SOLVED PROBLEMS

Q1) Determine the thevenin’s equivalent  circuit between the terminals A&B For the circuit shown in figure 11.1 .

First we are about to find the thevenin’s resistance. To find the thevenin’s resistance we remove the resistance RL and open circuit the AB terminals. Then we remove the voltage source and short circuit it. See figure 11.2

We can easily find the thevenin’s resistance now.

Rth = (5//10 + 3)

Now we have to find the thevenin’’s voltage. For this we remove the load resistance RL . See figure 11.3

Note that there is no current flow through 3ohms resistance as it is open ended from point A. Hence in nodal equation we don’t have to add the current through this resistance.
Applying nodal equation method to point c,

(V-30)/5  + (V-0)/10  = 0

V = 20V

From figure 11.3 you can see that,

VA = VC
VA = VTH

Therefore V = VTH
VTH = 20V

Q2) Find the equivalent e.m.f of the network when viewed from terminals A&B and the equivalent resistance of the network when looked from terminals A&B and the current through load resistance RL . Refer figure 11.4

Current in the network after load resistance has been removed (figure 11.5)

I = 24/(12+3+1)

I = 1.5A.

Applying Ohm’s law on 12  Ohm’s resistance,

Voltage across A and B = VTH = 12 Ohms x 1.5A

Now we are about to find the RTH

It is very easy. Just remove and short circuit the voltage source and find the equivalent resistance as seen from A&B terminals.

NOTE: YOU SHOULD LEAVE THE INTERNAL RESISTANCE OF THE VOLTAGE SOURCE THOUGH YOU REMOVE IT. SEE FIGURE 11.4.(a)

Rth  = (3+1)//12 = 3 Ohms

Hence we have find the thevenins equivalent circuit. So we can easily find the current through load resistance. See figure 11.5.(a)

Q3) determine the thevenin’s equivalent circuit between the terminals A&B. you may refer figure
11.6

This network contains a current source. So it will be important to you to understand how to deal with  a current source in thevenin’s theorem.

Figure 11.7 shows the circuit to find RTH. Note that two terminals between the current source has been open circuited while the voltage source do the opposite.

RTH =  5 + { [ (10//10 ) + 2 ] // 3 }
RTH = 7.1 Ω

From figure 11.8

Voltage at point C = Vc = Vth  (as there is no current through 5Ω resistor)

We have grounded the point B to make our calculation easier.

Nodal equation to point C

(VC – 0)/3 + (VC-VD)/ 2 = 0

5VC – 3VD = 0 -------------------------------------------(1)

Nodal equation to point D

(VD - VC)/2 + VD/10 + (VD -5)/10 – 1 = 0

7VD – 5VC = 15 ---------------------------------------(2)

By solving 1&2

Vc  = 2.25V

VC = VTH

VTH  = 2.25V

I’ll add more examples in my next post.

Pabindu lakshitha

1. What happens if there is a resistance in the path where current source is located?

1. in a current source, internal resistance is very high (infinite). so though we connect a resistance series with current source, it won't do a considerable change to the internal resistance of the current source.therefore we don't get that resistance in to our account.

http://eefundamentals.blogspot.com/2012/08/direct-current-5-voltage-current-sources.html

thank you.

2. In a current source the internal resistance is parallel to the current source not in series.
The parallel resistance ideally must be infinite.
So if we connect a resistor there will be a voltage drop of I x R

2. how about if the resistance is varied

1. This comment has been removed by the author.

3. (V-30)/5 + (V-0)/10 = 0

V = 20V

in Q1, V-30 and V-0. Would you like explain to me about this? I am not understand yet . Tq

1. it should be (30-V)/5+V/10 = 0'
if so, v = 60.
can it happen?

2. By nodal analysis...

4. in the place of RL if there is a voltage is that Voc?

5. how to choose R1,R2,R3

1. as your wish there is no important what you are give name resistance

2. as your wish there is no important what you are give name resistance in the circuit importance is the which resistor are in series or parallal ok

6. if resistance varies through in path of current how then we compare there.

7. Quite helpful....bt plz post some more solved problems based on circuit having current depending voltage source

8. Is the Vth value in 3rd question correct?? because i'm getting 0.75

1. I am getting same Vth=2.25volt

9. in Q 2) HOW thevinins voltage is calculated??????

10. how we solve when 10ohm and 3 ohm resistor are ground directliy plz reply

11. I didn't get how you wrote the equation at nodal D in 3 problem while finding out Voc

12. If a battery is connect near to point B (in series)??? What will happen???
Is Vth value changes????

13. i, understand topic easily but, i have
A small doubt, in finding the vth only
A method of nodal is done or not
And sent some more problems with examples

14. i, understand topic easily but, i have
A small doubt, in finding the vth only
A method of nodal is done or not
And sent some more problems with examples

16. 0-vc/3 + Vc-Vd/2 = 0

Is the correct equation no. 1 !

17. This comment has been removed by the author.

18. if there is no 10 ohm resistor in figure 11.6 then what are the two equation for solving Vth? can you please help me?

1. Wish you are mentioning the 10 ohm resistor at point D in parallel with the current source.
(VC – 0)/3 + (VC-5)/ 12 = 0

Solve this an find Vc.
Then
(VD - VC)/2 + (VD -5)/10 – 1 = 0
Find Vd.
You must re calculate Rth without 10 Ohms resistor.

19. what if in question 3 I add a voltage source in series with 5 ohm resistor which was neglected in calculating the Vth . Will the voltage source be neglected too and Vc=Va hold true?

20. Using Thevenin’s theorem, what is the load
current in Fig. 1-9a for the following values of RL: 2 kV, 6 kV, and 18 kV?
If you really want to appreciate the power of Thevenin’s theorem, try
calculating the foregoing currents using the original circuit of Fig. 1-9a and any
other method.

Rth=6kohm
24v

21. PAKISTAN is great

1. Thank you for selecting the best forum to say it

22. Why ideal current is subtracted in Q3?

23. Why ideal current is subtracted in Q3?

24. This has really been helpful. Please, what if the circuit only has a current source

25. Sir please solve the last sum in mesh analysis also

26. Sir please solve the last sum in mesh analysis also

27. This comment has been removed by the author.

28. (V-30)/5 + (V-0)/10 = 0

V = 20V [why is it (+), so does that mean both current enters the node? and (v/10) is a current that's from the ground??]

1. This comment has been removed by the author.

2. Hellow Hans Lim,
It seems you must study nodal equation method. Please follow below link which will lead you to my post on nodal equation method.
Take 30 mins and study this post. It will solve all your problems.
https://eefundamentals.blogspot.com/2012/08/network-analysis-3.html

Thank you.
Pabindu Gamage.
B.Sc Eng (Hons)

29. also from what i learned its higher potential to lower potential, (30-v)/5 :/ cauuse there would probably be a voltage drop. idk i wnna clear things up

1. Hellow Hans Lim,
It seems you must study nodal equation method. Please follow below link which will lead you to my post on nodal equation method.
Take 30 mins and study this post. It will solve all your problems.
https://eefundamentals.blogspot.com/2012/08/network-analysis-3.html

Thank you.
Pabindu Gamage.
B.Sc Eng (Hons)

30. send more problems with dependent sources

31. Hi, thanks a lot for this information. I loves to read this blog. Keep it up
By the way, visit the new post at my blog Best books for iit jee preparation to get free ebooks in reference of academic engineering field and highly competitive engineering exams.

32. how can we actually know where to use nodal analysis method?
we can also use the Kirchoff's laws to find Vth Rth right. So basically where exactly we can use nodal analysis method?

33. Why you minus by on the Vd nodal analysis

A.C Fundamentals (Solved Examples 2) 1) Find the R.M.S of, V (t) =     Sin 2πt +  Sin 6πt Answer First of all, We have to find t...