Reading Maximum Power Transfer Theorem Problems In Practical

Reading Maximum Power Transfer Theorem Problems In Practical

In This Below Network, Our Aim To Find The Value Of Resistance At Load i.e, R_L As Mentioned And Also Determine The Maximum Power Through Load Resistance R_L By Using "Maximum Power Transfer Theorem".

To Solve The Above Problem, We Will Follow Below Steps One By One:-

By Using The Definition Of "Maximum Power Transfer Theorem", As We Discussed The Load-Resistance Will Be Disconnected From The Circuits And The Point A & Point B Both The Ends Will Be Act As "Equivalent Resistance", For Which We Will Find The Answer By Using Our Previous Method Of Solving.

Now, After Removing Load-Resistance The Network Will Be Like As Below And We Will Solve This To Find "Equivalent Resistance".

(If You Miss To Learn, Kindly Have Revision By Reading Previous Posts)

Here Also Note That Voltage Source Will Be "Short-Circuited", Remember Current Source Will Be "Opened Circuited".

Now, Move Forward

Hence, The Load Resistance/Equivalent Resistance Is "3.77 Ohms"

Now, We Need Voltage Across Point A & Point B. After This Only We Can Apply Direct "Maximum Power Transfer Theorem" Formula.

Now, For Maximum Power through Point A & Point B,  Need to Find V_o.c.

Here, V_o.c is known as "Voltage" between Open Circuits (Point A & Point B).

So, Next Steps Is:-

By Applying "Kirchhoff's Voltage Law", We Can Easily Determine The Value Of Both Loop i.e, Loop 1 & Loop 2 As Shown In The Network As Above.

Applying "Kirchhoff's Voltage Law" in loop 1^st:-

6-6I₁-8I₁+8I₂=0

-14I₁+8I₂=-6         ∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙ (1)

Applying "Kirchhoff's Voltage Law" in loop 2^nd:-

-8I₂-5I₂-12I₂+8I₁=0

8I₁-25I₂=0             ∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙ (2)

By Solving Equation (1) & Equation (2), We Will Get Answer:-

I₁ = 0.524 A

I₂ = 0.167 A

Now, From the Circuit V_o.c is

V_A-5I₂- V_B = 0

V_o.c/ V_AB= 5I₂ = 5X0.167 = 0.835v

So, the "Maximum Power Transfer" Through R_L Will Be:-

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