Help finding emf and internal resistance?

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.

bonobo · Accepted Answer

V =  İ • R

        Vτ =    (66)•R  + 8.91

        Vτ = (1.98)•R  + 12.1
         ─────────────────<subtract>
         0 = (64.02)•R − 3.19

        R = 0.0498 Ω


  Vτ = (66)•(0.0498)  + 8.91 = 12.19866 = 12.2 volts

  Vτ = (1.98)•(0.0498)  + 12.1 = 12.19866 = 12.2 volts (check)

Anonym · Answer

You are correct on both. The emf of the battery is the "electromotive force" or voltage, so 1.5V is correct. And yes the internal resistance in this example is 149Ohms and yes this is incredibly high. Most alkaline AA cells get up to about 0.5 Ohm internal resistance near the end of their life-cycle. This is why, if you actually did connect a 1.5V AA battery across a 1 Ohm load, it would heat up very quickly and definitely pull more than 10mA.

Help finding emf and internal resistance?

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