00:01
Hi, here in this given problem, emf of the alkaline cell that is given as 1 .58 volt and its internal resistance that is small r is equal to 0 .100 oom.
00:38
Now in the first and one more thing, resistance of the volt meter which is put across this alkaline cell to measure the terminal voltage, its resistance is given as 1 .00 kilo -oom or we can say this is 1000 oom.
00:59
So in the first part of the problem, current passing through the voltmeter through this circuit that will be given by simply using om's law net emf divided by net resistance.
01:14
This is 1 .585 different.
01:19
Divided by 1000 ome plus 0 .1 om.
01:25
And finally here it will come out to be equal to 1 .58 into 10 dash to the power minus 3 ampere.
01:36
Which becomes the answer for the first part of the problem.
01:40
Then in the second part we have to find terminal voltage across the cell and that will be given by emf minus the voltage dropped across the cell, across the electrolyte of the cell, and that is i into r...