Kirchoff's current law and Kirchoff's voltage law

追加: 2026-05-11

[00:00:00]Welcome to this video. In this class, we'll be looking at how to compute the current passing through each resistor in the given circuit. Now, we have three resistors here. We have the 3 ohms resistor. We have the 4 ohms resistor and we have the 12 ohms resistor. Now this is the kind of complex circuit where we apply the ke loss and basically before we solve or calculate each current passing through this resistors we have to look at the concept of the cor law. Now we know that we have two kos law here. The first one we have the KCL which is the K current law.

[00:00:59]This K current law is just stating that the current entering a junction is equal to the current leaving that particular junction. And in physics or in circuit problem whenever we see a junction we are relating it to this point. Whenever you see something like this something of this nature this is a junction just like T junction. So he's stating that the current entering a junction is equal to the current leaving the junction just like the summation of the current entering a particular junction is equal to the summation of the current leaving the junction. So we can say if perhaps we have a current here this is I1 entering this particular junction and we have this I2 also this direction is telling us that we're entering this junction and we have something like this when we have something like this it means it's leaving this junction let's call this one I3 using the kos law that says that the total current entering a junction equals the total current leaving it we can see that what from that we can say I1 + I2 + I2 = what? I3 because that's what Kos point law is talking about and that's what junction is. Now the second Kos rule that we have to know and is very important is the KVL which is the Kos Kos volage node.

[00:02:38]Now when we say the ke voltage law voltage law states that the summation of the voltage around a loop equals zero.

[00:02:49]The summation of the voltage acro or around a loop equals z. When you are talking of a loop, a loop is simply you can take a loop in a clockwise or anticlockwise direction. What we mean by a loop is when we have a circute like this. Let me say we have a circute. I do something like this. This is a loop. So the voltage across or around this loop.

[00:03:18]Let's say we have a voltage here 12 volt and we have another voltage here 11 volt. So we summation of this voltage equal to zero that's what the K VL is talking about. So that's basically summation of the voltage around a loop equals zero. So that's the two takeoffs law we applying to solve this particular question. Now let's go back to the question. Now this question how do we calculate the current passing through each resistor?

[00:04:04]Now to calculate the current passing through each resistor. Let's say R1 will be 3 ohm here. R2 will be this one here.

[00:04:15]And R3 will be this one here. So we can say the current passing through this 3 ohms is I1. And for it to pass through it must go in this direction. Right? The current passing through these 4 ohms is I2. And for it to pass through it must go in this direction. And the current passing through this 12 ohms resistor is I3. And for it to pass through this 12 it must go in this direction.

[00:04:48]Now what we have to do is to take a particular junction. Let's consider this junction so that we can use our cable current law that says that the total current entering a particular junction equals the total current leaving that same junction. If you take this junction for example, we can now see that this I1 is entering this junction. This I2 and I3 they are leaving the junction because this I3 is just like coming like this and coming down. So I2 + I3 equals I1.

[00:05:22]that will be our first equation. So we can say using using the KCN um I2 or we could say I1 equals what?

[00:05:34]Because I1 is entering this particular we taking this jun and this I2 are leaving it. So we have the total current entry I1 equals the total point leaving that I2 + I3.

[00:05:51]Now this our equation one.

[00:05:56]Now that's the only way we can apply the K's current law. Now let's look at the KO's voltage law. How we can apply to get um other equation because when we have these three terms unknown we need three equation and we solve it simultaneously. Now let's use our using voltage law. How can we use voltage to get more equation here? Now we have two we have two space here. You can let me bring this one.

[00:06:32]Now let's draw our loop because they say voltage just say that the total voltage across the loop. So you just imagine let's say our loop is coming like because the current is flowing like this. Let's just get our loop like this and we draw this one like this.

[00:06:49]So let's calculate the summation of the voltage across this loop or around this loop loop. Now this is R1.

[00:06:58]For voltage don't forget that voltage is I. So whenever we have a resistance we we multiply this resistance three by the current passing through it which is I1 and that's same as saying voltage. So in place of voltage you can be using this resist this resistance which is three multiply by this I1. So we can say across around this loop here we have I1 and there's another thing to note before we continue.

[00:07:32]It is known that current flows from conventional current flows from higher potential that's from positive sign to negative sign that's the conventional current but in a real sense current actually flows from lower potential to higher potential but from this case that's why they won't assign they won't put the signs there but you have to know that current actually flows from positive to negative and for this one which is I Also it flows from what?

[00:08:02]Positive to negative. I3 also flows from what? Positive to what? Negative. And you can now see that when this current is flowing when we say our V= I. So in this if you looking at this resistor here the resistance was three. So the I is I1 the current that is flowing through it. Now we need positive or negative depends on the the sign for this one here. So we have that flowing from positive to negative. So it will be what minus the sign here will be minus because the current is flowing from higher potential to lower potential. So we have to put this minus here. So so that's about that. Now if this current is flowing through and we've calculated this voltage this one voltage because we see this and this don't for this another voltage across this loop and this loop is taken in a clockwise direction. Now another voltage you can see is this one here we have 4 ohms resistor here. So this 4 ohms resistor multiplied by the current passing through it. The current passing through this 4 resistor is I2 not I1 because at this junction I2 is entering. But when we get here the current passing through this 4 resistor is actually I2. We will not use I we will not use I1. H just say plus. Now note something that it is it is flowing from positive to negative which is for conventional current I2 multiply by the resistance which is what* 4 and let us look at. So the last voltage we are looking at is this one.

[00:09:47]But this one now if you are taking this loop is coming like this. We considering the direction of this that's why it's flowing from positive to negative. If you consider the direction that is this is the current here will be flowing from negative which is here to positive. So that's why this one will be what + 24 and every if you remember voltage does say that what summation of the voltage equals what zero and that's how we have this one here. So that let's say this is our one L1. So we try to sum up all the vol around this loop. So let's now get now we have let's simplify this one 3 I1 - plus* - 4 I2 + 24 = 0 you can say 3 I1 - 4 I2 = -4 this gives us our equation 2. Now let's look at this loop here. There's another loop here. Let's call this one f_sub_2 so that we can get one last equation because whenever you see three unknown for you to get the three you need three equation. So that's why let's say around L2. Now around L2 what do we have? Let's get all the voltage around L2 and we are taking this L2 also in clockwise direction. The current is flowing from here. So the first voltage let us look at this one here. It is I3 that is flowing through this resistor resistor here. So we can just say and it is flowing from as we all know conventional current flows from higher potential positive to lower potential negative.

[00:11:38]Hence we see what the current that is there I3 multiplied by the resistor which is 12 and that's one of the thing. So we want to use KVM here also summation of the voltage across around the group will be equal to Z. Now what other resistor we have here or voltage we also have this.

[00:12:02]So we only have two resistor here. So now let's look at this. How can we see the so from here now if you are considering this loop this current will be flowing from negative to positive in this case. Now if you are continuing this loop it will be from negative to positive and you'll be adding plus 4. Now let's say I right I* this what we have because it's going from negative to positive lower potential to higher potential that's why we have our positive sign then the I2 multip ohms and everything that's everything we have around this everything= 0 simplifying this we have - 12 I3 + 4 I2 = Z and that give us our equation 3. Now from this we can combine all equation.

[00:13:06]Let's combine all equation. Let's get there.

[00:13:18]So the equation one is I1 = I2 + I3.

[00:13:26]The equation two is this value - - 3 I1 - 4 I2 = -4 and the equation 3 = - 12 I 3 + 4 I 2 = 0 Now to solve this particular question is very straightforward. Since I1 = I2 + I3 in place of I1 in equation 2, we can substitute I2 + I3 there. Hence we have that we have that in place of I1 we have 3 - 3 I1 here. We substitute I2 + I3. Hence we have something to solve simultaneously. We have what -3 in place of I1 then you put I2 + I3 that I2 + I3.

[00:14:34]Don't forget we are trying to compute all the current passing through each resistor. So what do we have there? I2 + I3 - 4 I2 = -4.

[00:14:48]So what do you have here? - 3 I2 + 3 I3 - 4 I2 = - 24. So - 3 I2 from here you can see - 3 I2 - 3 - 3 I3 - 4 I2 = -4. Now - 3 I2 - 4 I2 can be what we can have - 3 I3 that is here.

[00:15:18]Then adding both these we have - 7 I2= -4.

[00:15:27]Let's call this equation four.

[00:15:30]So now since we have this we have an equation that has I3 I2 and I3 I2. We can then solve this simultaneously using our um cashio calculator. So let's let's see what we want to sol. Let's bring this equation three here. We have - 12 I3 + 4 I2 = 0 equation 3. So we want to solve equation four and equation three simultaneously to get our I2 and I3.

[00:16:08]Once we have our I2 and I3, we then substitute to this man here to get our I1. So solving this um so what do we have there?

[00:16:30]We have given the question we have - 3 - 7 - 24 - 4 + 4 and 0. So we have that from here when I punch this inside the calculator I have I3 = 1 1 amp. The unit for current is ampere and I2 equals what?

[00:17:15]3 amp. So that's our I3 and I2. is telling us that that the this I3 is the current passing through this 12 ohms resistor which is the R3 and this I2 3 ampere is the current passing through this 4 ohms resistor. Now to get our I3 I1 how do we get our I1? So we already know that I1 = = I2 + I3 and we can say what what's our I2 3 what's our 3 amp what's our I3 is 1 amp we can say what that total of 4 amp so I1 = 4 amp I2 = 3 ampere and I3 = 1 ampere and that's how to use kov's law kov's laws to compute the current passing through each resistor these are the kind of question that you can apply k's