00:01
Hello students, let's consider the emf of the battery is e and internal resistance is r.
00:08
So, we have e minus i times r where i is the current equals to the voltage across the battery.
00:16
Now, we are given that when i equals to 9 .6 ampere v equals to 25 .6.
00:28
So, we get the equation e minus 9 .6 times r equals to 25 .6.
00:38
Equation number 1 and another set of values are given as i equals to 12 .8 and v equals to 19 .2.
00:51
So, in this case e minus 12 .8 e minus 12 .8 r equals to 19 .2.
01:07
So, this is equation number 2.
01:09
Now, from 1 and 2 solving 1 and 2 we have emf of the battery e equals to 44 .8 volt and internal resistance equals to 2 ohm.
01:29
This is the answer for part 1.
01:31
Now, for part 2 we are asked to find the power dissipated by the internal resistance and external resistance.
01:39
So, power dissipated by internal resistance equals to i square r.
01:45
Now, for the first case it will be 9 .6 square into 2 watt which is equal to 184 .32 watt.
02:05
Now, in second case current is 12 .8 into 2 watt.
02:14
So, which is equal to 327 .68 watt.
02:21
Now, external resistance capital r is given by the voltage across the battery divided by the current...