- #1
Whakataku
- 12
- 0
Relevant equations:
V = I*R
ε = I*(R+r)
Then,
V = {ε/(R+r)}*r
V = potential difference, ε = electromotive force, R = resistance of external load resistor,
r = internal resistance of the battery.
So say the external load resistor's resistance is a constant for two batteries of the same make, difference being that one is good and the other is weak. Then since by intuition the internal resistance will be different, the good battery at room temperature having less r than the weak battery. The ε is constant.
Then is it correct to say that the voltage or potential difference will be smaller, or there will be voltage drop for the weak battery compared to the good battery?
V = I*R
ε = I*(R+r)
Then,
V = {ε/(R+r)}*r
V = potential difference, ε = electromotive force, R = resistance of external load resistor,
r = internal resistance of the battery.
So say the external load resistor's resistance is a constant for two batteries of the same make, difference being that one is good and the other is weak. Then since by intuition the internal resistance will be different, the good battery at room temperature having less r than the weak battery. The ε is constant.
Then is it correct to say that the voltage or potential difference will be smaller, or there will be voltage drop for the weak battery compared to the good battery?