\"\"

\

 

\

 

\

\

Maximum Power Transfer:

\

For any power source, the maximum power transferred from the power source to the load is when the resistance of the load \"image\" is equal to the equivalent or input resistance of the power source \"image\".

\

Apply Thevinins theorem:

\

Find Thevinins equivalent impedance.

\

1.Open the load resistor.

\

2.Open Current Sources and Short Voltage Sources.

\

3.Calculate the Open Circuit Impedance. This is the Thevenin Impedance (\"image\").

\

Redraw the circuit:

\

\"\"

\

Observe the circuit:

\

\"\"

\

\"\", \"\" and \"\" are in series.

\

\"\"

\

\"\"

\

 

\

 

\

 

\

 

\

(2)

\

Step 1:

\

Use Superposition principle to calculate volatge across A and B. \ \

\

Super Position Theorem :

\

Take one current source at a time and replace all other with short or internal resistance.

\

\"image\"

\

\"\" in complex form can be written as \"\".

\

Apply KCL

\

Apply current divider rule: \"\".

\

Find the total impdenace \"\".

\

\"\" and \"\" are in series. \ \

\

 

\

\"\"

\

\"\" is in parallel to \"\".

\

\"\"

\

\"\" is in series to \"\" and \"\".

\

\"\"

\

\"\"

\

Current divider rule:

\

\"\"

\

\"\"

\

Find the voltage drop across the impedance Z.

\

\"\"

\

\"\"

\

\

\"\" V.

\

Step 2:

\

Take one voltage source at a time and replace all other with short or internal resistance.

\

\"\" \ \

\

\"\" in complex form can be written as \"\"

\

Apply KVL to loop 1:

\

\"\"

\

Find the voltage drop across the impedance Z.

\

\"\"

\

\"\"

\

\

\"\" V.

\

Step 3:

\

By superposition principle : 

\

The Voltage acoss the terminal A and B is \"\".

\

\"\"

\

\"\"

\

Thevinins equivalent voltage is \"\"

\

Draw the equivalent circuit.

\

\"\"

\

Power in the circuit is \"\"

\

\"\"

\

Power in the circuit is 17.89 W.

\

Solution:

\

Power in the circuit is 17.89 W.