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ALL ABOUT ELECTRONICS, and today we will see Kirchhoff’s Current Law or KCL in the network

analysis. So, this Kirchoff’s Current Law also known

as Kirchoff’s first law or Kirchhoff’s junction law. So, let’s see this Kirchhoff’s Current law.So,

according to the Kirchoff’s current law, the algebraic sum of all the branch currents at

given node is zero at all instant of time. or, in a simple way we can say that at any

node or a junction in a network, the sum of the currents flowing into the node is equal

to the sum of the currents flowing out of that node. so let’s understand it by taking

a simple example. so let’s say we have a one in a network, which

contains five branches. the current in this five branches is i1 to

i5 respectively. so, here as you can see this current i1, i2

and i3 are entering into the node, while current i4 and i5 are leaving the node. so, here we

use the following sign convention that whenever the current is entering into the node we will

consider it as a positive current. while current is leaving the node, then we

will consider it as a negative current.So, here as you can see, current i1,i2, and i3

are entering into the node. So, we can consider them as a positive current,

while current i4 and i5 are leaving the nodes we can consider it as a negative current,

and according to the Kirchhoff’s law, the algebraic sum of all this current is zero. So, we can write i1+ i2 + i3 – i4 – i5=0. Or we can write it as i1+ i2 + i3=i4 +i5.that

is the current which are entering into the node equal to the currents which are leaving

the node. Now, earlier we had seen that current I can

also be defined as Q / t that is the rate at which charge is flowing. so, we can write this equation as (Q1/t) +(Q2/t)+(Q3/t)

=(Q4/t)+(Q5/t). So, here we had written this current in the

form of charge. and we can simplify it as Q1+Q2+Q3=Q4+Q5. So, we can see that the charge which is entering

into the node is equal to the charge that is coming out of the node. So we can say that this Kirchhoff’s Current

Law is nothing but Law of Conservation of Electrical charge. That means the sum of the charge that is entering

into the node is equal to the sum of the charge that is leaving the node. So, now let’s take one numerical and see the

application of this Kirchhoff’s Current Law. So, here we have given one circuit which contains

three nodes, A, B and C. And current I1 to I6 are the currents which are flowing through

this circuit. We had given some of the currents and we need

to find the current i6 that is entering into this node B. So, to find this current i6,

we need to apply KCL at node B. So, let’s apply KCL at node B.So, after applying KCL

at node B, we will get i3+i6=i1. As current i3 and i6 are the currents which

are entering into the node B, while current i1 is the leaving the node B. Now here we

had given i1=2 A. So, we can substitute this current i1 as 2 A in this equation. So, we can get i6+i3=2 or i6=2 – i3. So, let’s say this is the equation 1. Now, here still current i3 is not known, so

to find this current i3, we need to apply KCLat this node C. So, by applying KCL at

node C, we will get a i2 + i5=i3. Now, here had given i5 as -4 A. So, we can

substitute this i5 as -4 A in this equation. So, we will get a i3=i2 – 4. And let’s say this is the equation number

2. Now, here also we still don’t know this current

i2. So, to find this current i2, we need to apply

a KCLat this node A. So, let’s apply a KCL at node A. So, applying KCL at node A, we

will have a i1+ i4=i2. Now, here current i1 and i4 are the known

currents. That is 2 A and -1 A. So, we can write as

2 -1A=i2. Or, we can write, i2=1A. So, now as we know this current i2, we can

substitute the value of this current i2 in the equation number 2. So as we know the current i2, so we can put

this value in equation number 2. So, we will get i3 as, i3=i2 – 4. And i2 is 1 A. So, i3=-3 A. So, now we also

got this current i3. So, now we can easily find this current i6. So, we had seen that, equation 1 that i6 can

be given as 2 – i3. so, let’s write it. i6=2 -i3. and substiting the value of i3,

we will get i6 as 5 A. So, in this way, using the Kirchoff’s current law, we can find the

current currents in any circuit. So, I hope you understood what is Kirchhoff’s

Current law and using this Kirchhoff’s Current Law, how we can find different currents in

electrical circuit.

nice…

informative…(y)

Ekdam mast bhai

xcellent xplanation …..thank u i understood easily

Good explanation.

Have a look into one more way to explain Kirchhoff's Laws.

https://www.youtube.com/watch?v=9rCL6a-YKr8&t=8s

veryyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyy helpfullllllllllllllllllllllllllllllllllllllllllllllllllllll

U are the best

Very helpful to learns electronic basics. Thank you sir

How i1 is 2 A

Thank u sir.. Sorry 4 ask u silly qun.

You are great and helpful….

Thank u its helpful

you passed me sir luv u

Well explained

very nice

truely correct explaination

Very perfect sir

Excellent

ππ

Pls do video on Hartley oscillator and colpitts oscillator…

Helpful..

When the V/R is used to represent current , do you always assume that the node you are summing at is at a higher potential voltage than the other nodes ?

U r awesome sir

Thank you very much sir

It was an awsm video I totally understood the concept which I couldn't understand from just reading the theoryπ€π₯

Plot twist, add capacitor

Just a simple correction,if I'm not wrong,value of I5 is -4,so I3 must be =I2 +4

Sir thats great, but u need to make a script , so that the video is a little fast and not boring

Thank uππ

wow

Very useful , thank you

Super

thank you so much sir it helps me to solve in the metro maintainer exam

Nice teach sir

thnq sir…. u r a nice teacher..very helpful video

Tnq sir….it's such a great explanation…

From where will the currents entering leave?

Thankyou so much bro…

best explanation

It's so conprehensive thanks.