So far, we have considered circuits which consisted of resistors and independent current or
voltage sources only. However, we often come across circuits which contain both independent and
dependent sources or circuits which contain only dependent sources. Procedure for finding the
value of Vlhand Rlhin such cases is detalied below :
(a) When Circuit Contains Both Dependent and Independent Sources
(i) The open-circuit voltage Vocis determined as usual with the sources activated or 'alive'.
(ii) A short-circuit is applied across the terminals a and b and the value of short-circuit
current ilhis found as usual.
(iii) Thevenin resistance Rlh = vo/ish' It is the same procedureas adoptedfor Norton's
theorem. Solved examples 2.81 to 2.85 illustrate this procedure.
(b) When Circuit Contains Dependent Sources Only
(i) In this case, voe =0
(ii) We connect 1A source to the terminals a and b and calculate the value of vab.
(iii) Rlh=VaJ I 11
The above procedure is illustrated by solved Examples.
Example 2.81. Find Thevenin equivalent circuit for the network shown in Fig. 2.162 (a) which
contains a current controlled voltage source (CCVS).
Solution. For finding Voeavailable across open-circuit terminals a and b, we will apply KVL to
the closed loop.
.. 12 - 4 i + 2 i - 4 i = 0 :. i =2 A
Hence, Voe= drop across 4 11resistor =4 x 2 =8 V. It is so because there is no current through
the 2 11 resistor.
For finging Rlh, we will put a short-circuit across terminals a and b and calculate Ish' as shown in
Fig. 2.162 (b). Using the two mesh currents, we have
12- 4 i( + 2 i - 4(i\ - i2) =0 and - 8 i2 - 4 (i2- i\) =O. Substituting i=(i\ - i2)and Simplifying
the above equations, we have
12 - 4 i} + 2 (i} - i2)- 4 (i} - i2)=0 or 3 i} - i2=6 ...(i)
Similarly, from the second equation, we get i\ =3 i2. Hence, i2=3/4 and Rlh= Vo/lsh= 8/(3/4)
= 32/311. The Thevenin equivalent circuit is as shown in Fig. 2.162 (c).
Example 2.82. Find the Thevenin equivalent circuit which respect to tenninals a and b of the
network shown in Fig. 2.163 (a).
Solution. It will be seen that with terminals a and b open, current through the 8 11resistor is
vaJ4 and potential of point A is the same of that of point a (because there is no current through 4 11
resistor). Applying KVL to the closed loop of Fig. 2.163 (a), we get It is also the value of the open-circuit voltage voc'
For finding short-circuit current ish'we short-circuit the terminals a and b as shown in Fig. 2.163
(b). Since with a and b short-circuited, vab=0, the dependent current source also becomes zero.
Hence, it is replaced by an open-circuit as shown. Going around the closed loop, we get
12 - ish (8 + 4) = 0 or ish=6/12 =0.5 A
Hence, the Thevenin equivalent is as shown in Fig. 2.163 (c).
Example 2.83. Find the Thevenin equivalent circuit for the network shown in Fig. 2.164 (a)
which contains only a dependent source.
Solution. Since circuit contains no indepenedent source, i=0 when terminals a and b are open.
Hence, Voc=O. Moreover, ishis zero since Voc= O.
Consequently, RShcannot be found from the relation R'h=v~ish' Hence, as per Art. 2.20, we
will connect a I A current source to terminals a and b as shown in Fig. 2.164 (b). Then by finding the
value of v00' we will be able to calculate R'h =vmJl.
It should be noted that potential of point A is the same as that of point a t.e. voltages across 120.
resistor is vab. Applying KCL to point A, we get
2i-voo voo
-12+1 = 0 or 4i-3vab=-12
Since i = vmJl2, we have 4 (vmJI2) - 3 voo=- 12 or voo=4.5 V :. R'h =vmJI =4.5/1 =4.50..
The Thevenin equivalent circuit is shown in Fig. 2.164 (c).
Example 2.84. Determine the Thevenins equivalent circuit as viewed from the open-circuit
terminals a and b of the network shown in Fig. 2.165 (a). All resistances are in ohms.
Solution. It would be seec from Fig. 2.165(a) that potential of node A equals the open-circuit
terminal voltage voc' Also, i = (vs - voc)/(80 + 20) = (6 - voc)/Ioo.
Applying KCL to node, A we get other branch, then the same e.m.f. E acting in the second bamch would produce the same current I
in thefirst branch.
In other words, it simply means that E and I are mutually transferrable. The ratio Ell is known
as the transfer resistance (or impedance in a.c. systems). Another way of stating the above is that the
receiving point and the sending point in a network are interchangebale. It also means that interchange
of an ideal voltage sources and an ideal ammeter in any network will not change the ammeter
reading. Same is the case with the interchange of an ideal current source and an ideal voltmeter.
The new circuit having 2 -V battery connected in the branch BD is shown in Fig. 2.169. According
to the Principle of Superposition,the new current in the 1-volt battery circuit is due to the superposition
of two currents; one due to 1 - volts battery and the other due to the 2 - volt battery when each acts
independently.
The current in the external circuit due to 1 - volt battery when 2 - battery is not there, as found
above, is 0.0723 A.
Now, according to Reciprocity theorem; if 1 - volt battery were tansferred to the branch BD
(where it produced a current of 0.0049 A), then it would produce a current of 0.0049 A in the branch
CEA (where it was before). Hence, a battery of 2 -V would produce a current of (- 2 x 0.0049) =-
0.0098 A (by proportion). The negative sign is used because the 2 - volt battery has been so connected
as to oppose the current in branch BD,
