Help solving a battery problem?

Question

When a car's starter is in use, it draws a large current. The car's lights draw much less current. As a certain car is starting, the current through the battery is 66.0 A and the potential difference across the battery terminals is 8.91 V. When only the car's lights are used, the current through the battery is 1.98 A and the terminal potential difference is 12.1 V. Find the battery's emf and internal resistance.

DH · Accepted Answer

For the starter we have 
Emf - 66*r = 8.91  Eqn 1

and for the lights    Emf - 1.98r = 12.1  Eqn 2

Now take eqn 2 and subtract eqn 1 giving

64.02r = 12.1 - 8.91 = 3.19

So r = 3.19/64.02 = 0.0498Ω

Now using eqn 2 we get Emf = 12.1 +1.98*0.0498 = 12.2V

Help solving a battery problem?

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