The equivalent resistance seen from a–b (10 5)//(10 5) 7.5 Ω
Therefore, the Thevenin’s equivalent and Norton’s (determined by applying
source transformation on Thevenin’s equivalent) are as shown in Fig. 5.5-8.
=
+
+++
×=
105
105105
105 V.
6 A
a
b
5 Ω
(a)
5 Ω
10 Ω
10 Ω
10 V
+
–
a
b
5 Ω
(b)
5 Ω
10 Ω
10 Ω
Fig. 5.5-6 Single-source
Circuits for Finding
Contributions to
Open-circuit Voltage
Across a–b
a
b
5 Ω
5 Ω
10 Ω
10 Ω
Fig. 5.5-7 The Deactivated Circuit ...
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