00:01
So we have two identical registers, r1 and r2, connected in parallel and connected with the battery of voltage 25 volt.
00:11
So there's a battery that has voltage 25 volt.
00:16
All right.
00:17
At first, the total power delivered by the two resistors is 9 .6 watt.
00:22
But now the register one is heated so that the value actually doubles.
00:28
So that means i'm going to call this r3.
00:31
1 prime the value becomes double the initial resistance all right now from here we need to find out the initial resistance of each register all right so let's do that first a part a so initially r p which was the parallel combination of register r1 and r2 was r1 times r2 over r1 plus r2 and r1 and r2 is basically r initially.
01:09
So this is going to be r square over 2r, and that is r over 2.
01:15
So this is the initial value of the parallel combination of the rest of the rest.
01:19
And the power delivered was 9 .6 watt, right? so p was i times v equals 9 .6 watts, which means the current i, was 9 .6 watts over 25 volt equals 0 .38 amps...