Week 3
1. Compare the calculated and measured equivalent resistance values between the nodes A and B for three circuit configurations given below. Choose your own resistors. (Table)
Circuit
|
A
|
B
|
C
|
Measured
|
83.5 ohms
|
4.76k Ohms
|
4.99k ohms
|
Calculated
|
84 ohms
|
4.75k ohms
|
5k ohms
|
2. Apply 5V on a 120 Ω resistor. Measure the current by putting the multimeter in series and parallel. Why are they different?
~~When in series you can read the current, because the resistance isn't being bypassed.
~~When in parallel you can't read the current, because the resistance is being bypassed.
3. Apply 5 V to two resistors (47 Ω and 120 Ω) that are in series. Compare the measured and calculated values of voltage and current values on each resistor.
Measured Voltage
|
Measured Resistance
|
Measured Current
|
Calculated Current
|
Calculated Voltage
|
3.61V |
120 ohms
|
29.94 mA | 3.59 V | |
1.39V |
47 ohms
|
29.94 mA | 1.41 V |
Table: showing the values we obtained from the circuit in SERIES
4. Apply 5 V to two resistors (47 Ω and 120 Ω) that are in parallel. Compare the measured and calculated values of voltage and current values on each resistor.
Measured Voltage
|
Measured Resistance
|
Measured Current
|
Calculated Current
|
Calculated Voltage
|
4.78V |
120 ohms
|
41.67 mA | 5.0004 V | |
4.78V |
47 ohms
|
148.05 mA | 4.9975 V |
Table: showing the values we obtained from the circuit in PARALLEL
5. Compare the calculated and measured values of the following current and voltage for the circuit below: (breadboard photo)
Calculated Current
|
Measured Current
|
1.98mA
|
2.03 mA
|
Resisters
|
Measured Voltage
|
|
Picture of the circuit with 5v from power supply |
Same circuit as previously pictured, just a different angle |
6. What would be the equivalent resistance value of the circuit above (between the power supply nodes)?
~~The equivalent resistance value is calculated by taking the 5V and divding it by the calculated current which is 1.98mA. For us the equation will be 5V/0.00198A = 2525.25 ohms or 2.52525 K ohms.
7. Measure the equivalent resistance with and without the 5 V power supply. Are they different? Why?
~~The equivalent resistance with the power supply can't be found, because the multi-meter will display "OL" which means overload. This happens because the multi-meter has its own voltage and so does the power supply. These 2 voltages will trick the multi-meter and it won't be able to find the equivalent resistance.
~~The equivalent resistance without the power supply can be found, and that's because only a single voltage coming from the multi-meter is effecting it. The equivalent resistance from the multi-meter is 2.6 kohms, which is accurate based on previously calculated equivalent resistance (see question 6).
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)
~~When the potentiometer is turned to 0 ohms there will be no voltage because there is no resistance, but if the potentiometer is turned to any ohms greater than 0 then the voltage would be 5V, with a differing current.
Video demonstration of how the potentiometer works
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.
~~When the knob is turned to as low as possible the minimum value will be close or at 0V. When the knob is turned to its highest position the maximum value will be at or around 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)
Video demonstration on finding the voltage in a circuit with a resistor and a potentiometer
11. For the circuit below, YOU SHOULD NOT turn down the potentiometer all the way down to reach 0 Ω. Why?
~~If you were to turn the potentiometer all the way down to 0 you would short your circuit, because there would be no resistance in your circuit.
12. For the circuit above, 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).
Video demonstration of measuring current over a resistor and potentiometer
13. Explain what a voltage divider is and how it works based on your experiments.
~~A Voltage Divider is a simple circuit that will take a large voltage and turn it to a smaller voltage. In this experiment the potentiometer itself is a voltage divider.
14. Explain what a current divider is and how it works based on your experiments.
~~A Current Divider is a linear circuit that produces a current output (Ix) that is a fraction of its input current (It). It works for our experiment, because of the resister value. When the potentiometer's resistance decreases the current will go up.
~~A Current Divider is a linear circuit that produces a current output (Ix) that is a fraction of its input current (It). It works for our experiment, because of the resister value. When the potentiometer's resistance decreases the current will go up.