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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In Our Previous Post, We Know What Is Superposition theorem? And, Formula's To Used Finding Of Current In Any Resistance In A branch Of Network/Circuits.

AIM:- Network shown below is the example in which we will find current passes through 10Ω resistance by using Superposition Theorem?

Let's Start To Solve:-

To Determine Current In 10Ω resistance, We Will Used Same As We Learn In Our Previous Post "Superposition Theorem & Formula".

Here, We Move On With The Step By Step Solutions.

Above Network Have Two Sources "Voltage Source" & "Current Source".

Hence, We Activate One Source At a Time.

Here, We Will Activate 16V Source And Deactivate 16A Source As Shown In Below Arrangement:-

It Seems Clear After Deactivating "16A" Current Source, Two Boxes Are Act Like Active Circuit.

Now, By Using Mesh Analysis, We Can Easily Find Current Through 10Ω resistance when ‘16V’ voltage source is active. 

NOTE: Here We Can Use nodal analysis or ohm’s law with current division rule.

 
Let's Have A Look,

Here, We Will Going To Discuss About "How To Find Current Through Mesh Analysis". Click The Link To Revise.

{ Hints: -    

Equations are,

  -8I₁+4I₂=-16       ··············· (1)

  4I₁-18I₂=0          ··············· (2)      }

 So, after solving this circuit, we get

 I₁ =   2.25A          &          I₂ =   0.5A

Then, 

I_10Ω = I₂ = 0.5A

Activating ‘16A’ Source This Time. Voltage Source Shall Be Short-Circuited. As Per Below:-

Applying Current Division Rule, We Can Determine The Value of current in 10Ω resistance. 

By Using Current Division Rule, We Find The Value As:-

I₁ = 4A

I₂ = 12A

So, 

I_10Ω = I₁ = 4A

Now, Reactivate All Source As It Is To Evaluate The Exact Value. In Below Figure, The Current Flows With Their Determine Value In The Same Direction As We Get While Deactivating Each Sources At A Time.

At 10Ω Resistance Current Will Be As Follows:-

At last, the current through 10Ω resistance is 3.5A and greater current has their direction as shown below.

Answer: -    I_10Ω = 3.5A

Thanks For Reading & Learning With Us. Don't Miss To Comment Your View.

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For any circuits or network, if we are finding current/voltage through a circuit or any two points in any given circuits, it's important to know "How to find Equivalent Resistance For Those Two Points". Later, We Can Use Any "Ohm's Law" To Determine the voltage or current whichever required.

In this topic, we will use our previous post "Series- Parallel Formulas" To Identified The "Equivalent Resistance" Between Point "A" & Point "B" For the Below Given Network.

Let's Start The Approach:-

Step-1:- In This Below Network, Let Start To Minimize The Circuit From The Last Connection From The Equivalent Point "A" & Point "B".

In Below Network, We Mark Each Box As "1", "2" & "3" For Our Better Understanding. 

Now, We Start Solving Network From Box "3" To Box "1" By Using Our Formulas Which We Learn It Previous Topic.

Finally, Our Answer For The Network Is RAB  = ?

It's Simple Math Equation, Kindly Do Comment Your Answer.

Thanks For Reading & Learning With Us.

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AIM: To Find out Resistance Between Point “A” & Point “B” In Below Network By Using Star(WYE) Delta Transformation Methods.

 

Let’s Start:-

We have used before Nodal Method & Mesh Analysis Method, Now In This Below Network, We Can Easily Determine The Value Of equivalent resistance i.e, R_AB through Star-Delta conversion.

We have Below Network.

Let Simplify The Network For Better Understanding, By Trimming Some Lines, As Below:-

As We can see the last network forms like Delta, Then Our Approach To Transform This Delta Into Star/Wye Connection.

Just Draw The Dots Line of resistance from each Nodal area to the center of delta to form star/wye, as per below:-

 

For finding value of each new star connector resistance (RA , RB ,  RC),  We can used Direct formula as learn in previous post, As Shown Below:-

 

Now, The Circuit Simplified As:-

Here, We Can Use Mesh & Nodal Method For Simplify Our Equation For Next Level:-

(Note: Circuit Structure Are Same, Reform circuit for easy understanding)

So, R_AB / R_equivalent = R₁ + R₂ + R₃ = 4Ω + 3.88Ω + 1.77Ω = 9.65Ω Answer

 

Hope This Help For You To Solve, Any Types Of Equation Based On Star Delta Transformation. If Any Query, Let Us Know In Comments.

 

Thanks for Reading & Learning. Best Of Luck!

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In This Below Topic, We Will Discuss About “Reading Problem Based On Thevenin's Theorem In Practical”.

Let’s Start To Find Answer By Solving Through Thevenin's Theorem Method.

(For Solving This Question, We Will Used OHM's Law As Well As "Kirchhoff's Voltage Law To Simplified Our Equation)

Aim:- Find The Current In 4Ω Resistance In The Below Network Circuit By Using Thevenin's Theorem Methods.

Solution: - For Solving The Value Of Current Through 4Ω Resistance, We Will Follow The Same Step As Discussed In Our Previous Post.

(Click Here To Revise Your Practice)

FOR R_L: -

 HERE, R_L= 4Ω

For Rth:-

Follow The Below Steps For Finding Thevenin's Equivalent Resistance For This Network:-

As Below:-

Here, R_th/R_AB=1.5+10=11.5Ω

For Vth:-

Follow The Below Steps For Finding Thevenin's Equivalent Voltage For This Network:-

Finally, The Thevenin's Equivalent Circuit Is,

Note: Hope It Simplified Your Question And Answer Technique, Don’t Miss To Comments Your View. Thanks For Your Visit Our Website.

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In This Below Topic, We Will Discuss About “How To Solve Problem Based On Norton’s Theorem In Practical”.

Below Is The Example Of A Circuit Which We Will Find Answer By Solving Through Norton’s Theorem Method.

Let’s Start.

Aim:- We Have To Find The Current In 4Ω  Resistance In The Below Network Circuit By Using Norton’s Theorem.

Solution: - For Solving The Value Of Current Through 4Ω  Resistance, We Will Follow The Same Step As Discussed In Our Previous Post.

(Click Here To Revise Your Practice)

FOR FINDING R_L: -

 HERE, R_L= 4Ω

NEXT STEP TO FIND R_N: - 

For Finding R_N, Just Find-out Equivalent B/w Removed Load-resistance. 

As Below:-

NEXT STEP TO FIND I_N/I_S.C: -

Next Step Is To Short-circuited Pt. A And Pt.  B

Now Current In The Circuit Will Easily Pass Through The Short Circuited Path And Parallel Resistance Will Remove, As Shown Below:-

In Below We Have Two Loop Circuit For Which We Get Current Of Each By Using Kirchhoff’s Voltage Law(KVL).

APPLY KVL IN LOOP 1^ST:-

 8-2I₁-2I₁-2I₁+2I₂ =0

 -6I₁+2I₂ = -8 ··············· (1)

 

APPLY KVL IN LOOP 2^ND:-

 -2I₂-2I₂-2I₂+2I₁ =0

 2I₁-6I₂ = 0    ··············· (2)

RESULT ON SOLVING, WE GET

 I₁ = 1.5A & I₂ = 0.5A

 

HERE, I₂= I_S.C/ I_N = 0.5A

RESULT ON SOLVING, WE GET

 I₁ = 1.5A & I₂ = 0.5A

 HERE, I₂= I_S.C/ I_N = 0.5A

FINALLY, THE NORTON’S EQUIVALENT CIRCUIT IS,

Note: 

Hope It Simplified Your Question And Answer Technique, Don’t Miss To Comments Your View. Thanks For Your Visit Our Website.

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