Capacitor Function

Part I: Basic Capacitor Functions

Step 1: Open Circuit File

Open the circuit file Week_1_Class_Activity_1_Cap_updated utilizing MultiSim.

Step 2: Energize Circuit

Click the switch to turn on the circuit.

Step 3: Operation and Observation Procedure

Accomplish the following procedure to observe the operation of the circuit and to assist in the answering of questions at the end of Part I.

1. Press the space key to close the switch and observe the effect on the capacitor voltage.

2. Press the space key again to open the switch.

3. Press the A key to close the switch and observe the effect on the capacitor voltage.

4. Press the A key to open the switch.

Step 4: Answer Questions

1. When a voltage is applied across a capacitor (J2 closed), what takes place?

a. The voltage was lowered to .012mv

2. Once a voltage source has been placed across a capacitor, what happens if the voltage source is then removed (J2 opened)?

a. The voltage value remains constant at its current value which in this case was .

012 mv

3. How long will the capacitor remain this way?

a. The capacitor will remain at this value

4. If a capacitor is “charged” and a conductive path is placed from the side of the capacitor to the other what happens to the charge (J1 closed)?

a. The charge will be released if there is a charged capacitor placed upon a conductive path.

Part II: Series and Parallel Capacitors

Step 1: Calculation

Calculate the theoretical values of total capacitance for the following sets of capacitors in both series and parallel configurations:

a. 100 pF, 220 pF

Series

Ct = 1/ 1/C1 + 1/C2 = 1/ 1/100pF + 1/220pF = 1/0.01+0.00454545 = 68.75pF

Parallel

Ct=C1+C2 = 100pF + 220pF = 320pF

b. 100 nF, 220 nF, 470 nF,

Series

Ct= 1/ 1/100nF + 1/220nF + 470 nF = 1/.01+.00454545+.00212766 = 59.98nF

Parallel

Ct=100nF+220nF+470nF = 790nF

c. 10 uF, 1 uF, 4.7 uF

Series

Ct=1/ 1/10uF + 1/1uF + 1/4.7uF = 1/.1+1+.21276596 = 1.31uF

Parallel

Ct=10uF+1uF+4.7uF = 15.7uF

d. 100 pF, 22 nF, 4.7 uF

Series

Ct = 1/ 1/100pF + 1/22nF + 1/4.7uF = 1/.01+.04545455+.21276596 = 3.73nF Parallel

Ct=100pF+22nF+4.7uF = .

000000000000100+.00000000022+.00000047=.0000004702201 = 470.22 nF

Step 2: Circuit Construction

Arrange the capacitors in the previous step in series and parallel configurations in MultiSim and measure the total capacitance in each configuration.

Step 3: Answer Questions

1. What is the major difference between series and parallel configurations in regard to total capacitance?

a. Parallel configurations allow more charge to be stored than with series configurations.

2. Were there any differences observed between the theoretical and measured values?

3. What is a valid explanation for any differences observed? 

Part III: RC Time Constants

Step 1: Circuit Construction

Construct the circuit in Figure 1 with a 1k ohm resistor and a 1 uF capacitor utilizing a 1 Vp square wave at 50 Hz as the source, in MultiSim.

Figure 1: RC Time Constant Circuit

Step 2: Calculation

Substitute the following RC combinations in the circuit and calculate the theoretical values of the time constants. Measure the actual values from the circuit and note any differences between theoretical and measured values. The time taken for the waveform to go from zero to 63% of max value should be equal to one time constant.

1 k ohm, 1 uF t = RC = (1kohm)(1uF) = 1ms

2 k ohm, 1 uF t = RC = (2k ohms)(1uF) = 2ms

1 k ohm, 4.7 uF t = RC = (1k ohms)(4.7uF) = 4.7ms Step 3: Observation

Adjust the frequency of the source to 100 Hz and connect channel A of the oscilloscope to the source and channel B to measure capacitor voltage and measure the actual values from the circuit for each configuration.

1 k ohm, 1 uF

2 k ohm, 1 uF

1 k ohm, 4.7 uF

Step 4: Answer Questions

1. Were there any differences observed between the theoretical and measured values?

a. I didn’t see any differences between the values both were correct 2. What is a valid explanation for any differences observed?

a. If any differences had appeared I would have guesses that there would be either been a miscalculation or there was a faulty piece of equipment.

3. Do the capacitor charge and discharge waveforms appear as expected?

a. Yes the capacitor charge and discharge waveforms appear to be as one.

4. Do the observed waveforms correspond to the theoretical waveforms?

a. Yes the theoretical waveforms and the observed waveform appear to be going together.

5. What happens if the capacitor is not allowed to fully charge or discharge before the waveform changes by increasing the frequency of the source even higher?

a. I think the values of the waveform would change including the time constant and appear the oscilloscope would be slightly different than those shown.

Inductor Properties and Functions

Part I: Basic Inductor Functions

Step 1: Open Circuit File

Open the circuit file Class Activity 1 Ind utilizing MultiSim.

Step 2: Energize Circuit

Click the switch to turn on the circuit.

Step 3: Operation and Observation Procedure

Accomplish the following procedure to observe the operation of the circuit and to assist in the answering of questions at the end of the Part I.

1. Press the space key to energize the inductor and observe the effect for 30 seconds.

2. Press the space key to open the circuit.

3. Triple the supply voltage and repeat Steps 1 and 2.

Return the supply voltage to the original setting, double the inductance, and repeat Steps 1 and 2.

Step 4: Answer Questions

5. What happens to the inductor current as time passes?

a. The voltage is dropping at a steady pace. Current is slowly increasing. Voltage is now in the negative.

6. What happened to the inductor current when the supply voltage was tripled?

a. Current is increasing by a 1/3 at a rapid pace. When I switched to volts it was already showing in the negative so I stopped the simulation and restarted it in measuring volts. It stayed at 36 volts when I applied power. I switched to current and then back to volts and it was at a negative one and declining at a small percent.

7. What happened to the inductor current when the inductor value was doubled?

a. The current was displaying 779.895MA and was climbing at a slow steady pace. The voltage instantly showed a negative 30 number and started declining rapidly.

8. What would be the current if the inductor were allowed to energize indefinitely?

a. The inductor in the sim has been running for about 5 minutes now which allowed me to search online for this answer. I did not find a suitable explanation but I am sure the inductor will stop charging at some point and become full. It will hold the charge until voltage is applied forcing the plates to release the charge.

Part II: Basic Inductor Functions

Step 1: Open Circuit File

Open the circuit file Class Activity 2 Ind utilizing MultiSim.

Step 2: Energize Circuit

Click the switch to turn on the circuit.

Step 3: Operation and Observation Procedure

Accomplish the following procedure to observe the operation of the circuit and to assist in the answering of questions at the end of the Part I.

1. Press the space key to energize the inductor. Once the current exceeds 1 amp, press the space key and observe the inductor voltage on the oscilloscope. Repeat several times.

2. Repeat Step 1, but allow the inductor current to exceed 5 amps.

3. Reduce the resistor value to 500 ohms and repeat Step 1.

Step 4: Answer Questions

1. What happened to the inductor voltage when the circuit was switched after exceeding 1 A?

a. The voltage dropped suddenly on the oscilloscope but then came back up.

2. Was the value negative or positive? Explain.

a. Negative. The oscilloscope showed the same behavior for +1 amp and for +5 amps. The voltage went into the negative on both occasions.

3. What was the peak inductor voltage? Explain.

a. 23.763pv The voltage would change from 23.763 to -23.763 and I also seen it change to -71pv at one time whenever I opened and closed the switch.

4. What happened to the inductor voltage when the current exceeds 5 amps?

a. If I opened and closed the switch the inductor voltage would travel from +120.000 nV to -0.000V.

5. What happened to the inductor voltage when the resistor was reduced?

a. The voltage showed 60.000nV which is half of what it was before and when I toggled the switch open and closed the volts dropped to -3.285v and then back to zero.

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Part III: Resistors and Inductors in DC Circuits.

Step 1: Circuit Construction

Construct the circuit in Figure 1 with a 470 ohm resistor and a 10 mH inductor utilizing a 1 Vp square wave at 5 kHz as the source, in MultiSim.

Figure 1: RL Time Constant Circuit

Step 2: Calculation

Substitute the following RL combinations in the circuit and calculate the theoretical values of the time constants. .

a. 470 ohm, 10 mH t=L/R or 10mH/470 ohms =21.28ms

b. 220 ohm, 10 mH

T=L/R or 10mH/220 ohms = 45.45ms

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Step 3: Operation and Observation Procedure

Connect channel A of the oscilloscope to the source and channel B to measure inductor voltage and measure the actual values from the original circuit configuration. Measure the inductor voltage for the times given in the table below for the energizing portion of the wave starting when the inductor begins to energize.

Table 1 Inductor Voltage vs. Time

1. Calculate the theoretical inductor voltage values for the times in the table.

2. Reverse the position of the inductor and resistor in the circuit and measure the resistor voltage for the times given in the table below.

Table 2 Voltage vs. Time

3. Copy the inductor voltages from Table 1 into Table 2.

4. Add the two voltages to obtain the total series voltage.

Step 4: Answer Questions

6. What was the theoretical and measured time constant?

a. 21.3/45.5

7. What is a valid explanation for any differences observed?

a. Lack of rounding and viltage droppage across connections

8. What happens to the time constant when the resistance was decreased?

a. Decreases

9. What would happen to the time constant if the inductance were decreased?

a. It would be able to hold more charge

10. Where there any appreciable differences between the expected values and measured values?

a. Yes the 0 time value

11. Does the resistor voltage waveform appear correct for a series RL circuit?

a. yes it appeared correctly

12. Which law addresses the relationship of the series RL circuit?

a. Vl=Vsxe(-t/v) =Vl =1V x (e – 100us/100us)