Question

6) (15 Points) A typical AA size 1.5 V Energizer alkaline battery has an internal resistance of 0.2 ohms. We connect this battery to different loads (i.e. light bulb) with different resistances. What will be voltage across the load if it has a resistance of 1 kohms, 100 ohms, and 1 ohms? For which one of these cases the internal battery resistance is most significant? Compare the current-voltage (I-V) curve of this battery to that of an ideal 1.5 V battery? Do they deviate more at small or large currents?

Answer 1

Answer :-  For r = 0.2 ohm, and V = 1.5 V then,
for R = 1 kohm we get current, i = 1.5/(1000 + 0.2) A = 1.4997 mA. Thus voltage across load = 1.4997 x 1000 /1000 = 1.4997 V.

for R = 100 ohm we get current, i = 1.5/(100 + 0.2) A = 14.970 mA. Thus voltage across load = 14.970 x 100 /1000 = 1.4970 V.

for R = 1 ohm we get current, i = 1.5/(1 + 0.2) A = 1.25 A. Thus voltage across load = 1.25 x 1 = 1.25 V.

For the last case i.e. for the bulb of 1 ohm resistance the internal battery resistance is most significant. Since in this mode most of the battery power is delivered to load. As we know if the internal resistance is equal to load resistance then maximum power is delivered to the load. Here 1 ohm is very close to 0.2 ohm, compared to 100 ohm and 1000 ohm.

With ideal battery source the I-V graph will be a straight line with some slope and passing through origin.
But for the practical battery with some internal resistance, the grapgh will be falling down from high current value to zero.