00:01
Hi, there is a circuit diagram in this given problem.
00:05
First of all, we should redraw it here.
00:10
This is the uppermost resistance in this circuit 5 .00 om.
00:20
Then another resistance in this upper most branch.
00:26
This is 8 .00 om, then it goes down.
00:32
And here in the bottom most, branch there is a single resistance having a value of 10 .00 om now there are two cells the first one along with a resistance the cell is having a potential of 12 .00 volt and this resistance here this is of 1 .00 oom then it is shorted here and now another cell of 9 .00 volt and its internal resistance that is also 1 .00 own.
01:24
Now as per the given problem current passing through this cell of emf 9 volt that is given as i1 here this current this is given as i2.
01:42
And the current through the lower most branch, this is given as i3.
01:48
Now, if you consider these three loops, closed loops, as loop 1, this is loop 2, closed loop 2, this is closed loop 3.
01:57
And we apply kirchof's junction rule here.
02:01
Current i2 is entering here and i3 is going to the lower most branch.
02:06
So the current remaining here, it will be i2 minus i2.
02:14
I3 now if you look here at this junction i1 is coming here and i3 is coming from the lowermost branch this is i 3 so this time both of these currents will join to give net current i 1 plus i 3 that is also as per kirchop's junction rule now finally this i 1 plus i 3 reaches here and at this junction the current i2 minus i3 it is coming from the left hand side both of these currents will get joined and again it will become i2 plus i sorry it will become i1 plus i2 and then i1 and i2 will go to their respective branches back here we have to find the magnitudes of the current i 1 and i 2 and i 3 for which we have already used kirchop's junction rule we can mention it here using kirchhoff's junction rule we have shown electric current passing through all the branches of this circuit now to find i1 i2 and i3 we will use kirchof's loop rule in different closed loops.
04:00
So first of all, using kirchap's loop rule in the closed loop one.
04:21
Here the sign convention says that all the potential drops due to the current in clockwise direction will be taken as positive while those due to the current in counter clockwise direction will be taken as negative and the cells sending current in clockwise direction these will be taken as positive and the cells which are sending current in anti -clockwise direction counterclockwise direction they should be taken as negative so using these sign conventions in the closed loop 1 we conclude that the sum of potential drops so first of all this is 5 into i 2 minus i 3 potential drop taking place across this 5 worm plus i 2 into 1 the 2 potential drops and that should be equal to 12 volt.
05:09
So here this equation is modified to 6 i2 minus 5 i3 is equal to 12...