Nodal Analysis with Current Sources

Consider the network of Fig. 2.68 (a) which has two current sources and three nodes out of
which I and 2 are independent ones whereas No.3 is the reference node.
The given circuit has been redrawn for ease of understanding and is shown in Fig. 2.68 (b). The
current directions have been taken on the assumption that
I. both V) and V2are positive with respect to the reference node. That is why their respective
curents flow from nodes I and 2 to node 3.
2. V) is positive with respect to V2because current has been shown flowing from node 1 to
node 2.
A positive result will confirm out assumption whereas a negative one willr . dicate that actual
directionis oppositeto that assumed. 1. product of potential VI and (l/RI + 1- R3) i.e. sum of the reciprocals of the branch resistances
connected to this node.
2. minus the ratio of adjoining potential V2and the interconnecting resistance R3'
3. all the above equated to the current supplied by the current source connected to this node.
This current is taken positive if flowing into the node and negative if flowing out of it (as per
sign convention of Art. 2.3). Same remarks apply to Eq. (ii) where 12has been taken negative
because it flows away from node 2.
In tenns of branch conductances, the above two equations can be put as taken as the reference node or common ground for all other nodes. We will apply KCL to the three
nodes and taken currents coming towards the nodes as positive and those going away from them as
negative. For example, current going away from node No. 1is (VI - V2)/l and hence would be taken
as negative. Since 4 A current is coming towards node No. 1. it would be taken as poisitive but 5 A
current would be taken as negative.
Source Conversion
A given voltage source with a series resistance can be converted into (or replaced by) and
equivalent current source with a parallel resistance. Conversely, a current source with a parallel
resistance can be converted into a vaoltage source with a series resistance. Suppose, we want to
convert the voltage source of Fig. 2.75 (a) into an equivalent current source. First, we will find the
value of current supplied by the source when a 'short' is put across in terrnials A and B as shown in
Fig. 2.75 (b). This current is I = VIR. withit representsthe equivalentsource. It is shownin Fig.2.75(c). Similarly,a currentsourceof I
and a parallel resistance R can be converted into a voltage source of voltage V =IR and a resistance
R in series with it. It should be kept in mind that a voltage source-series resistance combination is
equivalent to (or replaceable by) a current source-parallel resistance combination if, and only if their
1. respective open-circuit voltages are equal, and
2. respective short-circuit currents are equal.
For example, in Fig. 2.75 (a), voltage across tenninals A and B when they are open (i.e. opencircuit
voltage Voc) is V itself because there is no drop across R. Short-circuit current across AB =I
= VIR.
Now, take the circuit of Fig. 2.75 (c). The open:circuit voltage across AB =drop across R =IR
= V. If a short is placed across AB, whole of I passes through it because R is completely shorted out.The two parallel resistances of 3 Q and 6 Q can be combined into a single resistance of 2 Q as
shown in Fig. 2.79. (a)
The two current sources cannot be combined together because of the 2 Q resistance present
between points A and C. To remove this hurdle, we convert the 2 A current source into the equivalent
4 V voltage source as shown in Fig. 2.79 (b). Now, this 4 V voltage source with a series
resistance of (2 + 2) =4 Q can again be converted into the equivalent current source as shown in Fig.
2.80 (a). Now, the two current sources can be combined into a single 4-A source as shown in Fig.
2.80 (b).