Wednesday, January 25, 2017

Blog 3 Group 1

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)

R1= 99.6 Ω
R2= 47.2 Ω
R3= 99.6 Ω
R4= 120.5 Ω

A: Measured: 24.9 Ω
     Calculated: 99.6 Ω // 47.2 Ω //99.6 Ω = 32.024//99.6 = 24.2326

B: Measured: 131.0Ω
     Calculated: 99.6Ω + (47.2Ω // 99.6Ω) = 99.6 +32.024 = 131.624

C: Measured: 137.7Ω
     Calculated: 99.6Ω + 47.2Ω //(99.6Ω +120.5Ω)
                        99.6Ω + 47.2Ω//220.1Ω
                        99.6Ω + 38.8654Ω                
Group Measured Rab 
Calculated Rab
A 24.9Ω 24.2326Ω
B 131Ω 131.624Ω
C 137.7Ω 138.465Ω

2.Apply 5V on a 120 Ω resistor. Measure the current by putting the multi-meter in series and parallel. Why are they different?
Parallel: 63.1 mA
Series: 41.8 mA
When measuring in parallel you are basically providing an alternate path of current through the power supply which happens to be less than the equivalent resistance of the circuit. Therefore, the new equivalent resistance of the power supply in parallel with the circuit decreases. Given V=IR if the resistance decreases the voltage is held constant then the current must increase to compensate.

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.
Calculated Values: V=IR
5V/(120.5+47.2Ω)=29.8 mA
29.8 mA*120.5Ω= 3.59 Volts
29.8 mA*47.2Ω=1.407 Volts
Measured Values:
120.5 Ω=3.59 Volts
47.2 Ω=1.39 Volts
Measured Current:
28.85 mA

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 Values:
V120.5=4.97 Volts
V47.2=4.97 Volts
Measured Current:
I120.5=39.41 mA
I47.2=94.6 mA
Calculated Values: V=IR
4.97V/(120.5//47.2)=4.97V/33.91Ω=146.5 mA
I120.5=41.21 mA
I47.2= 105.2 mA
Our values calculated values were relatively close to our measured values. However, there seems to be some significant error in our measurements possibly by using a 4W measurement we could have more accurately measured our resistances.

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, b. Voltage across both 1.2 kΩ resistors. 
a. Current on the 2kΩ resistor:
Calculated: V=I*Req
I=2.20 mA

b. Voltage across both 1.2kΩ resistors:
R2:=>1.984 mA (0.5197 kΩ /0.75454) = 1.383 mA
R2= 1.032 V -1.383 mA(0.1 kΩ) = 0.8133 Volts
V1=.926 Volts
V2=.785 Volts

Figure 1: Shows the breadboard circuit used.

6. What would be the equivalent resistance value of the circuit above (between the power supply nodes)?
The equivalent resistance across the circuit is 2,519.7Ω.

7. Measure the equivalent resistance with and without the 5 V power supply. Are they different? Why? 
Without the 5V connected, you get a value of 2.463kΩ.  When the power supply is attached you receive a value of 1.89kΩ.  The reason you get a difference in values is because the power supply acts like another resistor in parallel with the entire circuit which happens to be less than the resistance of the entire circuit. This brings down the resistance being read by the DMM.

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)

Video 1: Shows the different resistance readings of the different pins of the potentiometer.

Between the left and middle pins on the potentiometer, there was very minimal resistance reading. Then both, the left and middle pins, when put with the right pin came back with a resistance of 10kΩ. It works as a voltage divider.

 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.
Max voltage: 5V
Min voltage: 0V
By changing the position of the knob you are adjusting the amount of resistance that the potentiometer is outputting.  So the max amount of voltage is what is put into it, and if the resistance is turned all the way up then the voltage is being dissipated across the other two pins.

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)
V1: 5.0 volts.  There is no voltage drop since there is nothing to cause any power loss.
V2: 4.19 Volts

Video 2: Shows the potentiometer changing the voltage on the circuit.
The voltage does not change till with the potentiometer until after the resistor.

11. For the circuit below, YOU SHOULD NOT turn down the potentiometer all the way down to reach 0 Ω. Why?
You should not turn it all the way down because the current will become too high through the other branch and burn up the potentiometer.

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).
Through the 1kΩ Resistor we receive a current value of 4.96 mA.  The pot doesn't change the current since it is in parallel.  The 5kΩ pot received a current of .99 mA.

Video 3: The video shows how the pot changes the current values based on how you turn it.

13. Explain what a voltage divider is and how it works based on your experiments.
When two resistors are in series, they divide and change the voltage across the circuit.  With a potentiometer, you can change their resistance, which then would change the voltage drop across each resistor causing a different division of the voltage.

14. Explain what a current divider is and how it works based on your experiments.
A current divider is very similar to a voltage divider.   The difference between the two is that the resistors are in parallel instead of series.  The voltage would be the same, but they would have different currents leaving each resistor.  With the potentiometer, you can adjust the circuit based on current instead of voltage now.


  1. First thing I noticed is that the information is very compacted together. I assume it is still in the works. However put some spaces between, its ver hard to read. Also it seems like you missed the calculated measurement for number 5, R2.
    To comment on number 11, I believe that if you turn it down too low the bigger risk is blowing the entire circuit by creating a short circuit.
    Also I am curious if you had any troubles while trying to measure the current over the potentiometer. For me that simple task took some time and actually the assistance of the TA. It just seems counter intuitive at times, but I am sure with more time it will become second nature.

    1. Thanks Alec for the good comments! I have fixed the things you noticed and I it for pointing those out. In regards to measuring the current over the potentiometer, yes it took us a few minutes and a little help from a friend to get the correct values that were needed.

    2. We got a slightly higher equivalent resistance than you guys did on #6, but this may be because of rounding. Also I'm pretty sure on #5 our breadboard was set up differently. Were any of your theoretical values more off than 10% from the experimental values? In the theoretical values tolerance is not taken into account like it is when you do an experimental value so that is why ours were off but they still fit in the tolerance. Also on number 11 this makes total sense to me from how I learned it in calc 2, dividing by a 0+ Resistance will cause an infinite amount of a current causing a dangerous shortage. Nice job on the blog though I enjoyed reading it and comparing it to my own...

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  2. I noticed that you had no caption on the table in number 1. Other than that your captions were great and did a nice job explaining what is going on. with your color schematics it is easy to read the text and i like the overall template of your blog in general. For number 7 we had a different understanding of why the readings were different. In our opinion it was because the DMM and power supply both supplied a voltage that interfered with each other. We also had a different understanding for number 11. We thought that you shouldn't turn the potentiometer all the way down because it would short the circuit and allow to much current rather than none, furthermore damaging the potentiometer.

    1. Yes, you are completely correct about number 11 Joe it would essentially attempt to use the path of least resistance, in this case being through the potentiometer, hereby overloading the component. However, for number 7 yes we believe that for measuring the equivalent resistance with the power supply attached even off should provide an additional path with internal resistance provided by the closed circuit within the power supply. Hope this makes more sense!

  3. For question 2, how is it possible that you measured a tangible value for in parallel? As per class discussion, when measuring the current, one must break the circuit and place the DMM in series with the circuit. Without this, you would not get a reliable value.

    1. We believe that you are essentially measuring only a small percentage of the current because the DMM actually has an internal resistance and therefore the current has the choice of flowing through the multimeter or the circuit and will do so based off of whichever has less resistance. Based off of the data we accumulated we concluded that the internal resistance of the multimeter must be less than the the resistor, 120 ohm.

  4. You have good design.
    I tried to compere our calculations with yours and that is great to find most ot the questions have mostly same answers. But I noticed that we have different answers in question 1 and 2. You or we may have mistakes in the the calculations and we will try to check ours.
    Good job.

    1. It appears that you are not the only one who disagrees with our data for question number 2. I have provided reasoning in other problems but if there is a mistake it may be possible that we did not use a 120 ohm resistor otherwise I do not know what would have cause us to have such a high current when measuring in parallel.

  5. I think question 11 is a very interesting one. It is remarkable that electricity is so powerful (no pun intended) when left to its own that we usually have to design circuits with resistors so that other components don't get damaged from all the current flowing through. The marginal amount of resistance in wiring is usually not enough.

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  7. Table caption missing and some questions (#2, #5) could have had a few sentences of explanation.
    #2: I loved the discussion with your peers.