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?



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.
Please tell me the answer so that I can recheck mistakes. 
your answer is corrrect i.e., 19.1 ohms
Thank God

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Let the required resistance be r
ohm's law V = IR  ⇒ 110 = 5(r + 0.5) ⇒ r = 22.5 ohm is the answer
sorry 21.5 ohm is the answer
Bro, the answer is 19.1 and you should take the total potential difference, read the question again or refer to my answer.
oh yay you are right then also it is easy just substract 12 from 110 v and you can get the answer 19.1 ohm