7
\$\begingroup\$

This is a newbie question.

It is clear that if you put two (or more) DC voltage sources in series like this:

schematic

simulate this circuit – Schematic created using CircuitLab

And you measure total voltage between A and D you get the sum of the two voltages. I also realized that if you flip one of the sources, its voltage is to be subtracted from the total. So if I flip V1, I get a total voltage of 7,5 from A to D.

The point is, in short, is that I cannot understand why.

Voltage is the potential difference between two points. As far as I have understood; when you take a single cell into account, that voltage is given by the fact that there is a difference in charge between one pole and the other of the cell. So, when you close a circuit between A and B (or A and C whatsoever being nothing in the way from B and C), then electrical charge searching for its balance finds a way to, and it start flowing from one point to the other, according to the resistance it may encounter. Same goes for Battery #2 of course, between C (or B) and D.

But when you put the batteries in series, what is going on between B and C if you close the circuit between A and D (with eventually any resistance in the middle)? Why is the overall potential difference increased (or decreased, according to the polarity?). How are electrons flowing between all the points? Why?

\$\endgroup\$
1
  • \$\begingroup\$ By the way, the question is interesting to me as well (if not even more) also in case of DC coming from a AC adapter. \$\endgroup\$
    – Dakatine
    Sep 9, 2013 at 20:34

3 Answers 3

Reset to default
2
\$\begingroup\$

For this case you can visualize Voltage not as the amount of electrons flowing but the pressure behind them. In this analogy the electrons are like air, pumped around by the batteries. Note that in this case the batteries are constant-pressure-adding pumps, not constant-flow pumps.

In your case battery 1 pumps electrons from D to C, adding a pressure of 9V. If you wire a lamp from C to D, the battery will force electrons through the lamp with a pressure of 9V.

battery 2 will pump electrons from B to A, adding a pressure (at B) of 9V (relative to A). When we regards things relative to D, the electrons at C and D already had a pressure of 9V, so at A they will have a pressure of 18V (relative to D).

Note that we don't care what the absolute pressure at D (or any other point) is. Electrical Voltage is always relative to some assumed 0 point.

If you wire a lamp between A and D electrons will circle round the loop, from A through the lamp to D, pumped to C, through the wire to B, pumped to A, and back into the lamp.

I talked about electrons here to simplify things, the more accurate term is charge. Charge can also be conveyed by ions, and an electron that flows into a wire might is not likely to be the electron that goes out at the other end.

\$\endgroup\$
5
  • \$\begingroup\$ But what is happening when you flip battery #1 (its voltage is 1.5V in the example by the way)? Voltage is subtracted. So there is energy involved so that the effects of the first pump diminish? Both batteries are exhausting in the process I guess. But I can't see what happens to electrons when they exit C and reach the positive pole of battery one (assuming the flow in this direction as you pictured it) \$\endgroup\$
    – Dakatine
    Sep 8, 2013 at 17:55
  • \$\begingroup\$ A reversed battery is more or less a device that accepts energy, but a normal (non-rechargeable) battery is not very good at this. If it were rechargeable (and the voltage of the first battery sufficiently higher) it would be charged. \$\endgroup\$
    – Wouter van Ooijen
    Sep 8, 2013 at 18:08
  • \$\begingroup\$ When the electrons exit at C they are 'sucked into' B and pumped to A. \$\endgroup\$
    – Wouter van Ooijen
    Sep 8, 2013 at 18:08
  • \$\begingroup\$ I used non rechargeable cells here. So electrons are still pumped through the cell reversed. And this decreases the overall voltage by the voltage that the battery has when you take it into account alone. What is happening into the chemicals and why does this subtract from voltage unfortunately I still cannot understand. \$\endgroup\$
    – Dakatine
    Sep 8, 2013 at 18:16
  • \$\begingroup\$ That's chemistry, not EE. Basically, when you apply a voltage to an electrolytical cell (above its electrochemical potential) it will start dissolving the electrolyt (discharging the ions at the two poles). This chemical process consumes energy. \$\endgroup\$
    – Wouter van Ooijen
    Sep 8, 2013 at 18:44
3
\$\begingroup\$

It's not really a difference in charge, it's a difference in the energy of the charge. An increase in voltage between two points means that the charge (electrons, usually) at the point of higher voltage have more energy than those at a lower voltage. As an electron moves through a battery from the - terminal to the + terminal, a chemical reaction in the battery adds energy to the electrons. For some batteries, you can make electrons flow into the + terminal and exit at the - terminal. In that case, the electrons lose energy when they cause a chemical reaction that recharges the battery. Note that the electrons will try to move to a place where they have lower energy, so current always flows through a resistor from the terminal with high voltage toward the terminal with low voltage and the energy of the electrons is always expended as heat.

You ask "what is going on between B and C?". Nothing. They are electrically the same point in the circuit, and they have the same voltage by definition.

Also, just to be pedantic, there are no electrons flowing in the circuit you drew because there does not exist a complete path for their flow. Although no current is flowing, the charge at different points has different levels of energy.

\$\endgroup\$
8
  • \$\begingroup\$ That's not pedantic, that's perfectly fine (I mean it is pedantic :) but being pedantic is absolutely necessary). I had this clear, but I am going to state it explicitly in the question. \$\endgroup\$
    – Dakatine
    Sep 8, 2013 at 14:03
  • \$\begingroup\$ So, what is happening when the battery is not rechargeable? They still flow through the battery without recharging it. But how voltage is reduced? And when you close the circuit what is going on then? If I buy three cheap multimeters and read current in all couples of points what I would read in a running circuit? \$\endgroup\$
    – Dakatine
    Sep 8, 2013 at 18:24
  • \$\begingroup\$ Forcing a current into a non-rechargeable battery generates heat, and the that't how the voltage (energy) of the charge is lost. If by "close the circuit" you mean connect point A and pont D, then the answer is that you can't connect those two points and have a valid circuit anymore. You would have two ideal voltage sources in parallel but with different voltages which is a nonsensical circuit. For a circuit that comprises a single closed loop the current must be identical at all points in the circuit. \$\endgroup\$
    – Joe Hass
    Sep 8, 2013 at 18:35
  • \$\begingroup\$ It sounds strange to me how the loss of energy (converted into heat) is exactly equal to the voltage of the unit. I will try to light a led with such a circuit as soon as I'll get home. But since I see a difference of potential between the points A and D there's no reason to expect it won't work. I will also check if temperature raises and how. Thanks. \$\endgroup\$
    – Dakatine
    Sep 8, 2013 at 18:40
  • \$\begingroup\$ Adding an LED is not what you asked about...you asked about "closing the circuit" and didn't say anything about an LED. If you put an LED between points A and D then the voltage across the LED will be equal to the voltage between those points (assuming that you have ideal voltage sources). An LED converts some energy to light, which is different than forcing current into a non-rechargeable battery. If you don't think LEDs can get hot try replacing a 60W incandescent bulb with an LED equivalent. \$\endgroup\$
    – Joe Hass
    Sep 8, 2013 at 19:05
0
\$\begingroup\$

This is because you are referencing a potential difference. A battery will have a given potential difference. Assuming we are dealing with small potentials, you will not change the potential difference from one side of a battery to another. If you place the positive side of a battery (anode) at some positive potential the other side of the battery will be the battery's potential difference lower in voltage. It will appear as a voltage drop

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.