Current in a circuit is 50% lower than predicted by Kirchhoff's law

Question

I am a math teacher and in my Linear Algebra class we were solving systems of linear equations that come from Kirchhoff's laws. We were trying to use a physical circuit to validate our computations. I did not expect the values to match the prediction exactly, but I was surprized that the actual measured current was 50-70% lower than predicted.

Here is what I did.

First, to test my equipment, I made a simple circuit with a 9V battery (when measured, it was actually 8.3V because it was somewhat drained) and a 10 Ohm resistor connected by wires with alligator clamps. The current I measured was exactly 0.83. Even though I am using cheap school-grade tools, they seem to work fine.

Then, I used the following diagram to set up a system of equations and solved it for the currents.

The math was checked and also confirmed by a virtual circuit assembled here.

I then assembled the circuit according to the diagram with wires and alligator clamps and measured the current on all sections. All the values were consistently about 50-70% lower than predicted.

I reassembled the circuit using a breadboard. The measured currents were almost exactly the same as first time.

The resistors are (in the order) 47, 10, 10, 22, 10. I only turned it on for a few seconds each time, because the resistors get very hot very quickly, so it did not drain measurably between tests, definitely not by 50%.

Since I am new to this, I am wondering if this is typical and where is this extra resistance coming from once I moved from a single loop to a more complex circuit.

Answer 1

tl; dr: you have unaccounted-for resistances in the circuit, especially the 9V battery itself. These add to the network and thus reduce the real current you're seeing compared to your ideal model.

Let's start with the battery since this is the primary culprit. 9V types have a fairly high internal resistance, several ohms at least. For their intended application to power low-drain loads (tens of mA), this isn't a problem. With a 10-ish ohm load like yours the voltage will droop quite a bit. Things will get even worse as the battery ages.

Related: Why can't I power everything with a 9 V DC battery?

Knowing your 9V battery's weakness, you can do a couple of things:

1. Account for the droop: Measure the battery voltage under load and use that in your simulation.
2. Reduce the current: Consider scaling up your resistor network values by 10x or 100x to make them more reasonable for the 9V battery.
3. Reduce / eliminate the droop: Replace the battery with a regulated power supply.

For (1), it's best to measure the voltage right at the points where the current enters and exits the resistor network. This eliminates the voltage drop in the wiring.

For (2), your battery droop will be reduced a lot, but you will still have some, see (1).

For (3), an inexpensive power solution is a USB power supply (5V). These hold a steady 5V that's reasonably well regulated. If you have more money to spend, a lab supply with meters would be even better. Might be worth it if this exercise and others like it will be a regular part of your curriculum.

Besides the battery, there's a few more error points in your real vs. virtual components:

• Resistors have a tolerance. This is specified when you buy the resistor. Common tolerances are 5% and 10%.

• Wires have resistance. Figure about 0.25 ohm for a typical 24" meter lead. Two of them will add half an ohm or so.

• Multimeters have shunt resistance. They measure current by inserting a shunt resistor and measuring the voltage drop across it. For a Fluke 77 the shunt is either 0.5 ohm on the 400mA input or 0.037 ohm on the 10A input.

How to calibrate things? To see where you stand on accuracy, start by using your meter to measure the overall network resistance, minus the power leads. Then, apply your power and measure your current + voltage. Use this to calculate your 'real' resistance. The difference will be your unaccounted-for resistances in your battery, leads and meter shunt. Add this resistance to your sim and see what you get.

Somewhat unrelated, that’s a simulator I haven’t seen before. I tried it and, honestly, I found it to be a bit clunky. It wasn't wrong; it did give the right result for the 'ideal' case. But it seems lacking in information. And its library seems very limited.

Have you looked at the Falstad circuit simulator? Like yours, it's web-based, provides interactive visuals, and is easy to use. One really handy thing Falstad does that your sim doesn't: mouse-over any element and it tells you its current, voltage, power and other info.

Speaking of which, here's your circuit in Falstad (simulate it here):

Mouse-over the power source, it gives current. It also shows the resistance it's seeing in its load.

Answer 2

I made a simple circuit with a 9V battery (when measured, it was actually 8.3V because it was somewhat drained) and a 10 Ohm resistor connected by wires with alligator clamps. The current I measured was exactly 0.83.

Did you measure the 8.3 V across the battery before or after connecting the 10 ohm load?

If you did it after connecting the load, then you did not measure that the battery is "somewhat drained". What you measured is that 10 ohms is much too heavy a load for even a brand new 9 V battery to supply without its output sagging substantially.

If you did it before connecting the load, then something is funny about your measurement. We don't expect a 9 V battery to be able to supply such a high current (nearly a full amp) without its output voltage dropping considerably. And if it is depleted, then we expect its output to sag even more under load.

You should also expect the battery voltage to drop quite quickly over time when it is connected to such a heavy load. 830 mA is much to high a load to expect to draw from a 9 V battery.

Try the experiment again using a new battery and 1 kohm resistors and see if you get better results.

Answer 3

You used an ideal voltage source in the simulation. It can provide infinitely high amount of current while providing a constant voltage.

What you assumed that real world batteries are also ideal voltage sources.

They are not, especially 9V block batteries are very poor compared to ideal voltage sources.

The more you load the 9V battery, the less it will have output voltage, due to high internal resistance.

9V batteries are not intended to be loaded with e.g. a resistor as low as 10 ohms, as the battery cannot output 0.9A under normal conditions.

Answer 4

1. When resistors get hot, their resistance usually increases
2. When you measure the current, the current meter itself inserts series resistance
3. A 9V battery has an internal series resistance of about 1 or 2 ohms

I would bet that simulating these 3 effects will match your measured data.

Answer 5

A 10 ohm resistor can draw high current from a 9V battery. It is like 0.9A which is enough to drain a battery quickly. Perhaps while being drained quickly, it is not showing the actual voltage across its terminals. Typically these batteries can deliver 100 to 300mA with some bursts of higher amperes. Try to use resistors in kilo ohm range like 10k or 20k or higher which draw current in the range of milli or micro amperes.

If you want to keep your resistor setup as it is, try using a DC power source which can deliver 1A to 2A at 9V. You may need such power source to conduct experiments like above. Beware that higher amperes like 5A to 10A can burn wires which are not rated for higher power delivery. Check the power rating of the resistors too to see if it is greater than the product of potential across and current in the resistor as the power rating should be higher than the product.

Power rating of resistor > \$V_R * I_R \$

Answer 6

There's nothing wrong with your simulated circuit model, or the theoretical results, however they are not matched by the experimental data for three reasons:

1. the 9V battery is more accurately modelled by a voltage source and a series resistance. The more the current drawn from the battery, the more volts lost in the internal resistance, the lower the terminal voltage. Your circuit draws a significant current.

2. the resistors heat up, which changes their resistance. You are certainly operating them well over the power range they are designed to operate under (typically 1/8th Watt or less).

3. the shunt resistance in the multimeter is going to be 'of the same order as' the resistors in your circuit (low single-digit ohms), so your measurements will significantly alter the current through each leg of the circuit. You will also be heating the inside of the multimeter, and probably are close to blowing the safety fuse inside.

The solution to all of these is to multiply the resistances x1000 ohms, ie (47k, 10k, 10k, 22k, 10k), after which you will find the results much more closely follow the model.