Deriving Thevenin equivalents
Rick Nelson, Chief Editor -- Test & Measurement World, 5/1/2006
A book review in Test & Measurement World's March issue (Ref. 1) takes an author to task for, in a chapter covering Thevenin's theorem (Ref. 2), failing to provide the Thevenin equivalent of a sample circuit he presents. The circuit appears in the figure.
To illustrate the power of Thevenin's theorem, this circuit can be analyzed in terms of voltage dividers and superposition, as outlined here. The Thevenin equivalent resistance is simply R1 in parallel with R2 in parallel with R3:
With V2 shorted:
With V1 shorted:
By superposition:
The Thevenin equivalent resistance is R1 in parallel with R2 in parallel with R3:
This result can be verified by solving the node equation i1+i2= i3 where:
Solving for VT yields:
Find further information and discussion at http://www.reed-electronics.com/tmworld/blog/640000064/post/1710002771.html.
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