Week 3: 25-31 January, 2016

  1.)  Compare the calculated and measured equivalent resistance values between the nodes A and B  for three circuit configurations given below.

Experiment 1 Schematics



The Table above shows the measured resistance values
from Point A to Point B for circuits a b and c. 

All measurements were within 1 Ohm of the values
calculated using the formulas above.  



  2.)  Apply 5V on a 120 Ω resistor. Measure the current by putting the Multimeter in series and parallel. Why are they different?

     
    If you put the Multimeter in series like you are supposed to, you will get the current reading that is going through the resistor. If you put the Multimeter in parallel, the current will just go through the Multimeter, skipping the resistor, essentially shorting the circuit.

  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.

       

    By applying 5 Volts to both the a 47 and 120 Ohm resistor connected in series, we were able to collect the following voltage and current Data. 



Measured and Calculated Voltage and Current across
a 47 and 120 Ohm resistor connected in Series 

All Measured Voltages were within 1% of the calculated values

The Measured current was within 0.7 mA of the calculated value.



  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.

Experiment 4 Schematic

The Voltage Measurement across the circuit above were as follows

Voltage across the 120 Ω resistor:  4.98 V 

Current through the 120 Ω resistor:  39.8 mA

Voltage across the 120 Ω resistor:  4.97 V 

Current through the 120 Ω resistor:  94.8 mA

These readings are consistant with the expected values since voltage is constant in a parallel circuit. The voltage measured at any point in the circuit should be equal to the 5 V applied by the power supply.

According to Ohm's Law Current = Voltage/Resistance, as such

5 V divided 120 Ohms = 41.7 mA  which is very close to the measured 39.8 mA 

and

5V divided by 47 Ohms = 106.4 mA  which is also close the measured 94.8 mA 

The variations between the calculated and measured values can most likely be attributed to a higher than expected resistances caused by using multiple sets of alligator clips and meter leads to construct and measure the circuit. 

  5.)  Compare the calculated and measured values of the following current and voltage for the circuit below:

a.       Current on 2 kΩ resistor, 

b.      Voltage across both 1.2 kΩ resistors. 

Photo of the circuit as laid out on our breadboard

Alternate view of the circuit as laid out on our breadboard



Experiment 5 Schematic with labelled Mesh Loops



Theoretical Voltages and Current Readings calculated using Mesh Analysis on the TI-Nspire CAS student software.

A.)  The current measured across the 2 kΩ resistor was 1.83 mA  Which closely matches the calculated I1 current of 2.05 mA.

B.) The Voltages were as follows 

(R1 refers to the 1.2 kΩ resistor drawn horizontally in the schematic above)

(R2 refers to the 1.2 kΩ resistor drawn vertically in the schematic above)

Measured Voltage Across R1:  1.149 V      

Measured Voltage Across R2:  1.045 V

Voltage across both 1.2 kΩ resistor was somewhat higher than the 0.88 V and 0.75 V calculated above.

  6.)  What would be the equivalent resistance value of the circuit above (between the power supply nodes)?

    The Equivalent Resistance for the circuit is equal to the applied voltage over the calculated resistance.  Since we are reading at the power supply we can use 5V as the Test voltage and 2.056 mA as calculated above for the I1 current. Using these values the equivalent resistance is calculated as 2.43 kΩ by dividing 5V by .002056A.

  7.)  Measure the equivalent resistance with and without the 5 V power supply. Are they different?  Why?

    Without Power resistance was measured at 2.73 kΩ which is close to the 2.43 kΩ calculated in question 6.

    With power applied the meter read "OL", this is because the current flowing through the circuit was overloading the Multimeter rendering it unable to take an accurate reading.

  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)

This video shows us using the potentiometer, and demonstrating how you can use it to put out a

constant 10KΩ, or using the first and second, or second and third pin to put out a variable resistance. 

     The resistance of the Potentiometer is measured as follow

    

    - T-1  to T-3 will always read the Potentiometer's maximum

       resistance (in this case 10 kΩ)

    - T-1 to T2 will vary from 0 to the Potentiometer's max value

       based on the knob's position
    - T-2 to T-3 will also range from 0 to the Potentiometer's max value
       again based on the knob's position.

-   * Note:  (T1 to T2) + (T2 to T3) = Potentiometer's Max value

-

  9.)  What would be the minimum and maximum voltage that can be obtained at V1 by changing the knob position of the 10 KΩ pot? Explain.

    If you have the potentiometer set to 0 Ω , there will be no voltage drop across it, because there would be no resistance. However, as long as the potentiometer is at anything other than zero, the voltage drop will always be 5 volts, with a varying amount current.

Experiment 9 Schematic

  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)

Experiment 10 Schematic

A short Video showing how the potentiometer setting

affects voltage at V1 and V2

  11.)   For the circuit below, YOU SHOULD NOT turn down the potentiometer all the way down to reach 0 Ω. Why?



Experiment 11 Schematic

      If you turn the potentiometer all the way down to zero, turning it's resistance to zero, you would short the circuit, and all of the current would then go through the potentiometer since there is no resistance. We unfortunately did exactly this our first time doing this circuit, and fried the potentiometer.


    *  Question 11 was not added to this week's blog requirements until after we discovered the our potentiometer's inability to handle the circuit's maximum current.  

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).

Short Video showing the effects of the potentiometer setting on both the 1 kΩ resistor and 5 KΩ potentiometer's current readings. 

    As the video above demonstrates the potentiometer has no effect on the current running through the 1 kΩ resistor, this is because current is a function of Voltage and Resistance, and since voltage is constant throughout a parallel circuit and the resistor's value is fixed its current remains the same regardless of what value the potentiometer is set at.  The only exception to this rule would be if the potentiometer was set to 0 resistance in which case all current would bypass the 1 kΩ resistor and it would draw zero current.  This would be a bad idea however as this short circuit condition would damage the potentiometer as we discussed in question 11.  What the potentiometer setting does effect is the amount of current going through the potentiometer itself.  As resistance increases the amount of current decreases  and as resistance decreases the amount of current passing through the potentiometer will increase until a value of 0 Ω is reached at which point maximum current will pass through the potentiometer.



  13.)   Explain what a voltage divider is and how it works based on your experiments.

Voltage divider is the process of how voltage splits across different resistors in a series circuit as demonstrated above (See # 10).  For that circuit a 5 volt drop was observed across the two resistors. However, when the potentiometer was set to zero, the 1 kΩ resistor would drop the entire 5 volts, and when the potentiometer was to 10 kΩ, the 1 kΩ resistor about only about 0.45 volts  and the remaining 4.55 volts would drop across the potentiometer. 

You can think that the total voltage as being divided into 11 parts (since there was 11 kΩ of  total resistance), since 10 of the 11kΩ of resistance was provided by the potentiometer it receives 10 parts of the voltage,leaving only 1 part for the 1 kΩ resistor. 

  14.)   Explain what a current divider is and how it works based on your experiments.



    Current divider is the similar as voltage divider, but deals with current in a parallel circuit. This process can be seen in the circuit from #11. When two resistors are in parallel, the current wants to take the path of least resistance. In the circuit from #11, the current has two possible paths, the left loop through the potentiometer, or the right loop through the  1 kΩ resistor.  When the potentiometer was set to 10K, most of the current (about 90%)bypassed the potentiometer and travelled through the 1 kΩ resistor.  As the potentiometer's resistance decreased more current travelled through the potentiometer and less travelled through the 1 kΩ resistor 
    
    Similarly to Voltage Divider you can think that the current  as being divided into 11 parts (since there was 11 kΩ of  total resistance), however; when using the current divider rule as resistance increases current decreases so when the potentiometer is providing 10 parts of the total resistance it only receives 1 part of the total current, leaving 10 parts for the 1 kΩ resistor.

 
    Post contributors (C Ruby & K Pawlak)