00:01
Now in this particular problem, as per the given condition, the defective starter motor draws a current of 300 ampere with 12 volts car battery.
00:09
So first, we'll just try to plot a data that current denoted as capital i, which is given as 300 ampiers.
00:22
Emf of the battery is denoted as a capital e, which is of 12 volts.
00:32
And the terminal voltage of the battery is denoted as, say, capital v, which is of 6 volt.
00:41
Now, we'll try to solve this sum into two steps.
00:45
The first one is, we first need to get the internal resistance of the battery.
00:52
So as per the formula that we are going to use is capital e minus v is equal to i times smaller.
01:03
Where e is equal to emf, r stands for the internal resistance.
01:09
We'll just mention here, this indicates the internal resistance, and v stands for the terminal voltage.
01:21
Now moving ahead, we get, therefore, value of e is given as 12 volts, value of v, that is the terminal voltage, is given as 6 volt, value of the current is 300 ampers, and we need to get the value of r...