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
simply subscribe to be tSinghua Kedah me otherwise catch you in the next video For example academy this is an old drunk occur
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1. 1973jdmc says:

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

2. Cool Videos says:

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3. Md Yusuf says:

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4. Angella Basnet says:

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