How to Thevenize a Given Circuit?

It is clear from above that any network of resistors and voltage sources (and current sources as
well) when viewed from any points A and B in the network, can be replaced by a single voltage
source and a single resistance* in series with the voltage source.
After this replacement of the network by a single voltage source with a series resistance has
been accomplished, i~is easy to find current in any load resistance joined across terminals A and B.
This theorem is valid even for those linear networks which have a nonlinear load.
Hence, Thevenin' s theorem, as applied to d.c. circuits, may be stated as under :
The current flowing through a load resistance RL connected across any two tenninals A and B
of a linear, acti~e bilateral network is given by VocII(Rj + RJ where Vocis the open-circuit voltage
(i.e. voltage across the two tenninals when RL is removed) and Rj is the internal resistance of tM
network as viewed back into the open-circuited network from tenninals A and B with all voltage
sources replaced by their internal resistance (if any) and current sources by infinite resistance.
2.19. How to Thevenize a Given Circuit?
I A
B
Fig. 2.128 "
1. Temporarily remove the resistance (called load resistance RJ whose current is required.
2. Find the open-circuit voltage Vocwhich appears across the two terminals from where
resistance has been removed. It is also called Thevenin voltage Vth.
3. Compute the resistance of the whose network as looked into from these two terminals after
all voltage sources have been removed leaving behind their internal resistances (if any) and
current sources have been replaced by open-circuit i.e. infinite resistance. It is also called
Thevenin resistance Rthor Tj'
4. Replace the entire network by a single Thevenin source, whose voltage is Vthor Vocand
whose internal resistance is Rthor Ri.
5. Connect RL back to its terminals from where it was previously removed.
6. Finally, calculate the current flowing through RLby using the equation
polarity as shown in Fig. 2.129 (a).
.. VAB = V,h =+ 10 + 5/3 =35/3V.
Incid~ntly, we could also fmd VABwhile going along the parallel route BFEA.
Drop across 8 Q resistor =8 x 5/12 =10/3 V. VAB equal the algebraic sum of voltages met on the
way from B to A. Hence, VAB=(- 10/3) + 15 =35/3 V.
As shown in Fig. 2.129 (b), the single voltage source has a voltage of 35/3 V.
For findign R'h' we will replace the two voltage sources by short-circuits. In that case, R'h =RAB
=4118=8/3Q. Now, we will find Rthi.e. equivalent resistnace of the network as looked back into the opencircuited
terminals A and B. For this purpose, we will replace
both the voltage and current sources. Since voltage source
has no internalresistance, it would be replacedby a short circuit
i.e. zeroresistance. However, current source would be removed
and replaced by an 'open' i.e. infinite resistance (Art. 1.18).
In that case, the circuit becomes as shown in Fig. 2.133 (c).
As seen from Fig. 2.133 (d), Fth= 6 II3 + 2 = 4 Q. Hence,
Thevenin's equivalent circuit consists of a voltage source of
12V and a series resistance of 4 Q as shown in Fig. 2.134 (a).
When 4 Q resistor is connected a The negative sign shows that point A is negative with respect to point B or which is the same
thing, point B is positive with respect to point A.
For finding RAS=Rth'we replace the batteries by short-circuits as shown in Fig. 2.128 (c).
:. RAS = Rth =8 + 2 II5=9.43 Q
Hence, the equivalent Thevenin' s source with respect to tenninals A and B is an shown in Fig.
2.136. When 10Q resistance is reconnected across A and B, current through it is 1= 6.24/9.43 + 10)
= 0.32A (I) When galvanometer is removed from Fig. 2.139 (a), we get the circuit of Fig.
2.139 (b).
(ii) Let us next find the open-circuit voltage VDC (also called Thevenin voltage V,h)between
points B and D. Remembering that ABC (as well as ADC) is a potential divider on which a voltage
drop of 10 V takes place, we get
Potential of B w.r.t. C = 10 x 10/110 = 10/11 =0.909 V
Potential of D w.r.t. C = 10 x 4/54 = 20/27 = 0.741 V
.. p.d. between B and D is Vocor V,h=0.909 - 0.741 =0.168 V
(iil) Now, remove the 10-V battery retaining its internal resistance which, in this case, happens
to be zero. Hence, it amounts to short-circuiting points A and C as shown in Fig. 2.139 (d).