going to arbitrarily
call this zero volts. Then if I go through this voltage source, this node up here is
going to be at 10 volts. 10 volts. So here’s a little bit of jargon. We call this voltage here. The voltage goes up as we go
through the voltage source, and that’s called a voltage rise. Over on this side, if we are standing at this point in the circuit right here and we went from this
node down to this node, like that, the voltage
would go from 10 volts down to zero volts in this circuit, and that’s called a voltage drop. That’s just a little
bit of slang, or jargon that we use to talk
about changes in voltage. Now I can make an observation about this. If I look at this voltage
rise here, it’s 10 volts, and if I look at that voltage
drop, the drop is 10 volts. I can say the drop is 10 volts, or I could say the rise on
this side is minus 10 volts. A rise of minus 10. These two expressions mean
exactly the same thing. It meant that the voltage
went from 10 volts to zero volts, sort of going
through this 200 ohm resistor. So I ran a little expression for this, which is, v-rise minus v-drop equals what? Equals zero. I went up 10 volts, back down 10 volts, I end up back at zero volts,
and that’s this right here. This is a form of Kirchhoff’s voltage law. It says the voltage rises minus the voltage drops is equal to zero. So if we just plug our
actual numbers in here what we get is 10 minus 10 equals zero. I’m gonna draw this circuit again. Let’s draw another
version of this circuit. This time we’ll have two
resistors instead of one. We’ll make it… We’ll make it two 100 ohm resistors. Let’s go through and label these. This is again 10 volts. So this node is at zero volts. This node is at 10 volts. What’s this node? This node here is… These are equal resistors, so this is gonna be at five volts. That’s this node voltage
here with respect to here. So that is five volts. This is five volts. And this is 10 volts. So let’s just do our visit again. Let’s start here and
count the rises and drops. We go up 10 volts, then we
have a voltage drop of five, then we have another voltage drop of five, and then we get back to zero. We can write the sum of the rises and the falls just like we did before. We can say 10 volts minus
five minus five equals zero. Alright. So I can generalize this. We can say this is general
we can do the summation, that’s the summation symbol, of the v-rise minus the sum
of the v-fall equals zero. This is a form of Kirchhoff’s voltage law. The sum of the voltage rises minus the sum of the voltage falls is
always equal to zero. There’s a more compact way to write this that I like better, and that
is, we start at this corner… We start at any corner of the circuit. Let’s say we start here. We’re gonna go up 10
volts, down five volts, and down five volts. So what we’re adding is the voltage rises. We’re adding all the voltage rises. Rise plus 10. That’s a rise of minus five
and a rise of minus five. So I can write this with
just one summation symbol. The voltages around the
loop, where i takes us all the way around the loop, equals zero. So this means I start
any place on the circuit, go around in some direction,
this way or this way, up, down, down, and I end up back at the same voltage I started at. So let’s put a box around that too. This is Kvl, Kirchhoff’s voltage law. Now I started over here in this corner, but I could start anywhere. If I started at the top and went around clockwise, if I started here say, I would go minus five,
minus five, plus 10, and I’d get the same answer. I’d still get back to zero. If I start here and I
go around the other way, the same thing happens. Plus five rise, plus five rise, and this is a 10 volt drop, so it works whichever way
you go around the loop, and it works for whatever
node you start at. That’s the essence of
Kirchhoff’s voltage law. We’re gonna pair this
with the current law, Kirchhoff’s current
law, and with those two, that’s our tools for
doing circuit analysis.

31 thoughts on “Kirchhoff’s voltage law | Circuit analysis | Electrical engineering | Khan Academy

  • what happens when charges move flat I mean why does potential drop occurs only at resistors? I mean when an electron travels from negative end towards the positive end it continuously loses potential so why we don't consider potential at flat areas where no resistors are present ?

  • resistance only reduce current.. which mean input voltage does not change.if input is 10v and output is also 10v.

  • (For 2:19) If I connect one voltmetre across the battery and another voltmeter across the resistor will the voltmeter across the battery show +10V and the voltmeter across resistor show -10V ?

  • Thank you so much for this video! I have been really struggling with understanding when voltage and amp sources are equated as being negative verses positive, especially when there is a mix of both in the circuit!

  • Hey, in 3:03 there is a error is guess, look, u have taken voltage rise as -10 and voltage drop as +10. The formula u told is saying voltage rise minus voltage drop=0 . If i substitute that " minus 10 minus 10 which leads to -20. Can u xplain on this.

  • Pretty badly explained because 'rise' has nothing to do with the perception of view, it has to do with the travel of conventional flow. Just to further prove this, imagine the resistor is sideways, left to right instead of up and down. There is still a drop on that resistor even though it's going sideways. The problem with this instructors explanation is that he is inferring that voltage is a sort of travel, but it is not gradual travel, it is explicitly a measure of the difference between two points. Here's where this video fails and it is the problem with the conveyance of the teaching.

    His equation Vrise – V drop = 0 is the equation for the whole circuit. On the left side, where the battery is, there is a 10V rise….on the right, where the resistor is, there is a 10V Drop…this is where he is getting 10(rise)-10(drop)=0 | That rise= -10 is, like he said, just another way of writing a drop. I understand the confusion, there is not much point of writing that unless he wrote 0=sum of voltage rise … in which case that specific "rise" on the right would technically be negative due to the resistor, however that is often confusing and he just was using this as a way to get you to understand that this is a decrease in voltage. His math however is all correct.

  • This video is really informative unlike half of those Indian videos which one can't even understand the language

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