Page 181 - Basic Electrical Engineering
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We convert the current source into its equivalent voltage source of the
network and redraw it as
Figure 2.87
Remove the 6 Ω resistor across AB and determine V OC ,
Figure 2.88
We write,
+20 V − 2I − 2I − 6I − 12 V = 0
or, 10 I = 8 V
I = 0.8 A
V OC = +12 + 6 × 0.8 = 16.8 V
R across AB after short circuiting the source of EMFs,
eq
Thevenin’s equivalent circuit is
