Answer:

Maximum Power Transfer Theorem :-

• Maximum Power Transfer Theorem states that in a active DC network, the maximum power will be transferred from source to the load when the external load resistance equals to the thevenins resistance of the circuit.

Given

We have to replace the above circuit  by its thevenins equivalent circuit. To replace  above circuit by its thevenins equivalent circuit then  we have to find

1. Thevenins Resistance RTH.
2. Thevenins Voltage VTH.

Calculation of Thevenins Resistance RTH:-

• For finding Thevenins Resistance RTH , 10V & 5 V voltage sources are replaced by a short circuit and both 5A current sources are replaced by a open circuit then the modified circuit is as follows

$R_{TH} = \dfrac{2\times\left(1+\dfrac{2\times2}{2+2}\right) }{2+\left(1+\dfrac{2\times2}{2+2}\right)} = 1 \ \Omega$

Calculation of Thevenins voltage VTH:-

• As the given circuit consists of four sources i.e. two 5A current source, 10V voltage source and 5V voltage source. We have to apply superposition theorem.

• Superposition theorem states that the voltage across any circuit element is the sum of the voltage across the circuit element if each source is considered separately and other sources are replaced by their internal resistances.

Note:- Ideal voltage sources have zero internal resistance and  ideal current sources have infinite internal resistance.

• VTH is the thevenins voltage in the given circuit.VTH' is the thevenins voltage considering only 10 V voltage source, VTH'' is the thevenins voltage considering only 5 A current source, VTH''' is the thevenins voltage considering only 5A current source.VTH'''' is the thevenins voltage considering only 5 V voltage source.

Step 1: Considering 10 V voltage source:-

• While considering 10 V voltage source, both 5A current source has been replaced by infinite resistance i.e. open circuit and 5 V voltage source has been replaced by zero resistance i.e. short circuit The modified diagram is as follows

$I_T^{'} = \dfrac{10}{2+\left(\dfrac{2\times(1+2)}{2+(1+2)}\right)} = 3.125 \ A $

Applying current division rule the I2' is given by 

$I_2^{'} = \dfrac{I_T^{'} \times 2}{2+(1+2)} = \dfrac{3.125 \times 2}{2+(1+2)} = 1.25 \ A$

$V_{TH}^{'} = I_2^{'} \times 2= 1.25 \times 2= 2.5 \ V$

Step 2: Considering 5 A current source:-

• While considering 5 A current source, the another 5A current source has been replaced by infinite resistance i.e. open circuit and 10 V & 5 V voltage sources are replaced by zero resistance i.e. short circuit. The modified diagram is as follows
• Applying current division rule let us first find current I''1.2 which is as follows

$I_{1.2}^{''} = \dfrac{5\times 2}{2+1.2}= 3.125 \ A$

• Applying current division rule let us first find current I''2 which is as follows

$I_{2}^{''} = \dfrac{I_{1.2}^{''}\times 2}{2+3}= \dfrac{3.125\times 2}{2+3}= 1.25 \ A$

• VTH'' is the voltage across 2 ohms and is given by

$V_{TH}^{''} = I_{2}^{''} \times 2= 1.25 \times 2= 2.5 \ V$

Step 3: Considering 5 A current source:-

• While considering 5 A current source, the another 5A current source has been replaced by infinite resistance i.e. open circuit and 10 V & 5 V voltage sources a replaced are by zero resistance i.e. short circuit. The modified diagram is as follows