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 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

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 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.

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Pabindu lakshitha

B.Sc (Engineering Undergraduate)