## 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 R

_{L}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.

R

^{th }= (5//10 + 3)
Now we have to find the thevenin’’s voltage. For this we
remove the load resistance R

_{L}. 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,**

**V**

_{A }= V_{C}**V**

_{A}= V^{TH}
Therefore V = V

^{TH}**V**

^{TH}= 20V

**Q**2) 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 R

_{L}. 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

__= V__^{TH}= 12 Ohms x 1.5A
Now we are
about to find the R

^{TH}
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)**

__R__

^{th }= (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)

I

_{LOAD}= 18 / (3+15)

__I___{LOAD}= 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 R

^{TH}.**Note that two terminals between the current source has been open circuited while the voltage source do the opposite.**
R

^{TH}= 5 + { [ (10//10 ) + 2 ] // 3 }

__R__^{TH}= 7.1

__Ω__
From figure
11.8

Voltage at
point C = V

_{c}= V^{th }(as there is no current through 5Ω resistor)
We have
grounded the point B to make our calculation easier.

Nodal
equation to point C

(V

_{C}– 0)/3 + (V_{C}-V_{D})/ 2 = 0
5V

_{C }– 3V_{D}= 0 -------------------------------------------(1)
Nodal
equation to point D

(V

_{D}- V_{C})/2 + V_{D}/10 + (V_{D}-5)/10 – 1 = 0
7V

_{D}– 5V_{C}= 15 ---------------------------------------(2)
By solving
1&2

V

_{c }= 2.25V
V

_{C}= V^{TH}_{}

**V**

^{TH }= 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)

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

ReplyDeletein 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.

Deletefollowing 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.

In a current source the internal resistance is parallel to the current source not in series.

DeleteThe parallel resistance ideally must be infinite.

So if we connect a resistor there will be a voltage drop of I x R

thanks

ReplyDeletehow about if the resistance is varied

ReplyDeleteThis comment has been removed by the author.

Delete(V-30)/5 + (V-0)/10 = 0

ReplyDeleteV = 20V

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

it should be (30-V)/5+V/10 = 0'

Deleteif so, v = 60.

can it happen?

By nodal analysis...

Deletein the place of RL if there is a voltage is that Voc?

ReplyDeletehow to choose R1,R2,R3

ReplyDeleteas your wish there is no important what you are give name resistance

Deleteas 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

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

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

ReplyDeleteIs the Vth value in 3rd question correct?? because i'm getting 0.75

ReplyDeleteI am getting same Vth=2.25volt

Deletein Q 2) HOW thevinins voltage is calculated??????

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

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

ReplyDeleteIf a battery is connect near to point B (in series)??? What will happen???

ReplyDeleteIs Vth value changes????

i, understand topic easily but, i have

ReplyDeleteA small doubt, in finding the vth only

A method of nodal is done or not

And sent some more problems with examples

i, understand topic easily but, i have

ReplyDeleteA small doubt, in finding the vth only

A method of nodal is done or not

And sent some more problems with examples

in your next post please solve using kirchoffs law

ReplyDelete0-vc/3 + Vc-Vd/2 = 0

ReplyDeleteIs the correct equation no. 1 !

This comment has been removed by the author.

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

ReplyDeleteWish you are mentioning the 10 ohm resistor at point D in parallel with the current source.

Delete(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.

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?

ReplyDeleteUsing Thevenin’s theorem, what is the load

ReplyDeletecurrent 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