Adjust the value of the load resistor and pay attention

Assignment Help Electrical Engineering
Reference no: EM131211757

Procedure:

Open Multisim Software

Program Files -> National Instruments (folder) -> Circuit Design Suite -> Multisim

Maximum Power Transfer

- Make a Thevenin circuit with any voltage and resistance you choose with V between 1-12V and R between 1kOhm and 50kOhms.

- Attach a load resistor and place indicators or multimeter to measure voltage across the load and the current through it. You will use this to calculate power dissipated by the load.

- Adjust the value of the Load resistor and pay attention to how the current and voltage changes for that resistor.

- Take measurements of load resistance vs. power. Verify that using the Thevenin resistance gives you maximum power in the load.

Take enough data that you have 5 data points at R values less than the Thevenin value and 5 above (11 data points in all).

- Repeat this with the Norton equivalent of the Thevenin circuit you chose.

- For efficiency, remember that η = (PLoad/PSource ) * 100%

- PSource Th = Eth2/(RL+RTh ) and PSource N = IN2*(RL||RN )

Insert a screenshot of your loaded Thevenin circuit diagram with multimeters here:

ETh = _____________         RTh = _________________

Thevenin

Choose R values between 1kOhm and 50kOhms.

No units needed in table, use units designated. 

1

RL <RTh

2

RL <RTh

3

RL <RTh

4

RL <RTh

5

RL <RTh

Max Power

RL =RTh

6

RL >RTh

7

RL >RTh

8

RL >RTh

9

RL >RTh

10

RL >RTh

R Load (kOhms)

 

 

 

 

 

 

 

 

 

 

 

I Load (mAmps)

 

 

 

 

 

 

 

 

 

 

 

VLoad  (Volts)

 

 

 

 

 

 

 

 

 

 

 

PLoad (mWatts)

 

 

 

 

 

 

 

 

 

 

 

PSource (mWatts)

 

 

 

 

 

 

 

 

 

 

 

η(efficiency =)

 

 

 

 

 

 

 

 

 

 

 

Insert a screenshot of your loaded Norton circuit diagram with multimeters here:

IN = _________________RN = _________________

Norton

Use the same resistance values as above

No units needed in table, use units designated. 

1

RL <RTh

2

RL <RTh

3

RL <RTh

4

RL <RTh

5

RL <RTh

Max Power

RL =RTh

6

RL >RTh

7

RL >RTh

8

RL >RTh

9

RL >RTh

10

RL >RTh

R Load (kOhms)

 

 

 

 

 

 

 

 

 

 

 

I Load (mAmps)

 

 

 

 

 

 

 

 

 

 

 

VLoad  (Volts)

 

 

 

 

 

 

 

 

 

 

 

PLoad (mWatts)

 

 

 

 

 

 

 

 

 

 

 

PSource (mWatts)

 

 

 

 

 

 

 

 

 

 

 

η(efficiency =)

 

 

 

 

 

 

 

 

 

 

 

For each load resistor, was there any difference in Current, Voltage or Power between the Thevenin and Norton Circuit?

Which resistor gave you the highest efficiency (for Norton and Thevenin) and why?

Does the highest efficiency load resistor also give you the maximum power?

Graph load power (y axis) vs. load resistance (x-axis) in excel and copy the graph into your lab with Thevenin and Norton data on the same graph. Create a second graph with the x-axis converted to a logarithmic scale. Also graph efficiency for Thevenin and Norton vs. load resistance with a linear scale only.

o Your graph should be a scatter plot and show measured power or efficiency on the y- axis and Load Resistance on the x- axis
o Each axis is labeled and shows the units.

Insert graph of Load Power vs. Load Resistance in regular scaling with Thevenin and Norton on the same graph:

Insert graph of Load Power vs. Load Resistance in log scaling for x with Thevenin and Norton on the same graph:

Insert graph of Efficiency vs. Load Resistance for Thevenin and Norton on the same graph:

Verified Expert

From the table 1 and table 2 data, it is clear there is no difference in the current, voltage and power they are all the same. The Reason is that the Thevenin circuit and Norton circuit configuration is different in the sense that in Thevenin the Voltage

Reference no: EM131211757

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