# Combining Series and Voltage: How is energy conserved?

1. Jan 19, 2008

### kmarinas86

Say you have three AA batteries each 2.5 amp hours.

In the first configuration, all the batteries are in series. The voltage is 4.5V and the current capacity is 2.5 amp hours. (4.5*2.5=11.25)

In the second configuration, all the batteries are in parallel. The voltage is 1.5V and the current capacity is 7.5 amp hours. (1.5*7.5=11.25)

In the third configuration, two of the batteries are in parallel, with one battery behind them. The voltage is 3V and the current capacity is 5.0 amp hours. (3*5=15)

So how is energy conserved here?

2. Jan 19, 2008

### tslibertan

Amp hour is measure of charge (e.g. useful electrons) in battery. Energy in battery is voltage * charge.

consider,

power, P = V x I
energy, E=P x time = V x I x time = V x Amp Hours

Each battery has fixed voltage and total charge, regardless of configuration, so total energy is not a function of configuration.

Your interpretation of the third configuration is incorrect. The lone battery will be exhausted before the ones in parallel, then configuration becomes just two cells in parallel.

3. Jan 19, 2008

### kmarinas86

I'm trying to understand why that's the case. Would the current of the lone AA battery be the same as the sum of the currents of the parallel AA batteries? If not, to what degree would it drain faster? I believed if it was the same type of battery being put under the same voltage, in this case 3V, that it would put out just as much current as each of the batteries in parallel.

Voltage does not split itself across y-junctions, do they? I thought that only current does that.

Last edited: Jan 19, 2008
4. Jan 19, 2008

### kmarinas86

Now let's say I have four AA batteries each 2.5 amp hours.

First scenario:

One battery is connected behind three batteries in parallel.

Now I would guess then the current of the battery behind would be three times that of each of the ones in parallel. So for the first 20 minutes of the run, the batteries collectively are 3 volts and 7.5 amps. Then, for the last 20 minutes of the run, the batteries collectively are 1.5 volts and 7.5 amps.

Together we have:

(1/3) * 3 * 7.5
+
(2/3) * 1.5 * 7.5
=
15

Second scenario:

Two batteries in series are connected behind two batteries in parallel.

Now I would guess then the current of each of the batteries behind would be two times that of each of the ones in parallel. So for the first 30 minutes of the run, the batteries collectively are 4.5 volts and 5 amps. For the last 30 minutes of the run, the batteries collectively are 1.5 volts and 5 amps.

Together we have:

(1/2) * 4.5 * 5
+
(1/2) * 1.5 * 5
=
15

Is this right?

Last edited: Jan 19, 2008
5. Jan 19, 2008

### tslibertan

Correct. As a result it would be exhausted in half the time that the parallel set would.

6. Jan 19, 2008

### tslibertan

Now let's say I have four AA batteries each 2.5 amp hours.

First scenario:

One battery is connected behind three batteries in parallel.

Now I would guess then the current of the battery behind would be three times that of each of the ones in parallel.

Correct.

So for the first 20 minutes of the run, the batteries collectively are 3 volts and 7.5 amps. Then, for the last 20 minutes of the run, the batteries collectively are 1.5 volts and 7.5 amps.

Together we have:

(1/3) * 3 * 7.5
+
(2/3) * 1.5 * 7.5
=
15

Right. Check: total energy in cells is 4 lots of 1.5 x 2.5 = 15.

Second scenario:

Two batteries in series are connected behind two batteries in parallel.

Now I would guess then the current of each of the batteries behind would be two times that of each of the ones in parallel. So for the first 30 minutes of the run, the batteries collectively are 4.5 volts and 5 amps. For the last 30 minutes of the run, the batteries collectively are 1.5 volts and 5 amps.

Together we have:

(1/2) * 4.5 * 5
+
(1/2) * 1.5 * 5
=
15

Is this right?

Yes, I agree with that. I think you've got it. Regardless of how complicated the configuration is, don't forget that the total energy is just that of the individual cells summed. As you've noted, these hypothetical set-ups change their behaviour with time.

7. Jan 20, 2008

### rbj

you wouldn't bet 5 amp hours out of the single battery that is in series wth the pair.

some of the energy remains behind un the two parallel batteries which were not discharged to death as would the single battery after 2.5 amp hours.

8. Jan 20, 2008

### tslibertan

A good point, but no-one's claiming a single battery would produce yield 5 AHrs (or even pass such a charge, which is not the same as yielding), or that a real AA would happily provide 7.5 amps.

In reality the depleted battery would probably break the circuit. Implicitly I understood these hypothetical scenarios to mean that once depleted the dead battery is removed from the circuit. In this case I believe everything written is fine. The important issue was energy conservation, which becomes moot if practical circuit reality is invoked. It is important to recognise this difference, however, since reality would give the impression that energy is not conserved, simply because of circuit failure.

Last edited: Jan 20, 2008
9. Jan 20, 2008

### kmarinas86

Let's test my knowledge again.

Say I have two 4-AA Battery holders and that these connect the AA batteries in parallel. Obviously, together these can hold 8 batteries.

First case:

The batteries are connected in series. The voltage is 12V, and the amps per battery, we'll just call it x. The current throughout the whole circuit is x, so the power is simply 12x. As for the length of time, we'll just call it t. So the electrical energy input is 12tx.

Second case:

The batteries are put into the battery holders which are then connected in series. The voltage is 3V, and the amps throughout the whole circuit is 4x, because each voltage component has 4 batteries in parallel. While each component is draining four times as fast, each component is also made of 4x as many batteries, so from each component, each of its four batteries is being drained at the rate of 1x creating the 4x. Imagine tiny hourglasses draining sand at the same rate. Since the hourglasses, or in reality, AA batteries contribute to the circuit in proportion to their parallel configuration, they will last just as long as they would in 100% series. However, now the current is 4x as much, meaning that the electrical power input is 12x. The run time is still t.

Third case:

The AA batteries are now put in complete parallel. The circuit current is 8 times the first case, and the voltage is simply 1.5V. This results in an electrical power input that is 12x. Each battery contributes 1x of current, so the run time should be the same, t.

HOWEVER

A higher voltage coil can power faster changing magnetic fields. 1 volt can power a magnetic field that changes 1 weber per second. The first case, whose electrical power output is 12x, can handle 12 webers per second. The second case, whose electrical power is 12x, can handle 3 webers per second. The last case, whose electrical power output is 12x, can handle 1.5 webers per second. For a given permanent magnet rotor that produces the magnetic flux, the change of magnetic flux per second is proportional to its spinning speed. Consequently, its rotational kinetic energy is proportional to the square of this spinning speed. We will call the spinning speed due to 1 volt y. We will use a constant, p, that represents the energy of the rotor per spinning speed squared.

By subtracting the rotational kinetic energy from the electrical energy, we get:

12tx - 144py^2 = ?

12tx - 9py^2 = ?

12tx - 2.25py^2 = ?

Last edited: Jan 20, 2008
10. Jan 20, 2008

### tslibertan

Are you making a Newman machine?

11. Jan 20, 2008

### kmarinas86

Not right now, but right now I have four "replicas" of the Newman machine, as can be seen on youtube.

Just to save time for the curious, I have these videos on a playlist you can view here:

They're basically electromagnets that switch on and off as a permanent magnet rotor rotates. So I'm not just thinking anymore, but I'm doing things too.

But I am more interested in what you think about the previous post I just made:

Last edited: Jan 20, 2008
12. Jan 21, 2008

### kmarinas86

Consider a set of identical batteries with voltage V. The current that one battery produces through the load is I. The current through the load is proportional to the number of batteries feeding through the same load. Such batteries are arranged in parallel, and their contribution to the current is n*I, where n is the number of batteries arranged in parallel for each set of batteries arranged in parallel. It is desirable to drain the batteries at equal rates, so it is ok in this circuit for identical battery sets (consisting of batteries in parallel) to be arranged in series. Therefore, the voltage of the circuit is proportional to the number of sets, or m*V. Therefore the power of the circuit is equal to I*V*n*m, where n*m is the number of batteries. Because the draining of the batteries is the same for all batteries, regardless of n and m, this is more proof that electrical energy is conserved.

Electrical energy should be convertible to mechanical energy, and eletrical power should be convertible to mechanical power. Rotational mechanical power is equal to torque times change in angle per unit time. The magnetic torque of the circuit is related to moment of the circuit's magnetic field. The use of a coil maximizes this magnetic moment. The magnetic moment of this coil is equal to the current of this coil times the turns of the coil times the area of each turn. Increasing the number of batteries arranged in parallel increases both variable n and the current. For a given circuit voltage m*V, increasing the magnet's strength reduces its maximum rpm. Increasing voltage increase possible rpms. Therefore, we find that the rotational kinetic energy is proportional to B * IA * rpm or (magnet strength) * (n * turns of coil * area of each turn) * (m / magnet strength). But what we find now is that we may get more rotational energy out of the system if we increase the turns of the coil and the area of each turn. But...

Let's consider that this coil is a uniform solenoid, a conducting wire shaped like a slinky. The electrical resistance of the coil's wire is proportional to its length and inversely proportional to its cross-sectional area. Likewise, the current from the batteries increases in proportion to the cross-sectional area and inversely to the length. By increases in the current, we do increase the strength of the magnetic field, but we proportionately reduce the run time of batteries, which suggests the conservation of energy. If the size of the rotating permanent magnet does not change, decreasing magnetic flux density by increasing the size of the loop would be canceled out by the increase of the size of the loop, there by doing little to affect the maximum rpm of the magnets. The area of existing loops may be increased, at expense to either the turns or the current, or any intermediate position thereof (Example: if the area of each turn quadruples: turns can be reduced by a factor of 2 so that current stay the same, otherwise current will drop, by up to a factor of 2). Doing this, would in effect, increase magnetic moment, and therefore the torque of the machine while leaving the maximum rpm much alone. This would mean that the machine could do more work when attached to a load.

Shape

Turn 1
Turn 2
Turn 3
Turn 4
Turn 5

Length TurnsArea

Single ring

15 225

Flat spiral

1 1
2 4
3 9
4 16
5 25

15 55

Torus

2 4
3 9
3 9
3 9
4 16

15 47

Hollow Cylinder

3 9
3 9
3 9
3 9
3 9

15 45