00:01
Hello and welcome to this video solution of new method.
00:04
So here we are given six resistors that are connected in this circuit.
00:07
So let me just draw a simplified diagram of the circuit.
00:11
Right.
00:12
So here we have got the resistor r1 and r2 and this is connecting the radius r3 sorry the resistance r3 not radius and here we have got this is connected to a secondary circuit r4 and r5 and this is connected to the last resistor r6 right and this entire circuit is connected to this battery right.
00:52
Now here what you need to do is you need to have the potential v, i and r right.
01:01
This is battery this is r1, r2, r3, r4, r5 and r6.
01:15
So this is a table that's given right.
01:17
Now the battery potential is given to be 30 volts right.
01:22
The voltage across the resistance r1 is 6 volts right and the current is 3 ampere right and the overall current is not given.
01:32
Similarly current in this r2 is given as 2 ampere, r4 is 1 ampere, r5 is having a potential of 8 volts across it right and rest you have to calculate and r3 is having the resistance of 3 ohm and based on this you have to fill the table.
01:54
So we can simply have the resistance of r1 which is equal to v by i right.
02:01
We apply v equal to ir formula right.
02:04
So r will be equal to v by i.
02:06
This is 2 ohm right as our resistance r.
02:10
Now if you see r1 and r2 are in parallel right so the potentials will be same.
02:15
So this is also 6 volts.
02:17
Then here the resistance will be 3 ohm right 6 over 2 3 ohm...