00:01
So for this question we have a resistor big r that is attached to a battery with some internal resistance small r.
00:09
We want to find the power dissipated by the load resistor.
00:15
And this is actually goes to, you can calculate as i square r because we are very easily fine why it is i based on the emf and the total resistance of the circuits which is equals to e square over plus r.
00:35
Square times r and we want to find, suppose that this big r can vary what would be the value of r which will give it the maximum power dissipation.
00:51
So to find that we will have to differentiate p with respect to the resistance of the resistor big r and we want to find the maximum point, the point of maximum dispower dissipation, which is when the differentiation is a 0, right? it's equals to 0.
01:14
That's the turning point, which is also the maximum point.
01:19
So differentiating this, you're going to use chain rule.
01:27
So we differentiate the numerator first, and we are left with the denominator.
01:31
Then we differentiate the denominator, and we equate this to 0.
01:52
So simplifying this, we try to make it to the same denominator.
01:57
J have r plus r e square minus 2 e square r and simplifying this further we get e square so this is the term right and for this to be 0 that's only one possibility that is physical that is when r over here is equals to small r and then this term will be 0, so the numerator is 0 and everything turns to 0.
02:58
So therefore, pmax occurs when r is equal to small r, which is basically the internal resistance of the battery.
03:12
Now to put it into a numerical case, we have emf of 9 volts, and we have this is small r to be 1 oms...