How can i convert the potential energy of a mass into mAh for this reason:

How heavy would a mass have to be if I wanted to charge my Iphone (2800mAh) using a apple phone charger (5V 1000mAh). The mass would only be able to fall 0.5m over the course of the charge. The mass would pull a rope that is geared to a electric generator. Assume the efficiency of the generator to be 75%.

Any useful formulas would be greatly appreciated!

Thank You!

  • \$\begingroup\$ is this a school assignment? \$\endgroup\$ – jsotola Nov 6 '17 at 4:32
  • 1
    \$\begingroup\$ Shame it wasn’t how much time on an exercise bike to produce that amount of energy : get them off the couch.... \$\endgroup\$ – Solar Mike Nov 6 '17 at 6:21
  • \$\begingroup\$ 'cmon, man! mgh=E! \$\endgroup\$ – Voltage Spike Nov 7 '17 at 1:29

This is essentially a physics question, but I'll humor you.

The relevant capacity of the battery is its watt-hour capacity, not its milliamp-hour capacity. For a iPhone battery, this is roughly 10 Watt-hours. (You can get the exact Watt-hour capacity by multiplying the milliamp-hour capacity by the terminal voltage and dividing by 1000.)

One Watt-hour is 3600 Joules, so 10 W-h is 36 KJ. Divide by efficiency, distance, and gravitational acceleration to get the required weight. Since this seems like a homework problem, I won't give you the exact numbers from here on out, but my math says the answer comes out to roughly ten tons. (Hopefully this wasn't supposed to be a practical exercise.)


I rather doubt you can get 75%. But it doesn't matter. It's just a constant factor. Call this factor, \$\eta\$.

There is \$4.9\:\frac{\textrm{J}}{\textrm{kg}}\$ available near the Earth's surface for the short drop distance of \$\frac{1}{2}\:\textrm{m}\$ that you are talking about. That's at 100% efficiency, of course.

Your battery needs \$10080\:\frac{\textrm{J}}{\textrm{V}}\$ to be completely charged. You don't specify your battery's voltage, only the charger voltage which really doesn't tell me anything much, so I cannot figure out the total energy available when fully charged. But that figure allows you to apply the battery voltage once you discover it.

Combining the two, this works out to the following equation:

$$\approx \left(2057\:\frac{\textrm{kg}}{\textrm{V}}\right)\cdot \frac{V_{battery}}{\eta} $$

So, just multiply 2057 by the battery voltage and then divide by your efficiency figure (75% would mean to divide by .75) and you have the mass in kilograms.

For example, if the battery voltage is \$17\:\textrm{V}\$ (just to pick a number) and your efficiency is \$\eta=.75\$, then you need almost \$4.7\times 10^7\$ grams or close to 52 short tons of mass.

If this doesn't give you an appreciation for chemical energy stored in batteries, nothing will.

  • \$\begingroup\$ An iphone battery voltage is approximately 5.8 volts... \$\endgroup\$ – Solar Mike Nov 6 '17 at 5:33
  • \$\begingroup\$ @SolarMike The OP should thank you. ;) That's like 17.5 short tons. Still a fair bit. Side note: playing with the numbers I just discovered (I had no idea before) that a calorie (not a food calorie) of energy is approximately the same is one pound impacting the floor, dropped from 1 yard above the floor. Almost too close in value for comfort. :) British units and all. (Yes, I know it's not defined that way. Just weird coincidence.) \$\endgroup\$ – jonk Nov 6 '17 at 5:36
  • \$\begingroup\$ Re "If this doesn't give you an appreciation for chemical energy stored in batteries, nothing will." -> Failing that, "vent with flame" may :-). \$\endgroup\$ – Russell McMahon Nov 6 '17 at 11:52

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