Example 3.2
3.5 The principle of superposition 45Determine the currents 13 and 12 flowing in the circuit of Fig. 3.7.Solution
Using KCL at node A we see that I~ = 12 +/3 so that5 = 12 + 13 and 13 = 5 - 12 (3.9)
Applying KVL to the closed path BCDAB and taking the clockwise direction
to be positive, we have
6 + 813 - 412 = 0 (3.10)
Substituting for ~ from Equation (3.9) above we obtain
6 + 8(5 - 12) - 412 = 0
6 + 40 - 812 - 4/2 = 0
1212 = 46
12 = 3.83 A
It follows that 13 = 5 - 3.83 = 1.17 A.3.5 THE PRINCIPLE OF SUPERPOSITION
This principle applies to any linear system, and when used in the context of
electric circuit theory it may be stated as follows: in any linear network
containing more than one source of emf or current, the current in any element
of the network may be found by determining the current in that element when
each source acts alone and then adding the results algebraically. When
removing, in turn, all the sources except one, any voltage source must be
replaced by its internal resistance (or by a short circuit if the source is ideal) and
any current source must be replaced by an open circuit.Example 3.3
Calculate the current flowing in the 10 f~ load resistor (RL) in the circuit of
Fig. 3.8.
A
(~V1 = 36V (~V2 =12V 10~
(load resistor)~
R1 = 15~ ~]R2 = 10~ RL
B
Figure 3.8