The metals in the screw and the coin react with the acid in the lemon to create a flow of electrons. Lemons, in reality, make for poor batteries. One reason is that the zinc continues to react with the lemon without a circuit present.

## Pipe Pressure Drop Calculations

This means that the battery would only have a shelf life of a few hours. Another factor is the internal resistance of the lemon battery. We will discuss internal resistance in the next section and why it is important for batteries. In addition to these materials, you will need a way to measure voltage.

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A multimeter will offer the best accuracy, but you can also build your own voltmeter from parts found in the SparkFun Inventor's Kit. In addition, you will need a hobby knife to cut a slit into the lemon. When designing a circuit with a battery, we often assume that the battery is an ideal voltage source. This means that no matter how much or little load we attach to the battery, the voltage at the source's terminals will always stay the same. If we model this battery as an ideal voltage source, changing the value of R L does not affect the voltage between the battery's terminals.

In reality, several factors can limit a battery's ability to act as an ideal voltage source. Battery size, chemical properties, age, and temperature all affect the amount of current a battery is able to source.

As a result, we can create a better model of a battery with an ideal voltage source and a resistor in series. Batteries can be modeled as an ideal voltage source with a series resistor labeled R I.

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We can measure the voltage of a battery across its terminals without any load connected. This is known as the open-circuit voltage V OC. Measuring the voltage of a AA alkaline cell with no load attached. Note that because no current is flowing across the internal resistor, the voltage drop across it is 0 V. Therefore, we can assume that V OC is equal to the voltage of the ideal voltage source in the battery. This drop in voltage is caused by the internal resistance of the battery.

We can calculate the internal resistance if we take readings of the open-circuit voltage and the voltage across the battery's terminals with a load attached. Here is our circuit. We want to calculate R I. We can plug in the loaded voltage we measured V L and the value of the resistor R L into Ohm's Law to get the current flowing through the circuit I. We also need to get the voltage across the internal resistor.

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## Physics Laboratory 6

We can do that using Kirchhoff's Voltage Law. Simplified for this circuit, we can say that the voltage drop across both resistors must add up to the voltage of the ideal voltage source.

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Now that we know the voltage drop across the internal resistor and the current through it, we can use Ohm's Law again to find its resistance. From this, we can see that the internal resistance at this moment of the AA cell is 0. Having a hard time seeing the circuit? Click on the Fritzing diagram to see a bigger image. Once you are done, you should have 2 wires hanging out from the side of your breadboard.

These will be your probes for testing voltage across terminals. This will act as our load for the first test. Then, the equation becomes:. We will set up a circuit that contains the battery we want to study in series with a resistor. It doesn't matter if it is a different resistor each time or more resistors in series or parallel. What matters is that the overall resistance of the circuit changes so that the current is different each time.

Plot your data on a set of axes similar to this example. The blue crosses represent the measured data points, the gray, dashed line is the drawn straight line through the data points. The best fit line you draw doesn't need to go through all the data points, it should, in general, have as many points above and below the line. The intercept with the horizontal axis would give you the maximum possible current the battery could deliver. Practise now to improve your marks You can do it! Sign up to improve your marks. Determine how to approach the problem It is an internal resistance problem.

Within the first few microseconds no current flow is seen. After a few milliseconds the current starts to flow and will rapidly increase to a peak before falling back and eventually reducing to a value somewhere close to the nominal published value. The current will continue to fall and in some cases, may flow for several hours before reaching zero. Actual short circuit testing is interesting but they do not necessarily give repeatable value that can be used for comparison purposes.

Circuit resistance, temperature and actual state of charge all contribute to variable results. The point where the current should be measure has the largest variation. Measurements taken after 1 second will show a large variation when compared with those taken after 1 minute. Because the tests are based on the same datum points, comparisons to other products can be made with good accuracy.

Typically, one discharge will be at a moderately high current and after a pre-determined discharge time the actual voltage and actual current will be recorded. The pre-determined discharge time will be quite short to minimize the Ah capacity removed but will be sufficiently long for reasonably stable results to be obtained.

The discharge will be terminated and the product will be stood on open circuit and allowed to recover for a few minutes before a second discharge is carried out. This second discharge will be at a higher current of typically three times the value of the first current. Again, after a pre-determined time the actual voltage and actual current will be recorded and the test completed.

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The short circuit current is then calculated by ohms law where 2. Some manufacturers may use the typical open circuit voltage of the cell which will largely be dependent upon the specific gravity of the cell.

Values between 2.