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JWST499-Cetinkunt
JWST499-c05
ELECTRONIC COMPONENTS FOR MECHATRONIC SYSTEMS 251
The steps to follow to find the Norton’s equivalent circuit are as follows (Figure 5.4),
1. Short-circuit the load resistor, and then calculate the Norton source current (i )
N
through the short wire jumping across the open connection points where the load
resistor used to be (Figure 5.4b).
2. Short-circuit voltage sources and open circuit current sources.
3. Calculate the total resistance between the open connection points (excluding the load
resistance, Figure 5.4c).
4. Draw the Norton equivalent circuit with the Norton equivalent current source in
parallel with the Norton equivalent resistance (Figure 5.4d). The load resistor re-
attaches between the two open points of the equivalent circuit.
Example In the example shown in Figure 5.3, let the following values for the parameters
be
V = 10 V, R = 7 Ω, R = 3 Ω, i = 5 Amp (5.19)
s
2
s
1
Then, it is easy to show that (Figure 5.3b),
V = R ⋅ i + R ⋅ (i + i ) (5.20)
s 1 1 2 1 s
10 V = 7 ⋅ i + 3 ⋅ (i + 5) (5.21)
1 1
= 10 ⋅ i + 15 (5.22)
1
i =−0.5 A (5.23)
1
The voltage potential V is
ab
V ab = 3 ⋅ (−0.5 + 5.0) = 13.5 V (5.24)
The equivalent resistance of the circuit after the voltage sources are short-circuited, and
current sources are open-circuited (Figure 5.3c),
1 1 1
= + (5.25)
R eqv 7 3
7 ⋅ 3
R = (5.26)
eqv
7 + 3
= 2.1 Ω (5.27)
Thevenin’s equivalent circuit of this example is represented by a voltage source
V = 13.5 V and equivalent resistor R eqv = 2.1 Ω (Figure 5.3d).
T
For instance, the current on the load resistor R can be calculated as a function of the
L
load resistance,
V T
i = (5.28)
R
R + R
L eqv
13.5
= (5.29)
R + 2.1
L
i = i (R ; V , R eqv ) (5.30)
R
T
R
L
It is instructive to obtain the Norton equivalent of the same circuit and confirm that
V = R ⋅ i . Referring to Figure 5.4b, the current i is sum of the current due to current
T eqv N N
source and voltage source,
i = i + i 1 (5.31)
N
s