Battery and Resistor as Source and Load

Quick Summary:

A voltage source, like a battery, is loaded by the circuit connected to it.

The load can be simply represented for quick analysis as a single resistance.

When talking about voltage sources, you should intuitively know these concepts:

-Ideal voltage sources are assumed to have an infinitely low (<0.0000001Ω) internal or "output" resistance.

-A "heavy" load will have a relatively low resistance.

-A "light" load will have a relatively high resistance.

-A real world voltage source has a finite output impedance. For a 9V battery, assume 3Ω when making calculations.

Items needed:

9V battery

1k 1/4W resistor

Alligator clips

Voltmeter

This is a dirt simple experiment to familiarize yourself with the idea of a power supply, or voltage source, and a load.

A power supply is normally chosen to provide a specific voltage. A "9V battery" is expected to have a potential difference of 9 volts between its two terminals. One terminal is labelled "+" for positive, and the other terminal is "-" for negative.

A 9V battery gets its voltage from a chemical reaction inside the battery's casing.

If the terminals are connected directly together (warning: DON'T connect them directly together!), then a large current, or flow of electricity will occur as electrons from the negative terminals get attracted to the positive terminal.

Analogy: Electrical Voltage to Air Pressure

I would compare this 9V battery to a small balloon. Let's say that the 9V of potential energy in the battery is equivalent to an air pressure that makes our small balloon very full and taut.

What is air pressure anyways, and how exactly does a balloon work? Well, the air inside the balloon is denser than the air outside the balloon, so there is a difference in pressure... think of the air inside the balloon as "trapped" and "wanting" to spread out into the open air.

Directly connecting the trapped air with the outside air is simple: poke the balloon with a needle. What happens? POP. A quick and violent equalization of the two different pressures.

Let's go back to the 9V battery. A direct connection from the positive terminal to negative terminal results in similar violent and destructive effects. If a small copper wire, for instance, is directly shorting the 9V terminals, it may quickly start to heat up and even glow. The battery will quickly drain itself. This is similar to how a popped balloon deflates quickly.

So it is not good to put a direct short across power supply terminals because it creates violent, destructive, and not-very-useful energy. Instead, we want to harness that potential energy into a useful circuit. Any power supply and useful circuit can be thought of as a source and load, respectively. The terms source and load and very common electronics terms, and you should know what they mean or general discussions about circuits may "go over your head." It is common to speak of something "loading" something else.

A Practical Experiment: Putting a "Load" on a Battery

This simple experiment examines the idea of a power supply and load in the most basic sense, and illustrates the idea of how a power supply can be unloaded or loaded. It is a very important concept.

The illustrations show a 9V battery with alligator clips on the + and - terminals. If the battery is fresh, a voltmeter (reading dc voltage, not ac!) should read a voltage higher than 9V. This could be 9.3V, 9.5V, even 10V with some "9V" batteries. There is good reason for this, as we will shortly see.

This initial reading is the "unloaded" voltage reading. Technically, there is a load, and it is your meter's input, but more of that in a minute. For now, we will consider the meter to be a "light" load, and not really affecting our measurement.

Next, we "load" our battery with a simple 1k (1000Ω) resistor. Just about any resistor with a near 1k value will do (820Ω to 12k for examples) for our purposes. We are putting in a dummy load. This is a colorful name for a test load that doesn't "do anything" but "load" the source.

In the illustration, I show the "unloaded" battery voltage to be 9.5V, and the 1k "loaded" battery voltage to be 9.0V. This is just to illustrate that the loaded voltage reading should be less than the unloaded voltage reading. With a real battery I found on my bench, it measured 9.40V unloaded, and 9.31V when loaded with a 1.0k resistor. With a 100.0Ω resistor, the measured battery voltage became 9.0V. I should really fix that picture...

The Limits of Loading a Battery

What kind of load was our direct short example previously? A direct short represents a "heavy" load to the battery. If the load draws "a lot" of current, it is a "heavy" load. If the load draws "little" current, then it is a "light" load.

Since we now have a voltage (9V) and some currents to work with ("a lot" and "little"), we can now apply some Ohm's Law to get an intuitive feel for this "load" concept.

Let's pick some common current draws to calculate with. A very common draw in battery operated circuits is 1mA which is "one milliamp" or 0.001A. That's 1/1000th of an amp.

A 60W lightbulb on a 120V circuit will draw about 500mA or 0.5A. That's 1/2 an amp.

We'll use 1A as well, just because 1 is such a nice number for calculations.

Maybe 10A also for variety. Your kitchen appliances and power tools will draw amps of current when in use. Your circuit breaker for home wall outlets will "trip" at 15A unless a higher amperage breaker is installed. (20A and 30A breakers are common as well.)

Now that we have some standard current draws to consider, lets do some Ohm's Law:

Ohm's Law: Voltage divided by current equals resistance.

(9V) / (0.001A) = 9000Ω

What's the point? If our 9V power supply, or power source, is connected to a circuit, and we observe a current draw of 0.001A (1mA), then the circuit no matter how complex is "equivalent" to a 9000Ω (9k) resistor.

It is given, by me telling you, that 1mA is a common battery-powered-circuit current draw, so we are going to consider a 9k load to be a light load for the 9V power source.

Let's try another.

(9V) / (0.5A) = 18Ω

That's a heavy load for our 9V battery. 18Ω is getting closer to a direct connection. Let's think about this intuitively. I said that a common 60W lightbulb would have 500mA of current on 120V. A 60W bulb is usually warm or hot, especially to the touch. Does it make sense that, imagining that the 9V battery could cause such a current to light up a 60W bulb, that this would quickly deplete the 9V's energy?

Going to heavier loads, the 1A calculation is easy:

(9V) / (1A) = 9Ω

So the smaller the resistance the heavier the load.

(9V) / (10A) = 0.9Ω

At 9/10th of an ohm, we are now getting close to that direct short condition, and we can see that a simply application of Ohm's Law makes it appear that the 9V is really working hard (read: heavy load) at such low resistances (like a direct short). But are we really getting the right numbers here? What if we make the load resistance really small?

We're going to rearrange Ohm's Law to solve for current now:

Ohm's Law: Voltage divided by resistance equals current.

(9V) / (0.001Ω) = 9000A

Hm. Something isn't quite right about this answer. Intuitively, I know that 9000A is too much current than a 9V battery could make. In power, that's 9V x 9000A = 81,000W. That's obviously wrong. So what's wrong with our simple calculation?

Source/Output Impedance

As is turns out, our 9V battery is limited by it's own internal resistance, also known as source impedance. This is very important that we take this into consideration when our source drives a heavy load. This same concept can be called output impedance.

For a 9V battery, the chemical reaction makes the internal impedance about 3Ω. This set the minimum load to 3Ω, and this internal impedance gets worse (higher) with age/use of the battery. Ideally, a power supply should have as low of an internal impedance as possible.

So for our equations above, the 10A and 9000A answers were incorrect, because those "equivalent loads" of 0.9Ω and 0.001Ω were less than the minimum of 3Ω.

Additionally, the voltage across the external impedance (the "load") would not be 9V, even if we used a fresh 9V battery. As the load's equivalent resistance approaches the internal resistance of the battery, a voltage division effect occurs. This is the subject of the next experiment.

Schematic Symbols for Battery and Resistor