Given circuit can be resolved as shown below,

Vx = 2 + 3 + 10 = 15 V

Alternate method

To determine VOC, we open terminal a & b

 i = 0        i1 = 5 A        

By KVL,        

To determine ISC, Short terminals a & b

By KVL,        

Due to open circuit, there is no current through  resister with  

Hence,  

Apply KCL at V

Maximum power transistor of VI product is maximum. If draw the curve, it intersect (10,4) that will give maximum power.

The terminal voltage is 10 V (Load voltage) and current is 4A (Load current).

Load resistance =

Since the terminals A & B are extended beyond so it is already opened.

KVL in Loop – 1:

                      …………………..(1)

KVL in Loop – 2:

                ………..……..…..(2)

From (1) & (2),

25i = 4

= 3.36 V

Finding Thevenin equivalent across  resistor:

By Potential Division,

Current in resistor = 0 A.

Note: If we carefully observe the given circuit, it is nothing but a balanced Wheatstone bridge. Hence, current will be zero.

For maximum power transfer, the resistive load should have the magnitude equal to impedance magnitude.

To determine Norton’s Current, short circuit the XY terminals

Using Mesh analysis,

For first loop:        

2.5 = 5  – 2.5    ……………..(1)  

For second loop:        

2.5 = 7.5  – 5     …………….(2)

Adding (1) & (2),

5 = 5 

 = 1A and ISC=1A

Since, from maximum power transfer theorem for maximum power to be transferred in load

This circuit doesn’t have any independent source, so to find out thvenin’s equivalent resistance across 1 & 2 connect a 1 volt voltage source across the terminals and find the current (I) through voltage source. The circuit can be redrawn as,

Applying KCL at node A

                ---------- (1)

Apply KCL at node B

                -----------(2)

Using (1)

                -----------(3)

Applying KVL in the loop

                -----------(4)

Also

                ----------(5)

From 3, 4 & 5 -

Representing the two port network by its Thevenin Equivalent,        

Case-1: 

                ---------(1)

Case 2: 

                ----------(2)

From 1 & 2         

From equation (1)

The Thevenin Voltage will depend on the applied voltage. Since, the network is linear.

Hence,

Case-1:

                        ----------(3)

Case-2:

                ----------(4)

Solving (3) and (4) we get,

a = 0.5 and b = 4

When 8V DC Voltage is applied,

Since, DC Voltage is 10V so

Now if

Since, variable resistance is in source side and load resistance is fixed.

So for maximum power to be transferred in a fixed load  

resistance, the current should be maximum for maximum current to be delivered from 10V source variable resistance R should be zero.

R=0

Note: Maximum Power Transfer Theorem can only be applied when the load resistance is variable. Also, this can be derived by deriving the load power in terms of ‘R’ and then finding the maxima of the function but that would be a tedious process.

From equation

When i=0 (open circuit)

V=100Volts

Also when V=0 (Short circuit)

i=100A

Thevenin equivalent circuit is

Applying super-position theorem

(i) When only  is considered        (ii) When only  is considered

Since voltage source is shorted in case-2, so entire current will flow through SC

Total current thought  resistor

To double the current through  resistance,

Voltage source  should be doubled as the current source doesn’t have any effect on value of

For Vth, the load terminals must be open circuited as shown below,

Applying KCL at node with voltage ‘V’

 +  +  = 0                 -------- (1)

By KVL,

VAB + 3VAB = V

V = 4 VAB                                 --------(2)

from (1) & (2)

 +  = 0

VAB = 0.5V

For RTh

Calculate IN or ISC

Rth =  =  =

For Vth, the load terminals must be open circuited as shown below,

Applying KCL at node with voltage ‘V’

 +  +  = 0                 -------- (1)

By KVL,

VAB + 3VAB = V

V = 4 VAB                                 --------(2)

from (1) & (2)

 +  = 0

VAB = 0.5V

For RTh

Calculate IN or ISC

Rth =  =  =

Short circuiting voltage source

For the Thevenin Voltage,                                

Using voltage division rule,

Voltage across resistor,

For maximum power transfer finding Thevenin’s equivalent circuit across

For

For

Applying KVL in loop

The Thevenin Equivalent is shown in the figure.

For maximum transfer in         

Power absorbed by the load

The given circuit only has dependent source. Hence, in order to determine the Thevenin Impedance we need to connect a 1V source as shown below,

Current in 4Ω resistor =

Current in 2Ω resistor =

Applying KCL,

Thevenin Resistance,

When the circuit is converted to a voltage source in series with a resistance then it is called as Thevenin Equivalent.

For , voltage sources have been shorted        

For

Applying KVL in loop:-

Using KVL

The Thevenin equivalent circuit is shown below,

For : open circuiting the current source and short circuiting the voltage source. The circuit then is shown below,

To determine IN short circuit the terminals AB and apply source transformation we get the circuit as shown below.

Applying KVL in loop 1

                --------------(1)

Applying KVL in loop 2

                --------------(2)

Solving equation 1 & 2

Hence,  and

To determine the Thevenin Equivalent open circuit the terminals A and B.

Due to open circuit, current in 4Ω resistor is 0A.

Using KVL

                        -------------(1)

Applying KCL

                        -------------(2)

Solving equation 1 & 2

 and

For :Applying 1 volts source across terminals AB, and short circuiting independent voltage source.

Applying KCL

 and

From the V-I characteristics when terminals

are short circuited V = 0        

At this time

when A & B terminals are open, i.e. I = 0

Under this condition .

Thevenin resistance of the network

=

After connecting inductor across the Thevenin equivalent the circuit is shown below,

Time constant of the RL circuit =

Finding Thevenin equivalent circuit

For , short circuit all independent voltage sources.

The resistances 5Ω and 15Ω can be combined in parallel as,

The network then simplifies to a form shown,         

Hence,

For

Applying KVL in upper loop,

                --------------(1)        

Applying KVL in lower loop

                --------------(2)

Solving equation 1 & 2

,        

Using KVL

So, thevenin equivalent circuit will look like as shown,

So the current through resistor ‘R’

Superposition theorem is valid for all types of linear circuit having time – varying or time invariant elements. It is not applicable for non – linear elements.

For maximum power, load resistance should be equal to the venin resistance.

From the characteristics shown,

Open Circuit Voltage, I = 0,

Short Circuit Current,         V = 0         

           …………..(1)        

                 …………..(2)

For maximum power transfer load resistance,        

Power =