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joshanders_84
Nov23-04, 05:01 PM
Hey I am confused as to the answer to this question; not sure exactly how to figure out what the answer is because it is not a constant resistance. Any help appreciated:

The potential difference across the terminals of a battery is V_1 when there is a current of I_1 in the battery from the negative to the positive terminal. When the current is I_2 in the reverse direction, the potential difference becomes V_2. What is the internal resistance of the battery?

I tried adding and subtracting the two resistances (R_1 = V_1/I_1 and R_2 = V_2/I_2), but that isn't correct, and I'm not sure of what to do. Thanks
Josh

ceptimus
Nov23-04, 05:22 PM
Call the real voltage of the battery (when it's delivering no current) V, and its internal resistance R.

In the first case V = V_1 + I_1 * R

(Here, some of the battery's real voltage is wasted overcoming the internal resistance, so the measured voltage V_1 is less).

In the second case V2 - V = I_2 * R

(Here, the current is in the opposite direction, so both the battery's real voltage and it's internal resistance oppose the applied current).

So you have a pair of equations and a pair of unknowns. Should be easy...

marlon
Nov23-04, 05:34 PM
correct, i get the same answer...
marlon