Voltage distribution with two batteries [closed]

Question

I have a question regarding a simple circuit of two batteries (9V) in series with a single resistor connected to the outer contacts to prevent a short. When I'm simulating this circuit, a current also flows between the two batteries where "almost no resistance" (despite the tiny wire resistance) is present - the inner contacts. So when there's a current there should also be a voltage drop directly between the two contacts of the batteries. And therefore there's also a voltage over the wire connection U = R*I. Am I right? Hence the voltage provided by the supply of the two batteries (9V + 9V) should be increased/decrease by this voltage?

The motivation is to finally understand the problem of connecting two batteries in series and the question why there's no flow without the outer connection - no closed circuit, that I'm sure about. But I'm still struggling to understand how to the two potentials of the batteries are working together, as these are different energy systems and why people say that the potential of the inner connection is referred to GND (so no voltage dropped on the contrary to my upper assumption).

Answer 1

A common 9 volt battery is actually six 1.5 volt cells connected in series.

We normally assume that the wire between the two batteries is short and has no significant resistance, so there will be no voltage drop across the wire. If the wire is long or has significant resistance, the voltage developed across it will subtract from the total voltage of the batteries.

If we want an 18 volt power supply, we could call the negative terminal of the "lower" battery Ground/Zero volts.

If we want a bipolar supply, +9 volts and -9 volts, as may be required for op-amps, then we would call the connection between the two 9 volt batteries Ground/Zero volts.

In most electronics "Ground" is just the part of the circuit we will consider as "Zero Volts". It has no special properties - it is just the place we connect our black meter probe when measuring voltages elsewhere in the circuit.

Answer 2

Short answer

And therefore there's also a voltage over the wire connection U = R*I. Am I right? Hence the voltage provided by the supply of the two batteries (9V + 9V) should be increased/decrease by this voltage?

Let's consider this section of the circuit. It consists of three elements in series - two 9 V voltage sources (batteries) and a resistor (wire resistance) between them. Source voltages are gain; therefore they are added to the total voltage of the section. The voltage drop across the resistor is a loss; therefore it is subtracted from the total voltage of the section. So the total voltage is less than the voltage that would be with just two batteries in series (no resistor).

So the resistor "steals" voltage I.R from the source thus decreasing the total voltage. However, there are electronic circuits that add I.R voltage to the source, i.e. they behave like a "negative" resistor. For this purpose, they produce the same voltage but with an opposite polarity; that is why they are called "voltage-inversion negative impedance converters" (VNIC). If your internal resistor was such a negative resistor, the total voltage would be higher than the voltage across only two batteries in series.

CircuitLab experiments

I suggest you get an intuitive idea of these basic circuit concepts not through boring textbook explanations but through a series of simple step-by-step CircuitLab experiments shaped into a sort of "circuit comics" :-)

For the purposes of this conceptual explanation, I have slightly changed the quantity values to more convenient ones (e.g., 9 V to 10 V). I have also represented some resistors by measuring instruments having the same resistance (CircuitLab allows to change their internal resistance).

One voltage source

Electrical phenomena are invisible to our senses; that is why we understand them through similar but visible life phenomena (analogies). Perhaps the most famous of these is the fluid analogy. Thus, we can imagine a voltage source as some kind of "pump" that creates a constant "pressure" (voltage).

Perfect voltage source unloaded: When there is no load connected, it performs its task without any effort.

^{simulate this circuit – Schematic created using CircuitLab}

Perfect voltage source "loaded": So if we connect a (perfect) voltmeter to the voltage source, it will show its voltage even though no current flows.

^{simulate this circuit}

Imperfect voltage source "loaded": If we add a resistor R in series, the Vopen voltage remains equal to the input because no current flows, and no voltage is lost across the resistor.

^{simulate this circuit}

Imperfect voltage source shorted: We can consider the circuit of two elements (V and R) as an imperfect voltage source. When it is shorted, a current begins flowing. To measure this current, we have to insert an ammeter. We can simplify the circuit if combain the resistor and ammeter into one. To do it, we can remove the resistor and set the same resistance of the ammeter; hence the name R1k. It means "an imperfect ammeter with internal resistance of 1 kΩ".

^{simulate this circuit}

Note that although both the upper source and resistor terminals are positive, the current exits the source and enters the resistor. That is why the source current is negative and the resistor current is positive.

Two voltage sources

It is time to add another voltage source in series to the first one (OP's configuration). Think of it as another "pump" that adds its "pressure" to the existing. Now we have more choices about which circuit point to use as reference; let's play them out.

V2's negative terminal grounded: Let's start with the most traditional place - the V2 negative terminal. As you can see, the total voltage is 20 V.

^{simulate this circuit}

V1's positive terminal grounded: But nothing prevents us from choosing the V1 positive terminal as ground. Now everything is turned upside down, and the total voltage is negative (-20 V).

^{simulate this circuit}

Middle point grounded: It becomes most interesting when we choose the middle point between the two sources for ground. Then we get two identical (10 V) but with different polarity voltages. This is the so-called “split supply”.

^{simulate this circuit}

Middle resistance inserted

Finally, let's insert some low (e.g. 100 Ω) resistance between the "inner" source terminals. Because we need to know what the current is, let’s use the trick above by using a “bad” 100 Ω ammeter instead of a resistor, and replace the R1k resistor by a “bad” voltmeter with 1 kΩ internal resistance.

R1k loaded: So the two voltage sources are loaded with two resistors (R1k and R100) in series.

^{simulate this circuit}

Shorted: The protective resistor R1k is no longer so necessary and can be removed to simplify the circuit.

Transformed: We can draw the schematic in yet another way to show more clearly how the two voltage sources are loaded by the R100 resistor.

^{simulate this circuit}