## 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.

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?

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

ReplyDelete