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This question will probably sound easy to most of you, but to me, it is still a kind of magic :)

Taking in account all I know until know, I guess a battery has two sides:

  1. one that "emits" electrons (any technical term for that ?)
  2. one that "lacks" electrons

So how can it be that when I put two batteries in series, they voltage adds up ? To me it sounds like the electrons from one should flow directly among the protons of the other.

Obviously there must be some kind of black magic there ;) Will one of you sorcerers explain it to me ?

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A cell or a battery is essentially a charge "pump". Now, to help form an intuition for the answer to your question, fall back to the hydraulic analogy.

Two water pumps in parallel can produce twice the water flow of one (ideally).

Two water pumps in series can produce twice the pressure (or head) of one (ideally).

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  • \$\begingroup\$ I do understand the hydraulic analogy, but this means that electrons may only move in one way in the batteries. How is that physically/chemically possible ? \$\endgroup\$ Apr 28 '13 at 21:39
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    \$\begingroup\$ @Antoine_935, inside the battery, charge is transported by ions in the electrolyte. The chemical reactions of the electrolyte with the anode & cathode materials remove electrons from one and add electrons to the other. But, within the electrolyte, the electric current is due to ions, not electrons. \$\endgroup\$ Apr 28 '13 at 21:42
  • \$\begingroup\$ The terminals are connected to electrodes made of materials with different electronegativity. Technically there is some movement in both directions, but that in one direction outweighs the other by orders of magnitude. en.wikipedia.org/wiki/Galvanic_cell \$\endgroup\$ Apr 28 '13 at 21:44
  • \$\begingroup\$ @AlfredCentauri - yes, but ions are atoms or groupings thereof with either excess or insufficient electrons - so in the end it's all about electrons, though there are consequences of the nuclei which come along for the ride. \$\endgroup\$ Apr 28 '13 at 21:47
  • \$\begingroup\$ @ChrisStratton, protons, AKA hydrogen ions, would likely be offended by your comment that "it's all about electrons". \$\endgroup\$ Apr 28 '13 at 21:50
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I can see why it feels like black magic, since it seems like the electrons that leave the anode of one battery just go into the cathode of the next battery, and there seems to be no reason for the voltages to add up.

But let's look at it in a different perspective. Let's say you have a battery A with a voltage v0. Battery A is the only battery in a circuit with a resistor R also in the circuit. Resistor R's resistance is SO high, that battery A is hardly able to generate a tiny current through the circuit. v0 just isn't enough to produce a significant current through resistor R.

But now, let's say you add battery B and battery C in series to battery A, with resistor R still in the circuit, too. Batteries B and C also have voltages of v0.

In each battery, within a closed circuit, an electron really wants to leave the anode while another electron from the wire enters the cathode. When we just had battery A hooked up to resistor R, it's voltage v0 wasn't enough to make this happen, remember? But now, with all 3 batteries in series, not only does battery A's voltage want to naturally push an electron away from it's anode, but battery B's cathode also is exerting a pull on an electron from battery A's anode, too! And with just a voltage of v0, how is battery B able to give up an electron from its anode to so that it can receive one in its cathode from battery A? Because it also feels a pull from battery C's cathode, from battery C's v0! And last, you guessed it: battery C is willing to give up an electron from its anode to accept one in its cathode (with just the low, individual voltage v0) because it feels a pull from battery A's cathode as well! There's no first or last, it all happens at once, like a chain. Do you now see how the effects of each battery compound to produce a much higher voltage (aka a much higher motivation, or pull/push for electrons to flow) ?? Also, assume resistor R was between battery C and A. That part doesn't really matter.

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  • \$\begingroup\$ There is actually an ion current (in the electrolyte) to match the electron current, so there is stuff flowing inside the battery to generate the voltage \$\endgroup\$
    – Voltage Spike
    Jun 28 '17 at 6:42
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Excess electrons are unstable and want to fill in atoms that can accept electrons. Excess electrons are loaded in the negative side of a battery and stored because the electrolyte will not allow them to the positive side where they can be accepted. There has to be a conductor between the neg and pos to allow them to flow or they will stay there, stored until they leak out slowly and there is no longer a potential (difference) between the two. When you add two batteries in series the potentials (voltage) are added because since the same charge is moved twice each time thru the same voltage (potential) the total work done is 2 * V but the current flow remains the same. In Parallel, the charge is moved once through either battery so the voltages of the batteries are not added but with two batteries, the charge can do twice the movement because both batteries allow twice the movement and so the currents are doubled. Kind of like a river and dam. Each battery is a wall of a certain height (potential) and the water is the current flow. Each battery (wall) can only allow so much water to go through. The main large river split into two rivers with a dam on each allows twice the water (current) through at the same water height (Voltage). Whereas if you stack the dams up then they only will allow the same amount of water as one dam and the water will be double the height (voltage). And each dam (battery) only has so much water to provide and must be connected back to the empty portion of the dam because it is a sealed water system so water can’t flow unless water comes back. The empty portion of the dam is lower (positive terminal) and thus can collect water but can’t re-run it through the system unless it is pumped (charged) back up to the higher portion of the dam (negative terminal).

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Here is what I think, you guys tell me what you think.

Lets us think about this in the physics perspective.

  1. First law of thermodynamics

The law of conservation of energy:the total energy of an isolated system is constant; energy can be transformed from one form to another, but can be neither created nor destroyed (wikipedia)

  1. electric potential and electric potential energy

I'll go straight to the formulas

electric potential or V =-W/q = U/q is the work done per unit of charge or is the amount of electric potential energy per unit of charge

Since we already know the electric potential of the batteries

electric potential energy or U = qV, this will be the potential energy carried by an electron leaving the negative terminal

Now lets put three batteries in series say, A, B, C. So that the (-) of A connect the (+) of B, the (-) of B connects the (+) of C and (- )of C connects a resistor which connects the (+) of A. All of 1.5 V

When an electron with a potential energy of U pass through the resistor, part of its potential energy is transferred to the resistor, lets say the resistor dissipate the transferred energy as heat, the potential energy of the electron after passing the resistor must reduce (First law of thermodynamics).

But the question is, what is the value of the potential energy U of the electron passing the resistor?

Here my theory(so to speak)

A to B

When electrons leaves from the negative terminal of A to enter to the positive terminal of B, they have potential energy U of qV. This potential energy must be transferred to the battery B.

Now the potential energy in B is no longer U = qV, but U = qV + Ua where Ua is the potential energy transferred by eletrons entering B.

Since Ua = qV, then the potential energy in B increase to U = qV + qV = 2(qV) = q(2V)

B to C

When electrons leaves from the negative terminal of B to enter to the positive terminal of C, they have potential energy U of q(2V). This potential energy must be transferred to the battery C.

Now the potential energy in C is no longer U = qV, but U = qV + Ub where Ub is the potential energy transferred by eletrons entering C.

Since Ub = q(2V), then the potential energy in C increase to U = qV + q(2V) =q(3V)

C to A

And finally, When electrons leaves from the negative terminal of C to enter to the positive terminal of A, they have potential energy U of q(3V).

This is the potential energy carried by the electrons flowing through the circuit.

As V = U / q, you se that the voltage between C and A is V = q(3V)/q = 3V = 3*1.5 ... the same as 1.5+1.5+1.5

Tell me what you think guys, I'm trying to understand this stuff too ... cheers

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  • \$\begingroup\$ One small problem with your explanation is that you talk about what happens to electrons as they travel around the circuit. But the notion of voltage doesn't require any electrons to move or current to flow. The potential difference across 3 cells is what it is regardless of current flow. \$\endgroup\$ Jun 11 '20 at 13:59
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A definition of potential difference ( what is measured in volts and sometimes called voltage ) may be useful: it is the work the battery does on a unit charge moving it from one terminal to the other. Thus since the same charge is moved twice each time thru the same voltage the total work done is 2 * V. In a parallel connection the charge moves thru one battery or the other so the work done is just V. ( I ignore here arguments about some current thru one battery some thru the other, they do not really matter )

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