The multimeter was then switched to a voltage setting and the potential

The multimeter was then switched to a voltage setting

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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
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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).
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