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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.

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.