Finding internal resistance?

Question

A battery having an emf of 10.60 V delivers 102 mA when connected to a 60.0 ohm load. Determine the internal resistance of the battery.

How do I solve this?

Answer 1

The emf of the battery is the "open circuit" voltage, that is when no current is flowing. So figure the voltage across the 60 Ohm load which is .102 * 60 = 6.12 Volts.

Where did the remainder of the voltage go? That is 10.60 - 6.12 = 4.48 Volts ? The answer is that voltage is dropped across the internal resistance of the battery. So now we know the current and voltage across the internal resistance we can calculate the Ohms = 4.48 / .102 = 43.92 Ohms

Answer 2

Use ohm's law. V=I*R, which when rearranged gives R=V/I. We already know the voltage (10.6 v) and the current (.102 A). Dividing 10.6 by .102 gives an approximation of 104. The given resistance is 60. Subtract 60 from 104 to give 44 ( or 43 depending on if it isn't rounded), which is the internal resistance of the battery.

Answer 3

Call the 60 Ohms Rload. Call the internal resistance Ri. The current flowing flows thru both Rload and Ri (in series). So by Ohm's law:

emf = 102 ma*(Ri + Rload)

Or you could use the logic: Rload has a voltage of
102 ma*Rload volts
across it. The emf is 10.6 V. The rest of the voltage, Vlost, must be lost across Ri. And since you know the current thru Ri,
Vlost = 102 ma*Ri

You can do the math.

Answer 4

The total resistance in the circuit is E/I = 10.6/.102 =103.92 ohms. ubtract the load resistance so the internal resistance must be 103.92 - 60.0 = 43.92 ohms.