The superposition theorem - faculty.psau.edu.sa · Thévenin’s and Norton’s theorems: (i) The open-circuit voltage, E, across terminalsAB in Figure (5) is equal to 10V, since

The superposition theorem

The superposition theorem states:

‘In any network made up of linear resistances and containing more than

one source of e.m.f., the resultant current flowing in any branch is the

algebraic sum of the currents that would flow in that branch if each

source was considered separately, all other sources being replaced at that time by their respective internal resistances.’

Example: The figure below shows a circuit containing two sources of e.m.f., each with their internal resistance. Determine the current in each branch of the network by using the superposition theorem.

Procedure: 1. Redraw the original circuit with source E2 removed,

being replaced by r2 only, as shown in Figure1 (a).

2. Label the currents in each branch and their directions

as shown in Figure1 (a) and determine their values.

(Note that the choice of current directions depends on

the battery polarity, which, by convention is taken as

flowing from the positive battery terminal as shown.)

R in parallel with r2 gives an equivalent resistance of:

From the equivalent circuit of Figure1 (b) we calculate I1

Figure 1

From Figure1 (a):

And

3. Redraw the original circuit with source E1 removed,

being replaced by r1 only, as shown in Figure 2(a).

4. Label the currents in each branch and their directions

as shown in Figure 2(a) and determine their values.

r1 in parallel with R gives an equivalent resistance of:

Figure 2

Figure 1

From the equivalent circuit of Figure 2(b) :

From Figure 2(a):

5. Superimpose Figure 2(a) on

to Figure 1(a) as shown in Figure 3.

Figure 2 Figure 3

6. Determine the algebraic sum of the currents flowing in each branch.

Resultant current flowing through source 1, i.e.

Resultant current flowing through source 2, i.e.

Resultant current flowing through resistor R, i.e.

The resultant currents with their directions are shown

in Figure 4.

Figure 3

Figure 4

General d.c. circuit theory

The following points involving d.c. circuit analysis need

to be appreciated before proceeding with problems using

Thévenin’s and Norton’s theorems:

(i) The open-circuit voltage, E, across terminalsAB in

Figure (5) is equal to 10V, since no current flows

through the 2Ω resistor and hence no voltage drop

occurs. (ii) The open-circuit voltage, E, across terminals AB in

Figure 6(a) is the same as the voltage across the

6 Ω resistor. The circuit may be redrawn as shown

in Figure 6(b).

by voltage division in a series circuit, i.e. E = 30V

Figure 5

Figure 6

(iii) For the circuit shown in Figure 7(a) representing a practical

source supplying energy,

where

E is the battery e.m.f.,

V is the battery terminal voltage and

r is the internal resistance of the battery

For the circuit shown in Figure 7(b),

Figure 7

(iv) The resistance ‘looking-in’ at terminals AB in Figure 8(a) is obtained by reducing the circuit

in stages as shown in Figures 8(b) to (d). Hence the equivalent resistance across AB is 7 Ω

Figure 8

(v) For the circuit shown in Figure 9(a), the 3 Ω resistor carries no current and the p.d. across the

20 resistor is 10V. Redrawing the circuit gives Figure 9(b), from which

(vi) If the 10V battery in Figure 9(a) is removed and

replaced by a short-circuit, as shown in Figure 9(c),

then the 20 Ω resistor may be removed.

The circuit is then as shown in Figure 9(d),

which is redrawn in Figure 9(e). From Figure 9(e),

the equivalent resistance across AB,

Figure 9

(vii) To find the voltage across AB in Figure (10): Since the 20V supply is across the 5 Ω and

15 Ω resistors in series then, by voltage division, the voltage drop across AC,

Hence the voltage between AB is:

VA −VB = 15−4=11V

and current would flow from A to B since A has a higher potential than B.

Figure 10

(viii) In Figure 11(a), to find the equivalent resistance across AB the circuit may be redrawn as in Figures

11(b) and (c). From Figure 11(c), the equivalent resistance across AB:

Figure 11