Today we look at cut your solos of electrical

networks. Thanks for tuning in each kiss a more academic Kirchhoff’s laws of Electrical networks are

basically two We’ll explore one by one the first law of

electrical networks states that The sum of all currents that are entering

any junction is equal to the sum of the currents That are leaving the junction by a junction we mean a place where wires are connecting

where’s our meeting for example in this simple illustration here at This junction you have N. Currents I1 and

I2 moving in and that that same junction, you’re having currents I 5 I for I 3 so the law simply states that the sum of these

currents that are getting in which is I 1 plus i 2 Should be equal to the sum of the currents

that are moving out of that junction. Which is I 3 plus i 4 + i5 That is just one of the versions of that law Still looking at that same junction. We can still describe this law as That there are you break some of all the currents

at that junction is equal to 0 and? When you mean that you break some we mean

that the currents that are getting into the junction are positives Positive I 1 and positive item and the currents

that are living the junctions are negative currents So there are g brakes sum of all these currents

is equal to 0? that is the first law of Catch off so the second law of Catchers in

any closed loop The algebraic sum of all the Emf in that loop

should be equal to the algebraic sum of all the potential drops Alternatively we all know that the principle

of conservation of energy States that energy is neither created nor

destroyed, but can be converted from one form to another so In this alternative way of looking at it the

principles of energy Are implied and so the in principle of energy

that the directed some of the electrical potential? Differences or the voltages or in a closed

in an in any closed loop? Circuit is or is equal to 0 you can look at

it like that so it means that if you the summation of the

total potential differences or the total voltages in a closed circuit their Algebraic sum should be equal to zero now. What exact is the loop a Loop can either be clockwise or anti-clockwise

as long as it is in a closed circuit like we shall illustrate later So you need to know? That if you have chosen to make your loops

clockwise in a certain circuit Then they all have to be clockwise if you’ve

chosen that they should be anti-clockwise Then they should all be anti-clockwise for

a given particular circuit The next design allocated to the potential

difference and the Emf will depend on the direction of the flow of current in the resistor In relation to the direction in which the

loop is traversed Like I Shall illustrate Let’s look at the sine cone vision that we

shall be using the in our calculations and kirchoff’s laws of electrical networks these two are resistors and These two are sources of Emf In our calculations. We shall be considering the loop We shall ever draw a clockwise loop or an

anti-clockwise loop and it is the direction of that loop that is going to determine How we shall be? Assigning these signs the negative or the

positives on the potential drops that are within that loop? now if the loop is moving in a direction such

that if the direction is the same as The direction of flow of current through a

certain resistor it means that looking at this loop It means that current is moving in from me

in fear and qaeda getting out here It is assumed that current is going to move

from a position of higher potential to towards the position of lower potential So it means that as current is going through

this Potential this resistor as by our loop They’ll be a potential drop and because of

the potential drop it means that our PD across this will be a negative Coming to this loop This loop is moving in the opposite direction

to the flow of current Now because this loop is moving in the opposite

direction to the flow of current with respect to this loop We see that it is moving from the a position of lower potential the position

of Higher potential in as regards to The flow of current through this resistor

so as far as this direction of loop is concerned. There is a voltage gain as We move from this direction To that direction, and so whenever there is

those two moving in the opposite direction. We shall have a potential gain a Voltage Gain and so that will be a positive

ir Then coming to our sources of Emf We know that the general conversion is that

current will always move from the positive terminal? Will go through the circuit and will enter

this circuit the de sel into the neck Valley negative Stamina so if the loop is such that it is coinciding

with the correct direction of Flow of current within the cell which correct

direction is that current as it’s moving through the cell will be moving from the negative to the positive Then that Emf is positive, but if the loop

is traveling in a direction such that It is opposing the normal flow of current within the cell

then it means that the Emf in that cell will be regarded to as negative with in a previous session, we said That the sum of the potential drops of the

sum of the p dS Within a closed loop is equal to zero so this

here is a closed loop It means that when current gets out of this

cell here. It’s going to move across and it will come

back There so it means from the positive terminal

back to the negative terminal so it means that this is a closed loop and so all the potential the voltages within this loop

must be equal to zero so we shall start with the potential voltage across a which is a cell so the voltage across a will

be a positive Why is it a positive because we are moving

from? The loop is in the same direction as the flow

of current within the cell which is From negative going through the cell to the

positive side then it comes out where positive side When we come to this resistor this? Direction of current is moving in that Direction so it means that this side of the resistor

will be positive, or it will have a higher potential and This side will have a lower potential But and since our loop is traversing in that

direction In the same direction it means that as we

are moving in as our loop is in coming here It is going to experience a voltage drop across

this resistor and so it means we shall have a negative voltage

across here when we come As our loop traverses in this direction The loop the Direction of the loop is coinciding With the direction of the flow of current

through this cell as the current flowing through the cell is assumed to move from Negative to positive so that makes it a positive

voltage voltage VB Then as we are moving the current is moving

it is entering from here So this the point of entry is the point of

higher higher potential the point of exit of the current is the point of lower potential But then the loop is moving in that direction So it means that as current is moving through

here It is experiencing a voltage drop so as it’s

experiencing a voltage drop that makes this voltage across this a negative VD and that

is the same case with ve and that is equal to zero so we have the total

sum of all the potentials in this loop being equal to zero and if we are to make the Emfs to be on one side and the Resistances all the potential drops to be

on the other side then you will have Va Plus VB. Which are the Emfs in this loop being equal

to the sum of the potential drops? So that is definitely That is basically kirchoff’s laws that the

sum of the Emfs in the circuits as we have seen is Equivalent to the sum of the potential drops

in that same circuit in upcoming videos, we shall be looking at Examples that will illustrate that will require

us to use kirchhoff’s laws This brings us to the end of this video. Thank you for watching Somebody out there needs to watch this video Don’t forget to like and share for more videos

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anja

Thank you so much for a great and very clear explanation- Really appreciate it-

Nice

Really! What i was in search of!!

Thanks a lot for such an appreciable explanation. Really grateful !