1.) Compare the calculated and measured equivalent resistance values:
Figure 1: The pictures above give a visual representation of the circuit configurations accounted for in the table below (Figure 2).
Figure 2: The table above shows the calculated and measured resistance values for each resistor along with the difference percentages between measured and calculated values.
2.).Apply 5V on a 120Ω resistor. Measure the current by putting the multimeter in series and parallel. Why are they different?
- When measuring current in series a correct measurement of 41 mA was achieved.
- When measuring in parallel the resistor was shorted and an incorrect value was measured for the current because when the resistor is shorted, it does not affect the current as it does when measured in series.
3.)Apply 5 V to two resistors (47Ωand 120Ω) that are in series. Compare the measured and calculated values of voltage and current values on each resistor.
Figure 3: The table above displays the calculated and measured current and voltage values for the two resistors in series.
4.)Apply 5 V to two resistors (47Ωand 120Ω) that are in parallel. Compare the measured and calculated values of voltage and current values on each resistor.
Figure 4: The table above displays the calculated and measured current and voltage values for the two resistors in parallel.
5.)Compare the calculated and measured values of the following current and voltage for the circuit below: (breadboard photo)
a. Current on 2 kΩ resistor
- Measured current for the 2k resistor was 1.6mA
- Calculated current for the 2k resistor was 1.9mA
b. Voltage across both 1.2 kΩ resistors.
- Calculated Voltage for 1.2k (A) was 0.80 V.
- Measured Voltage for 1.2k (A) was 0.80 V.
- Calculated Voltage for 1.2k (B) was 0.74 V.
- Measured Voltage for 1.2k (B) was 0.74 V.
|Figure 6: Different angle of the same circuit shown in Figure 7.|
|Figure 7: Picture of the circuits that were built above.|
6.)What would be the equivalent resistance value of the circuit above (between the power supply nodes)?
Based on our calculations, the equivalent resistance value of the circuit above (Figures 5-7) is 2.5 kΩ.
7.)Measure the equivalent resistance with and without the 5 V power supply. Are they different? Why?
- Without the power applied the meter reads 2.56 kΩ. A resistance is given because there is no voltage, in this case the 5V, to disrupt the DMM.
- With the power applied the meter reads OL(Overload). OL occurs because there is voltage from both the DMM and the power supply which interfere with each other.
8.)Explain the operation of a potentiometer by measuring the resistance values between the terminals (there are 3 terminals, so there would be 3 combinations). (video)
Figure 8: In the video above, the operation of the potentiometer is explained.
9.)What would be the minimum and maximum voltage that can be obtained at V1 by changing the knob position of the 5 KΩ pot? Explain.
|Figure 9: Pictured is the circuit relevant to the problem.|
When the potentiometer is set to 0Ω there's no resistance so there will not be a voltage drop. If the potentiometer is set to something other than 0Ω the max voltage achieved will always be 5v.
10.)How are V1 and V2 (voltages are defined with respect to ground) related and how do they change with the position of the knob of the pot? (video)
|Figure 10: Pictured is the circuit being demonstrated in the video.|
Figure 11: In the video, the relationships between V1 and V2 are shown.
11.)For the circuit below, YOU SHOULD NOT turn down the potentiometer all the way down to reach 0 Ω. Why?
|Figure 12: Pictured is the circuit relevant to the problem.|
Because there's no resistance through the potentiometer all the current will flow through it causing it to overload and be damaged.
12.)How are current values of 1 kΩ resistor and 5 KΩ pot related and how do they change with the position of the knob of the pot? (video)
Figure 13: In the video the relationship between the 1K resistor and the potentiometer is displayed with respect to the position of the knob.