0:00
Hello everyone.
00:01
In this question we have given a circuit having resistors r1 is equal to 13 om, r2 is equal to 6 .1 om, r3 7 .1 om and r4 12 om.
00:13
The emf of the battery is 18 volt.
00:17
We have to find the current i1, i2, i3 and i4 by the methods of 5 first, by the method of combination of resistors.
00:40
Second, by the method of kirchof law.
00:45
Now first, by the method of combination of resistors.
00:50
Here, r2 and r4 are in series.
00:58
So, their equivalence resistance r24 is equal to r2 plus r4.
01:09
And this equal to 6 .1 plus 12 this equal to 18 .1 ome now r2 4 and r3 are in parallel so their equivalence resistance r2 4 3 can be find as 1 divided by r2 4 3 is equal to 1 divided by r2 4 3 is equal to 1 divided by r2 plus 1 divided by r3.
01:58
So on putting values and solving, we get r243 is equal to 5 om.
02:10
Now r2 .4 and r1 are in series.
02:17
In series.
02:20
So their equivalence r is equal to r243 plus r1.
02:31
Putting values and solving, we get r is equal to 18 o.
02:41
So the total current in the circuit is i is equal to v divided by r.
02:52
This equal to 18 divided by 18.
02:58
So i will be 1 ampere.
03:02
As r to 4 3 and r1 are in series.
03:05
So current in r1 is i1 is equal to 1 ampere and in r243 the current is i243 and this is also 1 ampere now voltage across r243 can be calculated as v243 is equal to current i243 multiply by the systems r2 -4 -3 by using oms law.
03:49
On putting values, we get 1 multiply by 5.
03:56
This equal to 5 volt...