How do I prove to my physics teacher that adding a battery in parallel doesn't double the current?

Question

^{simulate this circuit – Schematic created using CircuitLab}

My physics teacher said that the current through the resistor is 4A because each battery has a current of 2A if hooked up to the resistor on its own, and so they both have 2A of current through them so the resistor has 4A total through it because of the junction rule (this was the explanation she gave when I asked her why the total current wasn't 2A), however that isn't true because the current through the resistor is 2A when the voltage is 80 (these batteries are in parallel), and so there is 1A through each battery. How should I explain that her logic doesn't work, as current does not double when you add another battery?

Edit: Her response to me when I asked about ohm's law: each battery provides 2A of current on its own, so they combine because apparently, you can treat each loop separately, so then by the junction rule, the 2A currents join to become 4A.

Answer 1

Just ask her what the voltage across the resistor is

Answer 2

Method 1

^{simulate this circuit – Schematic created using CircuitLab}

Figure 1. A simple practical experiment.

Performing an experiment with the circuit of Figure 1 would demonstrate that parallel voltage sources don't change the current. You should get a reading of 9 mA with either or both batteries in circuit.

Method 2

A thought experiment:

^{simulate this circuit}

Figure 2. The battery box has two batteries and a switch whose position can't be seen.

• What is the terminal voltage where the wires leave the box?
• Does it change if I close the switch?
• What is the expected current for that voltage?

Answer 3

He said that

each battery has a current of 2A if hooked up to the resistor on its own, and so they both have 2A of current through them

Right. Both circuits have 2A through them.

^{simulate this circuit – Schematic created using CircuitLab}

so the resistor has 4A total through it because of the junction rule

But if we combine the above circuits into one, we get this instead of the original circuit.

^{simulate this circuit}

Both resistors having 2A throught them, 4A in total.

Update: Of course you can not just take two independent circuits, wire them together any way you like, and expect that they work the same afterwards. But it won't change a thing if you connect some points that are on the same potential.

Now, a basic question. What is the resultant resistance of the parallel connected R1=40Ω and R2=40Ω resistors?

20Ω, because \$\dfrac{1}{\dfrac{1}{40}+\dfrac{1}{40}}=20\$

therefore the equivalent circuit is rather

^{simulate this circuit}

Answer 4

Others have already abundantly pointed out the teacher's wrong reasoning. I want to mention another part of this where there also seems to be some confusion.

We all understand now that the current thru the resistor is 2 A. However, it is incorrect in the real world to say that each battery therefore supplies 1 A. The total supplied by the two batteries is 2 A, but in practice you can't really assume the batteries are sharing the current equally.

Batteries are quite complex electrically and chemically, and past history matters. In the real world, you can't ever assume two batteries are identical.

To a first approximation, you can think of a battery as a voltage source in series with a resistance. The voltage is what the chemical reaction causes. It is dependent on the exact chemical composition, which varies with time, past history, recent current demand, and temperature.

The series resistance in part models how easily the ions can diffuse thru the electrolyte of the battery, but also includes resistance of the connections, and varies significantly with how depleted the battery is.

Even using just this simple model of the batteries, you actually have this circuit:

Depending on the values of R1 and R2, and the exact internal battery voltages, the current supplied by one battery relative to the other can vary significantly.

However, Ohm's law still holds, and the current thru the resistor will be the voltage across it divided by its resistance.

Answer 5

The error is the misapplication of the Superposition Theorem.

The circuit does not meet the criteria for independent multiple sources. The test is shorting one voltage source to 0V ( which is often done in transformations ) and realize changing the voltage on one must not affect any others ( i.e. true voltage sources 0 ohms) to be independent.

Answer 6

Tell her that brain farts are okay to have. It happens to the best of us.

Just explain that with ohm's law it's \$I=\frac{U}{R}=\frac{80}{40}=2\text{ A}\$.

In order for it to be \$4 \text{ A}\$ then the batteries has to be in series.

More equal voltage sources in parallel = same voltage source = same current. If she can't accept that she farted, then ask her how the circuit is equivalent to the batteries being in series (which it isn't).

^{simulate this circuit – Schematic created using CircuitLab}

Just show her that image. Or send her this link.

Answer 7

Here is how the superposition principle is being misapplied.

When we apply the superposition method, we consider each energy source in the circuit in isolation, while "turning off" the other energy sources. Then we add the results. "Turning off" the other energy sources means reducing them to zero: 0V for voltage sources and 0A for current sources.

Now, (ideal) voltage sources have an impedance of zero. So when they are turned off, they become a short: a piece of ideal wire. Ideal current sources have infinite impedance. When they are turned off and generate 0A current, they are open.

Thus, in a nutshell: voltage sources not being considered are shorted; current sources open.

The teacher's mistake is replacing the excluded power source, a voltage source, with an open circuit: literally yanking it out of the circuit diagram. That is only correct for current sources.

However, when we do the analysis correctly, we instantly run into the problem that the battery we are analyzing is being short-circuited by the one we set to 0V which calls for the flow of infinite current. So what can we do is model the resistance of the wires with some negligible values, like 0.001 \$\Omega\$ so that we are then dealing with a finite (but large) current through those parts of the circuit.

^{simulate this circuit – Schematic created using CircuitLab}

Aha! And so now what happens is that most of the current action is flowing through the R2-R3 voltage divider. The circuit node between R2 and R3 is sitting at almost exactly 40V, and so R1 sees 1A of current.

Of course, the intermediate voltage is very sensitive to the values of R2 and R3 being exactly equal, which isn't realistic. This is not a problem.

Suppose that R2 and R3 are instead 1 and 3 \$\text{m}\Omega\$. Then we have a 1:3 divider, so the voltage at the given node is 60V. But in that case, when we analyze with opposite battery, the divider is reversed and we will get 20V. So we get 0.75A from one analysis and 0.25A from the other: they still superimpose to 1A through R1.

(To model this with greater realism, we have to include internal battery resistance. That is to say, we do not replace the batteries that we are not analyzing with short circuits, but with their internal resistance.)

Why the simplified voltage-divider reasoning applies: it is because the small R2-R3 values swamp the big R1 value. We can draw the analysis circuit like this:

^{simulate this circuit}

When the impedance through a voltage divider is less than around twenty times smaller than its load (1:20 rule), we can pretend that the load is not there when calculating the midpoint voltage. Here the difference is many thousands, by deliberate choice of R2 and R3.

Of course, instead of this short-cut reasoning, we can do the exact analysis whereby the current through R2 is equal to the sum of the currents through R3 and R1, and the midpoint voltage ends up being slightly less than 40V due to the tiny loading effect of R1.

Answer 8

The battery is not supplying current, it supplies voltage

Your teacher is going wrong at this point:

each battery provides 2A of current on its own

An ideal battery does not supply a fixed current, it supplies a voltage. The voltage is fixed. The current is not fixed. The current will be whatever is consumed by the rest of the circuit.

The easy way to explain to her is this: when one battery has to work on its own, it must supply 2A. But when we have two batteries working together, they share the work. And so the batteries only need to supply 1A each in the second case.

She will turn this around on you: how do we know it will be 2A? Because that is what that resistor will draw for that particular voltage. Ohm's Law cannot be cheated.

Answer 9

Your physics teacher is obviously not conversant with even rudimentary electronics, so she may not change her mind by argument alone. But she is a science teacher, and experimental results trump all logical argument.

How practical would it be for you to take in a small-scale demo comprised of 2 x 9V batteries in parallel, a suitable resistor (in my neighborhood, there's a plethora of discarded old electronic circuit boards) and a digital multimeter with a suitable current (mA) scale?

Seriously, if you're going to teach electronics in a physics class, a sprinkling of physical experiments/demos would be a good idea.

Answer 10

The lesson for the teacher is that you can treat each loop separately - but you MUST be careful about using the correct currents and voltages within that loop. If there are multiple voltage or current sources, this is a common source of error amongst students. Unfortunately it also seems to be a source of error for this teacher too.

As the example clearly shows, the current passing through the resistor is (I1 + I2). If you take either loop though, the equation is

80 - (40 * (I1 + I2)) = 0

I1 + I2 = 2

That is the equation according to Kirchoff's Law, and is the only solution according to Kirchoff's Law.

In theory there is nothing which stops one voltage source from delivering 0.1A and the other from delivering 1.9A - that would satisfy Kirchoff's Law perfectly adequately. In practise the voltage sources would deliver half each. But with further thought, in practise there will always be some small difference between the voltage sources, and if the top line is a short circuit then one voltage source will drive infinite current into the other voltage source! (This would lead to discussion of current balancing resistors, if you want to try the experiment for real with batteries and meteres.) However the current through the resistor will always be 2A, and will never be anything other than 2A.

^{simulate this circuit – Schematic created using CircuitLab}

Answer 11

As the others correctly stated, she is mixing up the junction rule and superposition, or voltage and current sources.

Since she already used the junction rule (knows as Kirchhoffs first law [1]), I'd add Kirchhoffs second law [2] to complete the explanation. Simplified, it says that the voltage drops in each closed loop of a circuit has to be equal to the voltage sources. So 40*2=80 in the right and left loop. If the current were indeed 4A then the second law is not satisfied loops (40*4>80, or 0<80 if one would decide to use the voltage drop of the resistor in just one loop).

If that is OK for your setting, you can support that argument with an example. Components for a direct proof (1.5V batteries, a resistor, a small multimeter) should be easy to obtain. You could even use a light bulb ("classic", not LED) to show that the brightness does not increase if you attach more batteries in parallel.

However, I'd not approach her in front of the classroom. She may be stressed out by being confronted in front of many people. Maybe phrasing the whole thing as a question would help: "If the current is 4A, how does this satisfy K's second law?".

Anyhow, I think this is a great example showing that one has to be really careful when and how to divide systems in smaller subsystems. Remember this, it may happen to you as well when things are more complicated (it most certainly happened to me).

References

• [1] https://www.miniphysics.com/kirchhoffs-first-law.html
• [2] https://www.miniphysics.com/kirchhoffs-second-law.html

Answer 12

Apparently your teacher intuitively fails to accept the fact that combining the batteries (in parallel) forces each to halve its output power. The hydraulic analogy may help.

• Each battery is a tank of water.
• The resistor is a narrow pipe (outflow).

Adding an additional battery in parallel is like adding a tank at the same height (as opposed to batteries in series, which is like stacking tanks). Adding a tank at the same height (or equivalently, widening the tank) does not increase the pressure on the pipe. Consequently, the current will not increase.

So if the extra battery does not affect voltage (= pressure) and current, then what is the effect? All it does is double the time it takes to drain the batteries. In other words, the power remains the same, but the total amount of energy doubles.

Another nice analogy is a traffic jam; traffic will not speed up by having more cars join the queue.

Answer 13

This is a poor example of a cicuit analysis problem.

Analytically, this is an under-determined system. Let I1 and I2 be the current from BAT1 and BAT2. From KCL, we have

I1 + I2 = 80/40 = 2

One equation, two unknowns, and an infinite number of solutions.

Superposition can't be used, because it requires that one of the voltage sources be set to zero, as a result, the voltage across the resistor must be 0V and 80V simultaneously.

Answer 14

Apparently, it is better to follow the logic proposed by teacher and find errors. Here, her logic of joining two circuits is perfectly correct but there is small mistake in implementation. She deserves much less disapproval than she's receiving.

In an apparent early modern example of urban legend, the invention of the pet door was attributed to Isaac Newton (1642–1727) in a story (authored anonymously and published in a column of anecdotes in 1893) to the effect that Newton foolishly made a large hole for his adult cat and a small one for her kittens, not realizing the kittens would follow the mother through the large one.

Random Readings: Philosophy and Common Sense

---

If someone is still searching for significance of above citation, for them, I'm trying to point out mistakes are integral part of human neural circuits.

Answer 15

A thought experiment using black boxes. We have two identical black boxes containing each two batteries with 80 V each. In one box only one of the batteries is connected to the terminals, in the other box both batteries are connected in parallel.

You got these two black boxes, a volt meter, a current meter and 40 Ohm resistor. Is it possible to to decide by measurement which box is the one with two parallel batteries?

You may measure the voltage with no load, no difference.

When you measure the current through the resistor you get the theoretical result using Ohm's law for both boxes. In both cases the voltage is 80 V and the resistance 40 Ohm.

You could not measure the short circuit current using the current meter only, there is no proper range and the fuse of the meter will melt if you try it with the first box.

Ask your teacher what measurement to be taken to distinguish the boxes. What should be in a third box to drive a current of 4 A through the resistor? Which voltage is necessary to drive 4 A through 40 Ohm?

Answer 16

Ask her what happens if you have a single battery with two sets of wires, then you cut the battery in half to form two separate batteries. Also, keep in mind that a battery is more than a voltage source; it has a small series resistance. You know enough to calculate everything if the series resistance is, for example, 0.01 Ohm. ( Calculate to 8 decimal places) We engineers would love to get batteries like the one in your problem with zero internal resistance!

Another idea that will help you solve this kind of problem is replacement of a voltage source in series with a resistor with a current source in parallel with that same resistor. Current sources in parallel add, just like voltage sources in series. To learn more, google “Thevenin-Norton”.

Answer 17

There are voltage sources and current sources.

Voltage sources supply constant voltage.

Current sources supply constant current.

Current sources constantly sense the amount of current they provide and adjust their voltage output (to meet the setup value) which according to Ohm's law will affect the current.

You can't "pump" current with constant voltage. It's fundamental!

If you want to prove her that she is wrong then go with a multimeter, 2 batteries, 1 resistor and a breadboard and ask her to prove current doubling. But probably she won't know how a multimeter works so it's a waste of time...

Power grid can supply thousands and thousands amps, does your device

Answer 18

Apply Faraday's law around any loop containing the resistance and the battery and since there is no changing magnetic fields in the circuit , the loop integral must be zero around any closed loop.after doing that , tell her that you have used one of Maxwell's equation's which can never be wrong and if she protested , tell her that this is violation of faraday's law which runs our economy.

Answer 19

She is wrong and right at same time.

Statement 1. If batteries are connected in parallel Current get high (2+2).

Statement 2. If batteries are in series, Voltage get high (80+80).

Ohm's law "(IF THE CURRENT IS 2A)" I = V/R. Which says "(Taking the given values)" I = 80/40 = 2A current.

If you take your teacher opinion (4A) and try again with V = IR.

IT IS NOT GIVEN ANY WHERE (below). I HAVE ASSUMED OHM's LAW for VOLTAGE V = 4*40 = 160V (HERE VOLTAGE HAS BEEN CHANGED DUE TO MORE CURRENT).

She is right because Batteries are in parallel. Value of Voltage or resistance need to be correct.