Problem — capacity of an AA battery

Problem

Capacity of an AA battery. An alkaline AA battery has 2900 mAh printed on its side. Read the label as “milliampere-hour”: it is the total charge it can deliver. (a) Express it in coulombs. (b) Estimate the total energy stored, knowing the average EMF is $\mathcal{E} \approx 1{,}5\ \text{V}$. (c) Compare it with the kinetic energy of a $100\ \text{g}$ apple dropped from $1\ \text{m}$: how many apple-falls does an AA battery “contain”?

Solution

(a) Charge in coulombs. An ampere-hour is the charge carried by $1\ \text{A}$ in an hour: $1\ \text{Ah} = 1\ \text{A} \cdot 3600\ \text{s} = 3600\ \text{C}$. So $Q = 2900\ \text{mAh} = 2{,}9\ \text{Ah} = 2{,}9 \cdot 3600 \approx 1{,}0 \cdot 10^4\ \text{C}$ (a bit more than $3\ \text{Ah}$ rounding the label).

(b) Stored energy. Energy is (charge) $\times$ (average potential difference): $E = Q\,\mathcal{E} = 1{,}044 \cdot 10^4\ \text{C} \cdot 1{,}5\ \text{V} \approx 1{,}6 \cdot 10^4\ \text{J} \approx 16\ 000\ \text{J}$

(c) Comparison with the apple. The kinetic energy of an apple ($m = 0{,}10\ \text{kg}$) dropped from $h = 1\ \text{m}$ is $E_\text{apple} = m g h = 0{,}10 \cdot 9{,}8 \cdot 1 \approx 1\ \text{J}$ So the battery contains $N = \frac{E}{E_\text{apple}} \approx \frac{16\ 000}{1} \approx 1{,}6 \cdot 10^4$ roughly sixteen thousand apple-falls. That’s a lot of chemical energy in a small object, but delivered at low power and low voltage.

$\ev{Q \sim 3\ \text{Ah} \approx 1{,}0\cdot10^4\ \text{C} \qquad E \sim 1{,}6\cdot10^4\ \text{J} \qquad N \sim 1{,}6\cdot10^4 \text{ apples}}$

Links

Topics: Circuiti elettrici Concepts: Forza elettromotrice Skills: Stime di Fermi