# Consider a battery charger that is supplying a voltage of 110 V. A cell of EMF 12 V is to be charged. Its internal resistance is 0.5 ohms. We have to limit the current to 5 A for a technical reason. In order to do this, how much resistance must be attached in series to the cell?

2
by trisha22
GIVE PICTURE OF CIRCUIT

2014-06-14T22:49:21+05:30
There are two ways of solving this problem, both having same calculations.
One is by Using Kirchoff's Law and one without. I am telling this without using the Law, because if you understand this, you can do it using the Law.
In the image attached, I have installed the internal resistance outside the ideal battery and limited the battery between the limits E and F. So, I can say my whole battery is between E and F. Now Since the battery is to be charged, it need to be placed in opposite way as the Charger's terminal's are connected with the series [Chargeable batteries need to reverse the chemical reactions to be charged]. So the effective Potential Difference will be 110-12 = 98 V. Now, there are two resistance and both in series [your question says so]. The total resistance is 0.5 +x.
Now using the formula V=IR, I given as 5 Amperes [Current to be restricted to 5 A], value of x comes to be 19.1 Ohms.