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Saturday, September 1, 2012

SOLVED PROBLEMS ON THEVENIN'S THEOREM (1)


SOLVED PROBLEMS


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



Answer:

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




Answer:

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)

ILOAD = 18 / (3+15)

ILOAD = 1A

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



Answer:


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.



PLEASE LEAVE YOUR PROBLEM AS A COMMENT. OR MAIL TO: pabindu@gmail.com


Pabindu lakshitha
B.Sc (Engineering Undergraduate)















11 comments:

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

    ReplyDelete
    Replies
    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.
      following link will direct you to the post about current sources.

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

      thank you.

      Delete
  2. how about if the resistance is varied

    ReplyDelete
    Replies
    1. This comment has been removed by the author.

      Delete
  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

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

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

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

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

    ReplyDelete