Question

A flashlight battery has a lifetime of 5 hours and it is rated for 0.6 ampere-hours.

(a) What is the current that can be delivered by the battery?
(b) How much power can be delivered if the terminal voltage is 8 V?
(c) How much energy is stored in the battery?

Answer 1

a) Current (I) will be a product of time t (in hours) and the number of ampere-hours.
\[I= 5(hours) \times 0.6 (I/hours)\]
(as you can see, the hours will cancel out, leaving you the current.

b) You are given the voltage (V).
Power (P) is given by the equation
\[P=IV\]

c) You are given the time t. Energy E is given by the equation
\[E=Pt\]

Answer 2

The answer in a) is incorrect, because ampere-hour is Ah, not A/h. Capacity = current*time, and therefore you can draw a current of
\[\frac{0.6\ \mathrm{Ah}}{5\ \mathrm{h}}=120\ \mathrm{mA}\]
for one hour.

Actually, the original question is a little stupid, because "the current that can be delivered by the battery" depends on the internal resistance of the battery, which is not given. I assume that the thing teacher is asking is the current that can be drawn for one hour of the he/she does not understand electric circuits :-).

(b) How much power can be delivered if the terminal voltage is 8 V? If the current is 120 mA, the power is 8V*120mA=960 milliwatts. If the current is something else, the power is something else (P=U*I).

c) This is the only question that makes sense. If the capacity is 0.6 Ah and the voltage is 8 V, the energy stored in the battery is E=P*t=U*I*t=U*capacity=4.8 joules.