To understand the effect of the internal resistance of a battery
A computer with the Internet connection, a calculator, paper, and pencil
An ideal voltage source is one that keeps a constant voltage regardless of the current it supplies. This is not the case for a real voltage source such as a battery. As the current demand from a battery increases, the voltage across its terminals drops more and more from its nominal voltage. This is simply because of the battery's internal resistance, r that becomes like a consumer in series with the external resistance (or load).
Suppose your car's battery shows 12.65 volts when it is not in use and you just measure the voltage across its terminals with a voltmeter. If you turn your car's ceiling light on and measure the voltage across the battery again, you may measure something like 12.58 volts. If you turn the headlights on, you may measure 12.2 volts or even under 12volts. The more current a battery has to deliver, the more voltage drop across its terminals occurs. When an external resistance R (load) is connected to a battery of nominal voltage V and internal resistance r , we may write Ohm's law as
Vnom = RI + rI, or Vnom = (R + r) I, or I = Vnom/(R+r)
where I is the current. Note that Vnom is the nominal voltage, that is the voltage measured when no external resistance is connected to the battery. RI or Vab is the voltage across the load R that is also the voltage across the battery when in use. See Fig. 1.
Click on http://www.walter-fendt.de/ph14e/combres.htm . The same applet that was used in experiments 2 and 3 will appear.
Click on the 100Ω-resistor to select it. The click on "Series Connection" to form 2 resistors in series.
Let the left resistor be the "Load" and the right resistor play the role of the internal resistance of the battery. Change the right resistor to 1.00Ω. This is the lowest this applet allows you to choose. Now, you have a battery with a nominal voltage of 12.0V and an internal resistance of 1.00Ω connected to a 100-Ω external load.
Click on the 100Ω load to select it and then click on "Voltage" to place a voltmeter across it. As you see the voltmeter measures a voltage of 11.9V across the load and not 12.0V. If the battery were an ideal one (for which the internal resistance r = 0 ), then the voltmeter would read 12.0V. Here; however, the internal resistance is 1.00Ω and not all the 12.0V of the battery drop across the load. Record the voltage across the load in Table 1.
Click on the "Amperage" to put an ammeter in the circuit for current measurement.
Change the external resistance to 50.0Ω. This is a smaller resistance and allows a greater current. Supplying a greater current means a greater load for the battery. You can see that when the battery has to supply a greater current, the voltage it can apply across the load becomes smaller (11.8V in this case). Record the voltage across the load in Table 1.
For each line of the Table 1, calculate the ideal current, the real current, and the % decrease in the current due to the internal resistance. Record all values in the Table 1. When you calculate the real current, you may compare it with the current that the ammeter reads in the circuit, just as a double-check.
Change the load's resistance to 30.0Ω, 15.0Ω and 5.0Ω respectively, and each time record the voltage across the load in Table 1. Also, calculate the rest of the items in each row including the % decrease in the current.
Repeat the experiment for the other 3 cases (batteries) as given in Table 1.
Voltage Across the
Refer to the above Table.
Comparison of the Results:
Conclusion: To be explained by students.
Discussion: To be explained by students.