How do we obtain an ampere-hour rating given voltage and watt-hour?

In summary: The solution provides a shortcut to calculate watt-hours from volts and amperes.The solution calculates watt-hours by multiplying volts (volts x amperes) and dividing by the battery's capacity in amp-hours.
  • #1
zenterix
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Homework Statement
A lithium-ion battery pack for a camcorder is rated as 7.2V and 5 W-hours. What are its equivalent ratings in mA-hours and joules?
Relevant Equations
This is an example problem from the book "Foundations of Analog and Digital Electronic Circuits" by Agarwal.

Let me show the solution in the book and then ask questions

Since a joule(J) is equivalent to a W-second, 5 W-hours is the same as 5 x 3600 = 18000J.

Since the battery has a voltage of 7.2 V the battery rating in ampere-hours is 5/7.2=0.69. Equivalently, its rating in mA-hours is 690.
The solution is thus 0.69 ampere-hours, or 690 mA-hours.

Now, as far as I can tell, an ampere-hour is unit of charge. That is we can convert it to coulombs (C).

For example, if we have a current of 1A for 1h, then we have 1 C/s x 3600 s = 3600 C = 1 ampere-hour. This is my guess since the book assumes this knowledge apparently.

The calculation in the solution does check out. A watt-hour is a unit of energy: 1 J/s x 3600 s = 3600 J. Thus, when we divide watt-hour by volt we get w-h/ volt = J/(J/C) = C.

Now let me show what I originally did before looking at the solution.

5 watt-hours means 5 times 1 watt x 1 hour = 5 x 1 J/s x 3600 s = 18000 J.

Work is being done at 7.2 J/C.

If we have a current of 1 A = 1 C/s, then we have a rate of work of 1 C/s x 7.2 J/C = 7.2 J/s.

For a store of energy of 18000J, this current can be sustained for 18000 J / 7.2 J/s = 2500s.

The amount of charge passing through the circuit is 2500 s x 1 C/s = 2500 C.

But we also know that 1 ampere-hour = 3600C.

Thus, the amount of charge passing through the circuit is (1/3600) ampere-hour/C x 2500 C = 0.694 ampere-h.

The solution in the book took a shortcut, but my question is if my solution (which seems to have reached the same result when we round) was arrived at with the correct reasoning about the underlying concepts.
 
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  • #2
It looks OK. You assumed a current of 1A and a corresponding time of 2500 s. But the results is independent of these values. The battery can use its capcity to provide various values of the current, during various time inetrvals. The quantities asked for are independent of the regime the battery delivers the current (ideally or as long as the current is within the limits designed for that battery). The solution provided is more like a general solution, you don't need to assume some regime. But your solution is not wrong and if it helps you understand better the situation, why not use it?
 
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  • #3
Watts = Volts times Amperes.

One joule is equal to 2.777778 times 10-7 kilowatt-hours.
 
  • #4
Right. Here is my interpretation of the much better solution provided by the book.

watt-hour is a unit of energy.
ampere-hour is a unit of charge.
potential difference is energy per charge.

Thus, potential difference times total charge moving across this potential difference equals total energy provided by the electric field generating the potential difference.

Thus, 7.2 J/C x Total Charge = 5 W-h

Total Charge = (5/7.2) (W-h)/(J/C) = 0.69 (J/s x h)/J/C = 0.69 (C/s x h) = 0.69 A-h
 
  • #5
It may be easier to write the formulas using math language rather than using words. And using quantities and not their units.
For example, electrical energy is
## E=\Delta V I \Delta t =\Delta V \frac{\Delta Q}{\Delta t} \Delta t=\Delta V \Delta Q##
and here you have the relationship between energy, voltage and the charge (battery "capacity").
 
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