00:01
Hello everyone, we are going to understand this question.
00:05
Here, given in the question, given value of r1, that is given 5 ome, and value of r2 that is given 15 ome, and value of r3 that is also 15 ome.
00:22
And applied potential means emf of the battery, that is 9 volt.
00:36
And internal resistance r is equal to 1 .5.
00:41
Now let's draw the circuit diagram for this.
00:51
Here r1 and r2 are in series so we can represent it like this.
00:58
R1 and this is r2 and this is the applied potential, means applied battery and here the internal resistance of the battery and for this this is r3.
01:19
Now let's grounding this point and taking the potential at the other end is v.
01:25
Current in this circuit in this while it is i1 and here it is i2 and in this it is i3 now applying kirch of current rule applying it's of current rule we know that the sum of total current at any junction always be constant so i1 plus i2 plus i3 is equal to now we can write v minus e upon small r plus v minus 0 upon r1 plus r2 plus v minus 0 upon r3 is equal to 0.
02:24
Now v minus applied potential that is 9 volt upon small r is 1 .5 plus v upon small r is 1 .5 plus v upon r1 plus r2 that is 5 plus 15 plus v upon value of r3 that is 15 is 15 is equal to 0.
02:53
Now doing further calculation here let's correct it.
03:06
We can write it as v minus 9 upon 1 .5 plus here v upon 20 plus v upon 15 is equal to 0.
03:19
Now we can write v .20 plus v.
03:25
15 is equal to 9 minus v upon 1 .5.
03:31
Now taking lcm so here we can write lcm is 60.
03:40
So 3 v plus 4v is equal to 9 minus 5, 9 minus v upon 1 .5.
03:48
Now we can write 7 v into 1 .5.
03:54
1 .5 is equal to 60 into 9 minus 60 v...