Ideal Constant-Voltage Source

Ideal Constant-Voltage Source
It is that voltage source (or generator) whose output voltage remains absolutely constant whatever
the change in load current. Such a voltage source muat possess zero internal resistance so that
internal voltage drop in the soruce is zero. In that case, output voltage provided by the source would
remain constant irrespective of the amount of current drawn from it. In practice, none such ideal
constant-voltage source can be obtained. However, smaller the internal resistance r of a voltage
source, closer it comes to the ideal sources described above.
Suppose, a 6-V battery has an internal resistance of 0.005 Q [Fig. 2.94 (a)). When it supplies no
current i.e. it is on no-load, Vo=6 V i.e. output voltage provided by it at its output terminals A and B
is 6 V. If load current increases to 100 A, internal drop =100 x 0.005 =0.5 V. Hence, Vo=6 - 0.5
=5.5 V.
Obviously an output voltage of 5.5 - 6 V can be considered constant as compared to wide
variations in load current from 0 A ot 100 A.
2.16. Ideal Constant-Current Source
It is that voltage source whose internal resistance is infinity. In practice, it is approached by a
source which posses very high resistance as compared to that of the external load resistance. As
shown in Fig. 2.94 (b), let the 6-V battery or voltage source have an internal resistance of 1M Q and
let the load resistance vary from 20 K to 200 K. The current supplied by the source varies from
6.l/1.02 =5.9 J.1A to 6/1.2 =5 J.1A. As seen, even when load resistance increases 10 times, current
decreases by 0.9 J.1A.Hence, the source can be considered, for all practical purposes, to be a constant-
current source.According to this theorem, if there are a number of e.m.fs. acting simultaneously in any linear
bilateral network, then each e.m.f. acts independently of the others i.e. as if the other e.m.fs. did not
exist. The value of current in any conductor is the algebraic sum of he currents due to each e.m.f.
Similarly, voltage across any conductor is the algebraic sum of the voltages which each e.m.f would
have produced while acting singly. In other words, current in or voltage across, any conductor of the
network is obtained by superimpsing the currents and voltages due to each e.m.f. in the network. It
is important to keep in mind that this theorem is applicable only to linear networks where current is
linearly related to voltage as per Ohm's law.
Hence, this theorem may be stated as follows :
In a network of linear resistances containing more than one generator (or source of e.m.f.), the
current whichflows at any point is the sum of all the currents which woludflow at that point if each
generator where considered separately and all the other generators replacedfor the time being by
I "4- 1-.'4- resistancesequalto their internalresistacnces
A seen, there are three independent sources are one dependent source. We will find
the value of v produced by each of the three independent sources when acting alone and add the three
values to find v. It should be noted that unlike independent source, a dependent source connot be set
to zero i.e. it cannot be 'killed' or deactivated.
Let us find the value of vI due to 30 V source only. For this purpose we will replace current
source by an open circuit and the 20 V source by a short circuit as shown in Fig. 2.99 (b). Applying
KCL to node 1, we get Solution. The given circuit has been redrawn in Fig. 2.109 (b) with 15 - V battery acting alone
while the other two sources have been killed. The 12 - V battery has been replaced by a short-circuit
and the current source has been replaced by an open-circuit (O.C) (Art. 2.19). Since the output
terminals are open, no current flows through the 4 Q resistor and hence, there is no voltage drop
across it. Obviously VI equals the votIage drop over 10Q resistor which can ,be found by using the
voltage-divider rule.
VI = 15 x 10/(40 + 50) =3 V
Fig. 2.110 (a) shows the circuit when current source acts alone, while two batteries have been
killed. Again, there is no current through 4 Q resistor. The two resistors of values 10Q and 40 Q are
in parallelacrossthe currentsource. Theircombinedresistancesis 10II40 =8 Q