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# Tag Archives: thevenin equivalent circuit

# Find Thevenin’s and Norton’s Equivalent Circuits

Find Thevenin’s and Norton’s Equivalent Circuits:

alt=”1254-1″ width=”300″ height=”156″ srcset=”http://www.solved-problems.com/wp-content/uploads/2013/10/1254-1-300×156.png 300w, http://www.solved-problems.com/wp-content/uploads/2013/10/1254-1-1024×534.png 1024w” sizes=”(max-width: 300px) 100vw, 300px” />

Suppose that , and .

# Solution

The circuit has both independent and dependent sources. In these cases, we need to find open circuit voltage and short circuit current to determine Norton’s (and also Thevenin’s) equivalent circuits.

Continue reading →

# Find Voltage Using Voltage Division Rule

Determine voltage across and using voltage division rule.

Assume that

, , , and

Solution:

Please note that the voltage division rule cannot be directly applied. This is to say that:

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# Thévenin’s Theorem – Circuit with Two Independent Sources

Use Thévenin’s theorem to determine .

Fig. (1-27-1) – Circuit with two independent sources

**Solution**

Lets break the circuit at the load as shown in Fig. (1-27-2).

Continue reading →

# Thévenin’s Theorem – Circuit with An Independent Source

Use Thévenin’s theorem to determine .

Fig. (1-26-1) – The Circuit

**Solution**

To find the Thévenin equivalent, we break the circuit at the load as shown below.

Continue reading →

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##### Basic Concepts DC Circuits EE Questions / Answers Electric Circuit Analysis How To

# Thevenin’s Theorem. Step by Step Procedure with Solved Example

**Thevenin’s Theorem in DC Circuit Analysis**

Step by Step Procedure with Examples.

**M.L Thevenin**, made one of these quantum leaps in 1893.

**Thevenin’s Theorem**is not by itself an analysis tool, but the basis for a very useful method of simplifying active circuits and complex networks because we can solve complex linear circuits and networks especially electronic networks easily and quickly.

**Thevenin’s Theorem**may be stated below:

Any Linear Electric Network or complex circuit with current and voltage sources can be replaced by an equivalent circuit containing of a single independent voltage sourceV_{TH }and a Series ResistanceR._{TH}

- You may Also Read: Norton’s Theorem. Easy Step by Step Procedure with Example (Pictorial Views)

**Steps to Analyze Electric Circuit through Thevenin’s Theorem**

- Open the load resistor.
- Calculate / measure the open circuit voltage. This is the
**Thevenin Voltage (V**._{TH}) - Open current sources and short voltage sources.
- Calculate /measure the Open Circuit Resistance. This is the
**Thevenin Resistance (R**._{TH}) - Now, redraw the circuit with measured
**open circuit Voltage (V**in Step (2) as voltage source and measured_{TH})**open circuit resistance (R**in step (4) as a series resistance and connect the load resistor which we had removed in Step (1). This is the_{TH})**equivalent Thevenin circuit**of that**linear electric network**or**complex circuit**which had to be. You have done.**simplified and analyzed by Thevenin’s Theorem** - Now find the Total current flowing through load resistor by using the Ohm’s Law :
**I**_{T}= V_{TH}/ (R_{TH}+ R_{L}).

- Also read: SUPERMESH Circuit Analysis | Step by Step with Solved Example

**Solved Example by Thevenin’s Theorem:**

**Example: **

**Find V _{TH}, R_{TH }**and the load current flowing through and load voltage across the load resistor in fig (1)

*by using Thevenin’s Theorem*.

*Click image to enlarge*

**Solution:-**

**Step 1.**

Open the **5kΩ load resistor** (Fig 2).

*Click image to enlarge*

**Step 2.**

Calculate / measure the open circuit voltage. This is the **Thevenin Voltage (V _{TH})**. Fig (3).

We have already removed the load resistor from figure 1, so the circuit became an **open circuit** as shown in fig 2. Now we have to calculate the Thevenin’s Voltage. Since **3mA** current flows in both **12kΩ** and **4kΩ** **resistors** as this is a series circuit because current will not flow in the **8kΩ** resistor as it is open.

So **12V** **(3mA x 4kΩ)** will appear across the **4kΩ resistor**. We also know that current is not flowing through the 8kΩ resistor as it is open circuit, but the 8kΩ resistor is in parallel with **4k resistor**. So the same voltage i.e. **12V will appear across the 8kΩ resistor** as well as 4kΩ resistor. Therefore 12V will appear across the AB terminals. So,

**V _{TH} = 12V **

*Click image to enlarge*

**Step 3.**

**Open current sources** and **short voltage sources** as shown below. Fig (4)

*Click image to enlarge*

**Step 4.**

Calculate / **measure the open circuit resistance**. This is the Thevenin Resistance (R_{TH})

We have removed the **48V DC source** to **zero** as equivalent i.e. 48V DC source has been replaced with a short in step 3 (as shown in figure 3). We can see that 8kΩ resistor is in series with a parallel connection of 4kΩ resistor and 12k Ω resistor. i.e.:

**8kΩ + (4k Ω || 12kΩ) ….. (|| = in parallel with)**

**R**

_{TH}= 8kΩ + [(4kΩ x 12kΩ) / (4kΩ + 12kΩ)]**R**= 8kΩ + 3kΩ

_{TH}**R _{TH} = 11kΩ**

*Click image to enlarge*

**Step 5.**

Connect the R_{TH}in series with Voltage Source V_{TH} and re-connect the load resistor. This is shown in fig (6) i.e. Thevenin circuit with load resistor. This the Thevenin’s equivalent circuit

*Click image to enlarge*

**Step 6.**

Now apply the last step i.e Ohm’s law . **Calculate the total load current & load voltage** as shown in fig 6.

**I _{L} = V_{TH }/ (R_{TH} + R_{L})**

= 12V / (11kΩ + 5kΩ) → = 12/16kΩ

**I _{L}= 0.75mA**

And

**V _{L} = I_{L}x R_{L}**

**V _{L}** = 0.75mA x 5kΩ

**V _{L}= 3.75V**

*Click image to enlarge*

Now compare this simple circuit with the original circuit of figure 1. Can you see how much easier it will be to measure/calculate the load current for different load resistors by *Thevenin’s Theorem*? Yes and only yes.

You may also read:

- SUPERNODE Circuit Analysis | Step by Step with Solved Example
- Maximum Power Transfer Theorem for AC & DC Circuits
- Kirchhoff’s Current & Voltage Law (KCL & KVL) | Solved Example

**JLCPCB – Prototype 10 PCBs for $2 + 2 days Lead Time**

China’s Largest PCB Prototype Enterprise, 300,000+ Customers & 10,000+ Online Orders Per Day

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Circuit Analysis DC Circuit Analysis Thevenin’s Theorem Thevinin

### Electrical Technology

- Website
- Google+
- Flickr
- YouTube

### What is an Electric Circuit? Types of Circuits, Network & Parts Of Circuit

### Norton’s Theorem. Easy Step by Step Procedure with Example (Pictorial Views)

### Related Articles

### Basic Concepts Library

### According to Ohm’s Law, Current increases, when Voltage increases,(I=V/R), But Current decreases, when Voltage increases according to (P=VI). Explain?

### How To Wire Two 24V Solar Panels in Parallel with Two, 12V Batteries in Series with Automatic UPS System (For 24 V System)? (OR) Parallel Connection of Solar Panel and…

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### 99 Comments

thanks, good easy to understand.but the voltage source have more than one. how to do it.

ReplyIt is simple example for ref & explanation purpose only…Wait for upcoming posts….Thanks

ReplySurely when the load resistor is replaced the circuit changes and the 8k and 5k are in parallel with the 4k causing the cu

ReplySurely when the load resistor is replaced the circuit changes and the 8k and 5k are in parallel with the 4k causing the current and voltage to change oops just figured it out the long way and the 5k gets 3.75 out of the 9.75 volts across that part of the circuit your way of showing thevenin is better thank you

ReplyThanks for your kind words…

Reply

If the voltage source is more than one you calculate two components of Nortons current that is I’ and I”, add them to get the total Nortons current and repeat the procedure explained above.

Reply

Admin. You are doing great work. Your posts related to electricals are easy to learn. . . Keep updating. . .

ReplyThanks for appreciation…

Reply

Thevenin's Theorem is not very intuitive for a newbie, good to have this nice easy to follow explanation.

Replysimple and good steps for easy understanding

ReplyGood one for basic understanding of theorems :)<br />

ReplyHow did u get 3mA?

ReplyI = V / R …<br />= 48V / (12kΩ + 4kΩ )<br />= 3mA <br /><br />So 3mA Current will flow in both 12kΩ and 4kΩ resistors as this is a series circuit because current will not flow in the 8kΩ resistor as it is open<br /><br />

ReplyIf the rounded off value of 3mA is used to find Vth you get 12V.

If the less rounded off value of 3.273mA is used we’d get 13V.Reply

when the load terminals are open then all the current passes through only first circuit and the current is from v=i/r can be calculated

Reply1st of all find req than by v=ir , i=v/r we can get the value of current

Reply

Thanku. <br />What happens when there is 2 voltage sources?

ReplyUse Kirchhoff law

Reply

i am proud that u can make this subject such intresting..<br /><br />hope many teachers of ur kind should be there..<br /><br />i am learning this during my job period ..if it was during my college it would have been a big deal for me..<br /><br />anyway thanks!!

ReplyThanks for your kind words…

Reply

hope many teachers of ur kind should be there..<br /><br />i am learning this during my job period ..if it was during my college it would have been a big deal for me..<br /><br />anyway thanks!!

ReplyThank you

Replytq admin..mechanical stdent frm malaysia 🙂

ReplyMost welcome dear….

Reply

step by step process is easy way for students to understand the problem

ReplyGreat job there, sometimes is easy to understand through other source.

Replythank you sir…. this is really a helpful information..

Replythank u very much sir……

Replyplesae provide telligan’s theorem

ReplyThanks for your very informative post. I had a very bad going with this topic, but now I can assure keep my hands dirty on complex circuits related to this phenomenon by using your idea.

Replyoutstanding sir

Replyit is very easy to understand the concept,but if more than one voltage sources are present then how will be the solution and procedure

ReplyIt is too much easy

Replyboss i love this your tutorials its the best i have ever seen you simplified the sturf to the last five star 4 u *****

Replybest tutors site

ReplyThank you…

Reply

sir this is awesome sir

i had clarify my doubts succesfullyReplyit is soo easy to understand

ReplyThank you very much!!!

Replycould you give another exercise with 2 suppliers? and show steps

ReplyThumps up to yo tutorials,now I understand the theorem way better than I previously did.Thanx so Much

ReplyThanks for appreciation….

Reply

why you solve 12 and 4 as parallel?

ReplyIf you see in step 4, 8kΩ resistor is in series with a parallel connection of 4kΩ resistor and 12k Ω resistor.

i.e. If you solve for the parallel connection of 4kΩ resistor and 12k Ω, It becomes in Series with 8kΩ.In step 2, The Circuit is Open, Hence, Current will not flow in 8kΩ, Therefore, That’s Why it is in parallel with 4kΩ.

Reply

will you please add step by step procedure for dual loops that will help alot

Thanx in advanceReplyGreat job. you make the determination of thevenin’s voltage is very very easy. Thank you…

ReplyI could not solve even a single problem .But this helped me a lot .Thank u

ReplyGlad it helps You….

Reply

Thanks…

Replycan i get more relevant problems using thevinin’s theorem & may i know that thevinin’s theorem can use for AC supplied circuits.

ReplyYes…. It can be used on both AC and DC Circuits,,, We will post about Thevinin’s theorem about AC Circuits in coming days… Thanks

Reply

Told how to solve thevinins them in ac by using kvl.

Replyplease inform more and more about electrical and electronics

Replysir I have a confusion when you calculate Rth because I dont think that Rth=8+4||12. But It should be Rth=12+4||8. Am I wrong Sir ?, if so then please guide me.

Thanks

AnkitReplyHi Ankit,

As mentioned above that We can see that 8kΩ resistor is in series with a parallel connection of 4kΩ resistor and 12k Ω resistor. i.e.:

8kΩ + (4k Ω || 12kΩ)ReplySaqib Javeed is Right…. Thanks

let me know if you want to know more… Thanks

Reply

simply superb

ReplyThank you very much,

this was the best explanation available on the internetReplyThanks for appreciation …

Reply

Was very helpful…thanks a billion times ?

ReplyMost Welcome….

Reply

Awesome explanation. It’s very useful to us, Thank you!!

ReplyMost welcome… And thanks for appreciation….

Reply

have we already been provided with 3Am or you have solved it? if you have solved it how have you done it?? a ain’t sure if I have followed you on that one

ReplyIt is not provided.

In Step 2, You may see that this is an Open Circuit, I.e. Current will not flow in the 8kΩ.

So the circuit of 48V, 4kΩ and 12kΩ becomes a series circuit.

By using Ohms Law, (I = V/R), We found the value of current as 3mA.

ThanksReply

what will be total R in the circuit if 4k shud the load resistance in revolving towards the source

ReplyYou may just change the value of the resistance and solve again same as above…. No Difficulties…. 🙂

Reply

you said 8k resistor is parallel to 4k resistor in step 2.in step 4 you said 8k resistor is sereis to 4k.why?

ReplyYes Dear…. If you see in step 4, 8kΩ resistor is in series with a parallel connection of 4kΩ resistor and 12k Ω resistor.

i.e. If you solve for the parallel connection of 4kΩ resistor and 12k Ω, It becomes in Series with 8kΩ.In step 2, The Circuit is Open, Hence, Current will not flow in 8kΩ, Therefore, That’s Why it is in parallel with 4kΩ.

Thanks.

let me know if you want to know more… ThanksReply

wow. thank you. now i understand. 🙂

Replythank you for this. but how about 3 sources using this theorem?

ReplyPlease how was the 3mA current gotten ?

Replycan you explain to me about how did you get 12V as it is 48V?

ReplyIT IS GOOD BUT I WILL LIKE IT IF IT SOLVED USING SMALL NUMBERS LIKE 3,4,2 TO MAKE IT LOOK MORE SIMPLER

ReplyThevenin made simply. thanks dear.

ReplyA/C to my point of view, Its very simple and easy method for understanding..

ReplySimplified example…great work

Replyif I find current across 5 ohm resister there is a greater value then the total load current that is 0.75 can anyone answer

Replythanks for the example.

Replythank you dear

ReplyA really Understanding work!

ReplyThanks for keeping the simplicity in the explination I thoroughly understood the conpect.

ReplyThanks soooooo much, it really help me out.

ReplyThank you so much to help me in easy way

ReplyAnd if there is an ammeter in parallel with the resistance, then what to do

ReplyReally good

Replythnks sir

ReplyIf there are a voltage or current source between A & B terminals or in the place of load resistor, Then how can i solve this????????

ReplyVERY SIMPLY AND THOROUGHLY EXPLAINED THE CONCEPT OF THEVENINS THEOREM. PLEASE INCLUDE THE NOTES OF TWO POLE NETWORKS ALSO

ReplyAm happy for the steps of thevenins i cam now calculate but following the steps

Replysir please tell me….is this the right and only method to solve problems by thevenin’s theorem?

ReplyYes! But you can analyze the same problem by different methods as well.

Replycan you please tell me the different methods sir?

Replyplease respond quick!!

Reply

Very good explanation. I tend to solve complex circuits using Kirchoff’s laws…Are there particular cases when one method is favourably used instead of the other? ( I have not read all posts…maybe you have already gone into this)

Replyhow did u get 3mA

Reply3mA current will flows in both 12kΩ and 4kΩ resistors as it is a series circuit because current will not flow in the 8kΩ resistor as it is open (Fig 2)

Reply

thankyou its help full.

request.. how can I calculate if there are voltage source in the circuit?Replythank u so much

Reply

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##### Basic Concepts DC Circuits EE Questions / Answers Electric Circuit Analysis How To

# Thevenin’s Theorem. Step by Step Procedure with Solved Example

**Thevenin’s Theorem in DC Circuit Analysis**

Step by Step Procedure with Examples.

**M.L Thevenin**, made one of these quantum leaps in 1893.

**Thevenin’s Theorem**is not by itself an analysis tool, but the basis for a very useful method of simplifying active circuits and complex networks because we can solve complex linear circuits and networks especially electronic networks easily and quickly.

**Thevenin’s Theorem**may be stated below:

Any Linear Electric Network or complex circuit with current and voltage sources can be replaced by an equivalent circuit containing of a single independent voltage sourceV_{TH }and a Series ResistanceR._{TH}

- You may Also Read: Norton’s Theorem. Easy Step by Step Procedure with Example (Pictorial Views)

**Steps to Analyze Electric Circuit through Thevenin’s Theorem**

- Open the load resistor.
- Calculate / measure the open circuit voltage. This is the
**Thevenin Voltage (V**._{TH}) - Open current sources and short voltage sources.
- Calculate /measure the Open Circuit Resistance. This is the
**Thevenin Resistance (R**._{TH}) - Now, redraw the circuit with measured
**open circuit Voltage (V**in Step (2) as voltage source and measured_{TH})**open circuit resistance (R**in step (4) as a series resistance and connect the load resistor which we had removed in Step (1). This is the_{TH})**equivalent Thevenin circuit**of that**linear electric network**or**complex circuit**which had to be. You have done.**simplified and analyzed by Thevenin’s Theorem** - Now find the Total current flowing through load resistor by using the Ohm’s Law :
**I**_{T}= V_{TH}/ (R_{TH}+ R_{L}).

- Also read: SUPERMESH Circuit Analysis | Step by Step with Solved Example

**Solved Example by Thevenin’s Theorem:**

**Example: **

**Find V _{TH}, R_{TH }**and the load current flowing through and load voltage across the load resistor in fig (1)

*by using Thevenin’s Theorem*.

*Click image to enlarge*

**Solution:-**

**Step 1.**

Open the **5kΩ load resistor** (Fig 2).

*Click image to enlarge*

**Step 2.**

Calculate / measure the open circuit voltage. This is the **Thevenin Voltage (V _{TH})**. Fig (3).

We have already removed the load resistor from figure 1, so the circuit became an **open circuit** as shown in fig 2. Now we have to calculate the Thevenin’s Voltage. Since **3mA** current flows in both **12kΩ** and **4kΩ** **resistors** as this is a series circuit because current will not flow in the **8kΩ** resistor as it is open.

So **12V** **(3mA x 4kΩ)** will appear across the **4kΩ resistor**. We also know that current is not flowing through the 8kΩ resistor as it is open circuit, but the 8kΩ resistor is in parallel with **4k resistor**. So the same voltage i.e. **12V will appear across the 8kΩ resistor** as well as 4kΩ resistor. Therefore 12V will appear across the AB terminals. So,

**V _{TH} = 12V **

*Click image to enlarge*

**Step 3.**

**Open current sources** and **short voltage sources** as shown below. Fig (4)

*Click image to enlarge*

**Step 4.**

Calculate / **measure the open circuit resistance**. This is the Thevenin Resistance (R_{TH})

We have removed the **48V DC source** to **zero** as equivalent i.e. 48V DC source has been replaced with a short in step 3 (as shown in figure 3). We can see that 8kΩ resistor is in series with a parallel connection of 4kΩ resistor and 12k Ω resistor. i.e.:

**8kΩ + (4k Ω || 12kΩ) ….. (|| = in parallel with)**

**R**

_{TH}= 8kΩ + [(4kΩ x 12kΩ) / (4kΩ + 12kΩ)]**R**= 8kΩ + 3kΩ

_{TH}**R _{TH} = 11kΩ**

*Click image to enlarge*

**Step 5.**

Connect the R_{TH}in series with Voltage Source V_{TH} and re-connect the load resistor. This is shown in fig (6) i.e. Thevenin circuit with load resistor. This the Thevenin’s equivalent circuit

*Click image to enlarge*

**Step 6.**

Now apply the last step i.e Ohm’s law . **Calculate the total load current & load voltage** as shown in fig 6.

**I _{L} = V_{TH }/ (R_{TH} + R_{L})**

= 12V / (11kΩ + 5kΩ) → = 12/16kΩ

**I _{L}= 0.75mA**

And

**V _{L} = I_{L}x R_{L}**

**V _{L}** = 0.75mA x 5kΩ

**V _{L}= 3.75V**

*Click image to enlarge*

Now compare this simple circuit with the original circuit of figure 1. Can you see how much easier it will be to measure/calculate the load current for different load resistors by *Thevenin’s Theorem*? Yes and only yes.

You may also read:

- SUPERNODE Circuit Analysis | Step by Step with Solved Example
- Maximum Power Transfer Theorem for AC & DC Circuits
- Kirchhoff’s Current & Voltage Law (KCL & KVL) | Solved Example

**JLCPCB – Prototype 10 PCBs for $2 + 2 days Lead Time**

China’s Largest PCB Prototype Enterprise, 300,000+ Customers & 10,000+ Online Orders Per Day

Inside a huge PCB factory

Circuit Analysis DC Circuit Analysis Thevenin’s Theorem Thevinin

### Electrical Technology

- Website
- Google+
- Flickr
- YouTube

### What is an Electric Circuit? Types of Circuits, Network & Parts Of Circuit

### Norton’s Theorem. Easy Step by Step Procedure with Example (Pictorial Views)

### Related Articles

### Basic Concepts Library

### According to Ohm’s Law, Current increases, when Voltage increases,(I=V/R), But Current decreases, when Voltage increases according to (P=VI). Explain?

### How To Wire Two 24V Solar Panels in Parallel with Two, 12V Batteries in Series with Automatic UPS System (For 24 V System)? (OR) Parallel Connection of Solar Panel and…

### Integration of Renewable Energy with Grid System

### Insulation Resistance of a Cable | Why Cables are insulated?

### What is the difference between real ground and virtual ground?

### 99 Comments

thanks, good easy to understand.but the voltage source have more than one. how to do it.

ReplyIt is simple example for ref & explanation purpose only…Wait for upcoming posts….Thanks

ReplySurely when the load resistor is replaced the circuit changes and the 8k and 5k are in parallel with the 4k causing the cu

ReplySurely when the load resistor is replaced the circuit changes and the 8k and 5k are in parallel with the 4k causing the current and voltage to change oops just figured it out the long way and the 5k gets 3.75 out of the 9.75 volts across that part of the circuit your way of showing thevenin is better thank you

ReplyThanks for your kind words…

Reply

If the voltage source is more than one you calculate two components of Nortons current that is I’ and I”, add them to get the total Nortons current and repeat the procedure explained above.

Reply

Admin. You are doing great work. Your posts related to electricals are easy to learn. . . Keep updating. . .

ReplyThanks for appreciation…

Reply

Thevenin's Theorem is not very intuitive for a newbie, good to have this nice easy to follow explanation.

Replysimple and good steps for easy understanding

ReplyGood one for basic understanding of theorems :)<br />

ReplyHow did u get 3mA?

ReplyI = V / R …<br />= 48V / (12kΩ + 4kΩ )<br />= 3mA <br /><br />So 3mA Current will flow in both 12kΩ and 4kΩ resistors as this is a series circuit because current will not flow in the 8kΩ resistor as it is open<br /><br />

ReplyIf the rounded off value of 3mA is used to find Vth you get 12V.

If the less rounded off value of 3.273mA is used we’d get 13V.Reply

when the load terminals are open then all the current passes through only first circuit and the current is from v=i/r can be calculated

Reply1st of all find req than by v=ir , i=v/r we can get the value of current

Reply

Thanku. <br />What happens when there is 2 voltage sources?

ReplyUse Kirchhoff law

Reply

i am proud that u can make this subject such intresting..<br /><br />hope many teachers of ur kind should be there..<br /><br />i am learning this during my job period ..if it was during my college it would have been a big deal for me..<br /><br />anyway thanks!!

ReplyThanks for your kind words…

Reply

hope many teachers of ur kind should be there..<br /><br />i am learning this during my job period ..if it was during my college it would have been a big deal for me..<br /><br />anyway thanks!!

ReplyThank you

Replytq admin..mechanical stdent frm malaysia 🙂

ReplyMost welcome dear….

Reply

step by step process is easy way for students to understand the problem

ReplyGreat job there, sometimes is easy to understand through other source.

Replythank you sir…. this is really a helpful information..

Replythank u very much sir……

Replyplesae provide telligan’s theorem

ReplyThanks for your very informative post. I had a very bad going with this topic, but now I can assure keep my hands dirty on complex circuits related to this phenomenon by using your idea.

Replyoutstanding sir

Replyit is very easy to understand the concept,but if more than one voltage sources are present then how will be the solution and procedure

ReplyIt is too much easy

Replyboss i love this your tutorials its the best i have ever seen you simplified the sturf to the last five star 4 u *****

Replybest tutors site

ReplyThank you…

Reply

sir this is awesome sir

i had clarify my doubts succesfullyReplyit is soo easy to understand

ReplyThank you very much!!!

Replycould you give another exercise with 2 suppliers? and show steps

ReplyThumps up to yo tutorials,now I understand the theorem way better than I previously did.Thanx so Much

ReplyThanks for appreciation….

Reply

why you solve 12 and 4 as parallel?

ReplyIf you see in step 4, 8kΩ resistor is in series with a parallel connection of 4kΩ resistor and 12k Ω resistor.

i.e. If you solve for the parallel connection of 4kΩ resistor and 12k Ω, It becomes in Series with 8kΩ.In step 2, The Circuit is Open, Hence, Current will not flow in 8kΩ, Therefore, That’s Why it is in parallel with 4kΩ.

Reply

will you please add step by step procedure for dual loops that will help alot

Thanx in advanceReplyGreat job. you make the determination of thevenin’s voltage is very very easy. Thank you…

ReplyI could not solve even a single problem .But this helped me a lot .Thank u

ReplyGlad it helps You….

Reply

Thanks…

Replycan i get more relevant problems using thevinin’s theorem & may i know that thevinin’s theorem can use for AC supplied circuits.

ReplyYes…. It can be used on both AC and DC Circuits,,, We will post about Thevinin’s theorem about AC Circuits in coming days… Thanks

Reply

Told how to solve thevinins them in ac by using kvl.

Replyplease inform more and more about electrical and electronics

Replysir I have a confusion when you calculate Rth because I dont think that Rth=8+4||12. But It should be Rth=12+4||8. Am I wrong Sir ?, if so then please guide me.

Thanks

AnkitReplyHi Ankit,

As mentioned above that We can see that 8kΩ resistor is in series with a parallel connection of 4kΩ resistor and 12k Ω resistor. i.e.:

8kΩ + (4k Ω || 12kΩ)ReplySaqib Javeed is Right…. Thanks

let me know if you want to know more… Thanks

Reply

simply superb

ReplyThank you very much,

this was the best explanation available on the internetReplyThanks for appreciation …

Reply

Was very helpful…thanks a billion times ?

ReplyMost Welcome….

Reply

Awesome explanation. It’s very useful to us, Thank you!!

ReplyMost welcome… And thanks for appreciation….

Reply

have we already been provided with 3Am or you have solved it? if you have solved it how have you done it?? a ain’t sure if I have followed you on that one

ReplyIt is not provided.

In Step 2, You may see that this is an Open Circuit, I.e. Current will not flow in the 8kΩ.

So the circuit of 48V, 4kΩ and 12kΩ becomes a series circuit.

By using Ohms Law, (I = V/R), We found the value of current as 3mA.

ThanksReply

what will be total R in the circuit if 4k shud the load resistance in revolving towards the source

ReplyYou may just change the value of the resistance and solve again same as above…. No Difficulties…. 🙂

Reply

you said 8k resistor is parallel to 4k resistor in step 2.in step 4 you said 8k resistor is sereis to 4k.why?

ReplyYes Dear…. If you see in step 4, 8kΩ resistor is in series with a parallel connection of 4kΩ resistor and 12k Ω resistor.

i.e. If you solve for the parallel connection of 4kΩ resistor and 12k Ω, It becomes in Series with 8kΩ.In step 2, The Circuit is Open, Hence, Current will not flow in 8kΩ, Therefore, That’s Why it is in parallel with 4kΩ.

Thanks.

let me know if you want to know more… ThanksReply

wow. thank you. now i understand. 🙂

Replythank you for this. but how about 3 sources using this theorem?

ReplyPlease how was the 3mA current gotten ?

Replycan you explain to me about how did you get 12V as it is 48V?

ReplyIT IS GOOD BUT I WILL LIKE IT IF IT SOLVED USING SMALL NUMBERS LIKE 3,4,2 TO MAKE IT LOOK MORE SIMPLER

ReplyThevenin made simply. thanks dear.

ReplyA/C to my point of view, Its very simple and easy method for understanding..

ReplySimplified example…great work

Replyif I find current across 5 ohm resister there is a greater value then the total load current that is 0.75 can anyone answer

Replythanks for the example.

Replythank you dear

ReplyA really Understanding work!

ReplyThanks for keeping the simplicity in the explination I thoroughly understood the conpect.

ReplyThanks soooooo much, it really help me out.

ReplyThank you so much to help me in easy way

ReplyAnd if there is an ammeter in parallel with the resistance, then what to do

ReplyReally good

Replythnks sir

ReplyIf there are a voltage or current source between A & B terminals or in the place of load resistor, Then how can i solve this????????

ReplyVERY SIMPLY AND THOROUGHLY EXPLAINED THE CONCEPT OF THEVENINS THEOREM. PLEASE INCLUDE THE NOTES OF TWO POLE NETWORKS ALSO

ReplyAm happy for the steps of thevenins i cam now calculate but following the steps

Replysir please tell me….is this the right and only method to solve problems by thevenin’s theorem?

ReplyYes! But you can analyze the same problem by different methods as well.

Replycan you please tell me the different methods sir?

Replyplease respond quick!!

Reply

Very good explanation. I tend to solve complex circuits using Kirchoff’s laws…Are there particular cases when one method is favourably used instead of the other? ( I have not read all posts…maybe you have already gone into this)

Replyhow did u get 3mA

Reply3mA current will flows in both 12kΩ and 4kΩ resistors as it is a series circuit because current will not flow in the 8kΩ resistor as it is open (Fig 2)

Reply

thankyou its help full.

request.. how can I calculate if there are voltage source in the circuit?Replythank u so much

Reply

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