00:01
All right, so for this question, we know that the current on the capacitor can be right as q equal to the capacitance times the emf times 1 minus 8 to the negative t over rc.
00:16
And in meanwhile, the current can be right as the emf over r times into the negative t over rrc.
00:26
So we know at this moment the current equal to 3, 3mf.
00:33
Right and the chart is equal to 40 times 10 to the negative 6 right so this is a good this is m from the second equation we can know that from the this is first equation second equation from second equation we can know that the emf times each of the negative t over r c is equal to 3 which is 36 right 36 votes and from the the first equation, we can know that, so if we divide the capacitance on both sides, so the emf minus emf times negative 8 over our c is equal to 40 times 10 to the negative 6 over the capacitance, which will give us 8 both.
01:31
So emf minus this part, so this part, which is 3r, which is 36, which is 36.
01:42
Equal to v8 volts, which is going to give us dmf equal to 8 minus plus 36 equal to 44 volts.
01:52
So that is the emf of the battery.
01:56
Okay, that's for question a.
01:59
For question b, again, from equation 2, we say that emf 44 times into the negative d over rc equal to 36...