**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

138.465Ω

Group | Measured R_{ab } |
Calculated R_{ab} |

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:

V

_{120.5}=4.97 Volts

V

_{47.2}=4.97 Volts

Measured Current:

I

_{120.5}=39.41 mA

I

_{47.2}=94.6 mA

Calculated Values: V=IR

120.5Ω//47.2Ω=33.9Ω

4.97V/(120.5//47.2)=4.97V/33.91Ω=146.5 mA

I

_{120.5}=

_{41.21 mA}

I

_{47.2}= 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*R

_{eq}

5V=I*2.5197kΩ

I=5V/2.5197kΩ=1.984mA

Measured:

I=2.20 mA

b. Voltage across both 1.2kΩ resistors:

Calculated:

R1=(1.032V/1.3kΩ)*1.2kΩ=.9526Volts

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

Measured:

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.**

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.