 Terminal Voltage of a Lantern Battery Printer Friendly Version
In this lab you will be using resistors, a multimeter, and a circuit board to discover the internal resistance of a 6-V lantern battery.

 Initially, use the voltmeter to record the emf, in volts, of the battery without any electrical loads.

Next, using collections of 100- and 50-ohm resistors, measure and record the voltage lost across 10 different resistance combinations and the current flowing through each one. On your data sheet, neatly draw the circuits for each resistance combination (battery, resistors, voltmeter, ammeter).

Data Table

 resistance voltage current trial (ohms) (volts) (amps)
 1
 2
 3
 4
 5
 6
 7
 8
 9
 10

Analysis

Once your data has been collected, use EXCEL to graph V vs I.

Theoretically, the voltage lost across each combination of resistors represents the terminal voltage of the battery. This voltage can also be calculated with the equation V = ε - Ir where r is the internal resistance of your battery.

Rearranging the equation for terminal voltage, V = ε - Ir, leads to the expression V = -Ir + ε

Consequently, your graph of voltage vs current should have a negative slope whose numerical value represents the internal resistance of the battery while the line's y-axis intercept represents the emf of the battery.

 What is the filename for your EXCEL graph?

 What is the equation of your line?

 What is the percent difference between your measured emf (step 1 above) and the y-axis intercept of your line?

Conclusions

Now we will test your equation with a new combination of resistors. Set up an 11th combination of resistors and measure the voltage across them and the current following through them.

 resistance voltage current trial (ohms) (volts) (amps)
 11

Using the equation, V = ε - Ir, substitute in the internal resistance as the value of the slope of your graph, the emf as the y-axis intercept of your graph, and calculate the voltage that should have theoretically been lost across this final combination of resistors. Give a percent difference between this predicted value and the voltage actually measured.

 The theoretical voltage across this new combination should have been

 The percent difference between these two voltage values is Related Documents