resistor, and back to the negative terminal. The multimeter was then switched to a voltage
setting and the potential difference was taken at points A to B, B to A, A to C, C to A, C
to D, and D to C.
Part 2:
1.
A circuit similar to the one from part 1 was set up again, but this time two batteries were
used instead of just one. To do this, another jumper cable was connected between the
negative and positive terminals of two batteries. The multimeter was again placed in
series and the settings turned to become an ammeter. The current was then observed and
recorded in the data sheet.
2.
Similar to the second step of part two, the multimeter was removed from the circuit and
placed in parallel. The settings of the multimeter were then changed to record the voltage
of the circuit. This voltage was then placed in the data sheet.
Part 3:
1.
Just as a second battery was added to the original circuit in part two, a third battery was
added to the circuit in this step. To do this, another jumper cable was placed between the
negative and positive terminals of the second and third battery. The multimeter was then
placed back in series, the settings were changed to record current, and this value was
placed in the table.
2.
Finally, the voltage of the three-battery circuit was observed. To do this, the multimer was
removed from the circuit, placed in parallel to the circuit, and changed to voltammeter
settings. The voltage was them observed and recorded in the data sheet.
Observations:
Part 1:
Resistance of resistor using multimeter: 99 Ω.
Current at point P and Q = 0.02A
Voltage from A to B: 1.59V
Voltage from
B to A: -1.59 V
Voltage from A to C: 0.01V
Voltage from C to A: -0.00V
Voltage from C to D: 1.59V
Voltage from D to C: -1.58V
Calculations:
V=IR
3.15 V = 0.03 A * R
R = 3.15 V / 0.03 A = 105 Ω
(These calculations are shown using data collected from the 3V circuit. Calculations for
the 1.5V and 4.5V circuits are identical to these calculations.)
Graphs:
Discussion/Questions:
Part 1:
When the multimeter was moved from point P to Q, the current was the same (0.02A).
