00:01
Hello everyone, in this question we have given two resistors in parallel each of 10 oam, three resistors each of 12 ome connected in parallel and five resistors each of 15 om connected in parallel.
00:19
All these three combinations are connected in series with a battery v is equal to 60 volt.
00:29
The circuit diagram is shown.
00:33
We have to find first the current in each arm equal to the 15 om resistors.
00:44
Suppose this current is i -dash and second we have to find the potential difference across each of the 12 -oom resistor.
00:57
Suppose this potential difference is v2.
01:02
The equivalent resistance is shown here the equivalence of both 10 -oom resistors will be r1 is equal to 5 -oom and the equivalence of all 3 12 -home resistors will be r2 is equal to 4 -oom and the equivalence of all 5 registers of 15 -oom will be 3 -oom these all are connected in series now the equivalence resistance of the circuit will be 5 plus 4 plus 3 and this equal to 12 om so the net current in the circuit can be calculated as v is equal to i r so i is equal to v divided by r i is equal to value of v is 60 and r is 12 so we get net current in the circuit in the circuit it is 5 ampere.
02:14
Now a clearly current in each resistors of 15 om will be i -dash is equal to 5 divided by 5 because there are 5 equal resistors.
02:34
So all this 5 -amper current will be divided equally in all 5 equal resistors...