00:01
In the given problem, emf of the given battery is epsilon is equal to 1 .5 volt and its internal resistance means small r is given as 1 .0.
00:35
In the first part of the problem, we have to charge this battery with an external emmmm.
00:45
Power supply whose value we have to find.
00:49
So suppose emf of power supply is e, capital e.
01:07
So net emf of the circuit comes out to be e minus epsilon.
01:21
E is the emf of the charging source and epsilon is the emf of the source which is being charged as the charging source always should have a voltage greater than that being charged.
01:36
Hence this is e minus e.
01:40
So the current passing through the circuit is given by i is equal to using omslaw net emf divided by net resistance.
01:51
So here it will be e minus epsilon divided by r means the internal resistance.
02:01
So plugging in all the values, it is given that the current in the circuit should be 0 .75 amp is equal to e minus 1 .5 divided by 1 .0.
02:18
Hence, we can say e is equal to 0 .75 plus 1 .5 volt.
02:33
Or we can say this enf of the external power supply comes out to be 2 .25 volt, which is the answer for the first part of this problem...