# Theoretical question about Ampere hour and Watt hour

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#### patchling

##### New Member
Battery capacity is either designated by Ampere/hour or Watt/hour. It is said that you can simply convert between the two by multiply W/h by Voltage.

Imagine a battery charging a capacitor to deliver a set voltage for a given time.

If we decide to double the voltage delivered by the capacitor, then we must double the amount of charge (current) delivered to it which doubles the current drain on the battery.

I would have thought that this would double the rate of battery drain; but the power equation P = V^2/R suggests that doubling the voltage squares the amount of work delivered by the capacitor (and implies that the amount of work needed to charge it was a squared increase), even though the current delivered is only double the original amount.

So a 10A/hr battery, load 5A/hr will last 2 hours --- double the voltage delivery, should last 1 hr. (i.e. 10A drawn in the hour).

But: 10W/hr doing 5W of work per hour will last 2 hours -- double the voltage delivery, you square the work done, i.e. 25W (therefore will last 15 minutes)

Therefore I cannot understand how the two measures can be converted between each other. It seems that the current delivery capacity (A-hr) is dependent on factors other than 'stored current'.

I know I've gone wrong somewhere, just can't see where. Thank you.

#### crutschow

##### Well-Known Member
You just didn't imagine enough. What you neglected is that the capacitor voltage varies as it's charged but the battery voltage is constant. The voltage difference (and thus energy) is dissipated by the series resistance between the battery and the cap (including the batteries internal resistance) as the cap is charging. Thus the energy delivered to the cap will always be less (the calculated value turns out to be 1/2) as compared to the energy extracted from the battery.

(There is an interesting question as to what happens if you could design a superconducting battery and capacitor with no resistance and perform the same experiment, then where does the energy go? One theory is that when the connection is made the circuit will oscillate due to the tank circuit consisting of the capacitance and parasitic inductance. This will generate radio wave radiation, as all oscillating circuits do, which will carry away the excess energy. It will then stop oscillating when 1/2 the energy has radiated away.)

#### patchling

##### New Member
Battery Capacity

Thank you, that does help with part of the problem; but I'm still a little unclear about measures of battery capacity.

Imagine ( ) a battery with a capacity of x ampere hours (the current reservoir) or y watt hours (the power reservoir). I understand that these two measures are largely equivalent.

The circuitry allows the battery to charge a capacitor to charge Q, at which point the circuit is closed (current stops flowing) and the capacitor discharged. The cycle then repeats with a regular frequency.

Now the programming of the circuitry is altered to allow the capacitor to charge to 2Q, before closing and allowing it to discharge. (The discharge cycle occurs with the same frequency as in the 1st case)

So the current drain doubles (2Q), but the energy drain quadruples (2Q)^2. It would seem that the current 'reservoir' is draining at a faster rate than the power 'reservoir'. But of course that can't be because they are just different ways of looking at a batteries longevity.

Does this make it clear? I'm not sure if I'm explaining my problem adequately. Thank you.

#### dknguyen

##### Well-Known Member
You have a problem with your unit defintions here that must be corrected.

It's not Amps/Hour. It's Amp-Hours. Amps/Hour would mean that the more hours you run the more amps you can pull from the battery before it runs out. If you have Amp-Hours and divide by amps consumed, the "amps" divides out of the equation leaving you with just "hours" the battery will last for which is what you want.

It's similar to how torque is Newton-Meters, not Newtons/meter. Newtons/meter would mean that if you applied torque to a lever arm, more force would appear at the the tip if it was longer, when in fact the force at the tip is less if the arm is longer.

The same thing is true for Watt-Hours. It's Watt-Hours, not hours per watt. A device will consume X Watts/Hour which is a measure of power consumption. But a device stores Watt-Hours which is a measure of energy storage:

Watts * Hours = Joules/Second *Hours = Joules/Second * 3600 Seconds = 3600*Joules
The time units cancel out leaving the final number in Joules- a measure of energy.

Is that clear?

So a 10A/hr battery, load 5A/hr will last 2 hours --- double the voltage delivery, should last 1 hr. (i.e. 10A drawn in the hour).

But: 10W/hr doing 5W of work per hour will last 2 hours -- double the voltage delivery, you square the work done, i.e. 25W (therefore will last 15 minutes)
5 Watts of work per Hour does not make sense. Work is Joules (energy). Watts is power. You could say 3600 Joules of work per hour which is basically saying 1W of power (Power = 3600 Joules/3600 seconds). Power output (Watts) is the same whether or not you output it for 1 second or a million years- what will be different is the total energy.

Power (Watts) = Energy (Joules) * Time (Seconds)

For current draw saying 5A/hour is incorrect (saying it draws 5A for one hour is a different thing). 5A is 5A, regardless of whether it exists for one second or a million years. Just like how Watts is Joules per second, amps is a measure of how much charge which is Couloumbs/second. Saying That Watts Per Second or Joules Per Hour doesn't make sense. You don't say that you are driving at 50km/h per hour, do you?

Do not move on until you understand this because it is part of the problem (and will cause many MANY problems elsewhere).

So a 10A/hr battery, load 5A/hr will last 2 hours --- double the voltage delivery, should last 1 hr. (i.e. 10A drawn in the hour).

But: 10W/hr doing 5W of work per hour will last 2 hours -- double the voltage delivery, you square the work done, i.e. 25W (therefore will last 15 minutes)
What's the problem? They are both measures of battery capacity but not equivelant. Amps-Hours (not amps/hour) is not a direct measure of energy storage since it only accounts for current (or more accurately stored charge since Voltage*current make power, not energy), not voltage. Watt-Hour (not watts/hour) is a direct measure of energy storage since it accounts for both Voltage and Current. Furthermore, 5 Watts is not the same thing as 5A so there is no reason that doubling the current draw should have the same effect as doubling the power draw. To convert between the two you NEED to work in that voltage in Amp-Hours, effectively converting it to Watt-Hours.

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#### dknguyen

##### Well-Known Member

What did I say the first time about the equation? If it was incorrect, I must have fixed it without even thinking about it because I don't remember it.

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#### patchling

##### New Member
Wouldn't the Voltage always be constant in my example, so I felt it could be taken out of the calculation?

Perhaps simplistically I think of the battery capacity as being equivalent to a gas tank (i.e. a total amount of charge or a total amount of energy stored) - but for a given increase in current drain, you are effectively squaring the energy drain (P = I^2.R) so the two reservoirs are being drained at different rates.

But obviously that can't be.

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#### dknguyen

##### Well-Known Member
Voltage is constant true. But it is not accounted for with a Amp-Hour rating. It is, however, accounted for in the Watt-Hour rating. Therefore to convert between the two you must take it into account.

Using the gas tank analogy-
1. Amp-Hours is like saying mL-hours (how many mL of gas you can constantly and continuously pull and burn from the gas tank for one hour before it is empty)
2. Watt-Hours is...Watt-Hours (how much power you can continuously get from the gasoline in the tank for one hour until it is empty)

1 does not take into account the energy stored in the gasoline (or whatever fuel is actually in the tank for that matter), while 2 does. In this case, changing fuels (diesel, nitromethane, gasoline) is like changing the battery voltage. A higher energy fuel will produce more energy per mL of fuel burned. A higher battery voltage will have more power per amps of current draw from it.

To convert 1 to 2, you need to know how much energy is actually gained from burning the gasoline (as opposed to some other fuel). Similarily, you need to work in battery voltage into amp-hours to convert it to watt-hours to actually compare it to a watt-hour rating.

You can see how one 12V and one 24V battery, both with 100Amp-hour capacities are providing different power outputs if you are drawing the same amount of current from them- but the 100Amp-Hour capacity says nothing of this. However, if both their capacities were given in the same Watt-Hours and you were using Watts drawn from the battery as your discharge rating, then the power levels and run-time would be the same, but the current draw would not be. The 12V battery would be outputting twice as much current as the 24V battery to produce the same power levels. ANd if you converted the Watt-Hour rating for the two batteries into Amp-Hours...the 12V battery rating would be twice as high as the 24V battery since the batteries had the same Watt-Hour rating which meant they could provide the same amount of power for the same amount of time except the power is balanced between voltage and current differently between the two.

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#### patchling

##### New Member
Ok, I can see the distinction between the two, and the way in which you convert from one to the other. In my example though I am keeping the gas (chemical reaction) constant, and the voltage constant, therefore the total amount of energy the battery/motor can produce should be constant.

But I am changing the way in which the gas is used.

For each cycle (charge capacitor, discharge capacitor, wait until end of cycle, repeat) I am either charging the capacitor with Q or 2Q. So, charge drain is only doubling. But by the energy equation, power drain is squaring.

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#### dknguyen

##### Well-Known Member
I was editing my post still! Please reread (now you can see the reasons for my signature lol)

#### patchling

##### New Member
Sorry about that . I've got to be a bit more patient.

Ok I think I'm up to speed with you. But...

In my example I'm not changing the voltage, just the time spent charging the capacitor. (Let's say the battery is Lithium bat, ~ 2.8V open circuit voltage).

To double the charge (Q) on the capacitor a significantly longer time will be spent but the voltage throughout is the same.

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#### dknguyen

##### Well-Known Member
What voltage is going to be the same? The voltage of the capacitor when it is charged to Q vs 2Q? As a capacitor gains more charge it's voltage will rise. The larger a capacitor is, the more charge is needed for the same rise in voltage (like a large or small swimming pool's water levels and the amount of water needed to get it to that level). As capacitors gain more charge they rise in voltage and decrease in current as they approach the source voltage. So if the Q you picked is a large enough charge such that it will make the capacitor voltage equal to battery voltage...there is not going to be a capacitor that can be charged to 2Q since battery voltage is the maximum. Similarily, if Q causes the capacitor to be higher than half the battery voltage, then the capacitor charged for a longer period of time is not going to charge up to 2Q since that would exceed battery voltage- it's going to gain enough charge to get to whatever the battery voltage is.

Therefore, if it's the same capacitance, they are NOT going to be the same voltage. The 2Q capacitor will have a voltage double that of the Q capacitor (assuming the chosen charge Q causes the voltage to be less than half battery voltage. If not, the capacitor charged for a longer period of time will max out at battery voltage, never reaching 2Q charge levels.

The charge and discharge graphs are most important on this link:
http://hyperphysics.phy-astr.gsu.edu/Hbase/electric/capchg.html

...Done editing. Now you go.

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#### patchling

##### New Member

Ok I've waited a few minutes.

Yes, I mean the battery voltage - this will remain constant while the capacitor is charging. And for my example the voltage on the capacitor is only charged to a level below the battery voltage.

To provide some more information I am trying to explain this, from a cardiac pacing textbook:

(In a discussion of doubling the charge stored on a capacitor...)

"Doubling of the amplitude (voltage) doubles both the current and the total charge delivered to the heart. It might seem that, because the charge per pulse is doubled, the average pacing current drawn from the battery would also be doubled. However, the impact on the pacing current is much larger than that as seen from the following argument....derives energy equation E = V^2/R... From this equation, it is readily apparent that energy consumption increases with the square of the output voltage."

I'm trying to understand how the impact on the pacing current can be larger than the predicted effect of drawing double the current.

So, the ***average pacing current drawn from the battery is doubled*** but the effect on the pacing current is ***much larger than that***.

That seems contradictory, like saying the amount of gas drawn from the tank is doubled, but the effect on the outflow of gas is greater than that.

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#### dknguyen

##### Well-Known Member
Actually the statement says that that doubling the voltage would intuitively cause the voltage to double, but that's not what happens. What really happens is that doubling the voltage causes the current to MORE than double. The keywords being "It might seem that...However". So what they're saying isn't contradictory. But, I'm going to have to say...not true.

You seem to have omitted their following argument. What do they argue?

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#### patchling

##### New Member
..argument. The energy per pacing pulse is defined by the following equation -
E = V . I.t

In this equation, V is the average pacing stimulus output voltage of the pulse generator, I is the average pacing current delivered to the heart, t is the pulse width. If the lead-electrode heart interface is considered to be mainly resistive, Ohm's law (i = V/R) can be substituted in Equation (above) making it

E = V^2/R

Here, R is the effective ohmic load of the heart and lead. From this equation it is readily apparent that energy consumption increases with the square of the output voltage.

#### dknguyen

##### Well-Known Member
Their argument has nothing to do with the current by increasing more than double. It has everything to do with the power/energy not doubling (but quadrupling which is understandable since you are doubling both voltage and current which make up power). I'd say you can ignore their statement there. I do not know where it comes from. I am inclined to agree with you that they do not make sense and that doubling the battery voltage charging a capacitor doubles the current.

Also, Power = V*I and Energy = Power * t = Energy = V*I*t which is correct.

However, Power = V^2/R, not energy. Energy would be V^2/R * t. You or they have a typo there.

THat probably concludes that unless you have something new. BUt I'm going to bed- will check again in the morning.

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#### patchling

##### New Member
Mathematically

Mathematically I can understand it. Conceptually...

The energy delivered to the heart is quadrupled; but the current drawn from the battery is only doubled. Are we effectively reducing the Ampere hours of the battery? That is, the reservoir is 'dynamic' depending on how the battery is operating.

Battery ==> (Charge)==> Capacitor ==> Discharge to heart.

Thanks for all your help. I'm getting there.

Cheers

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#### dknguyen

##### Well-Known Member
Voltage is doubled. The nature of the circuit means double the voltage causes double the current. BOth current and voltage make up power/energy, so power gets a two doublings which means a quadrupling.

Original Power = V*I
New Power = 2V*2I = 4VI = 4*Original Power

We are talking about the capacitor voltage here. You are either ending the the capacitor charge process prematurely to produce a voltage less than the battery voltage, or you are increasing the battery voltage itself and charging the capacitor battery voltage. Either way, doesn't matter.

THat's all. Reducing amp-hours of the battery? That would mean changing the battery's total capacity. Do you mean reducing the run-time of the battery due to increased power consumption? If so, then yes.

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#### patchling

##### New Member
Final..

So during the charging of the capacitor, equivalent proportions of the current reservoir and power reservoir are discharged, for any given charge on the capacitor?

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#### crutschow

##### Well-Known Member
As I previously noted if you directly charge a capacitor from a battery, half the energy will be dissipated during the charging process. To minimize this loss you could use a switching regulator in a current regulated mode. That way you can typically charge the capacitor with 80-90% efficiency and the power stored on the cap would more nearly equal the power drawn from the battery. (Note that the charge drawn from the battery will now be less then the charge transferred to the cap since the switching regulator can have a higher current output than input).

#### patchling

##### New Member
Wrapping it up

So, by way of conclusion - though correct me if wrong.

1. The relative current drain and energy drain from the battery must remain constant - current drain * voltage = energy drain. This is a truism, because there is only one "gas tank".

2. You cannot infer the current drain from the battery by measuring the current flow in another part of the system - there are methods for increasing current flow or reducing it in different parts of the system, without affecting the current drain from the battery; charging and discharging a capacitor for instance, effectively loses a great deal of the original charge that the battery drained.

3. The original example where current is doubled, but energy is quadrupled, demonstrates one instance where doubling the current in part of the system, actually has a substantially greater effect on current drain from the battery than you might think.

4. Perhaps best to consider battery drain directly in terms of energy and power. You can't create energy in the system that the battery hasn't output. Therefore in our example, the energy delivered is quadrupled, though current has only doubled - the effect on battery drain is better estimated from the energy delivered than the current delivered.

Correct where you think I have gone wrong.

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