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Thevenin and Norton Equivalent

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Date Posted: 7/22/2008 1:48:52 AM  Status: Live
Thevenin and Norton Equivalent
Course Textbook Chapter Problem
N/A N/A N/A N/A
Question Details:
The way we measure the Thevenin & Norton Equivalent of a circuit is to look at the
open-circuit voltage, Voc = Vt, and short-circuit current, Isc = In. The Rt and Rn can
then be found using Ohm's law:
Rt = Rn = (Voc/Isc)

In a practical application, a more realistic model of an Ammeter and a Voltmeter are
shown in the figure below. The internal resistances, rm, on the Ammeter is in the order
of rA ˜ 0.1[Ω] and for the Voltmeter is in the order of rV  ˜ 1[MΩ].

For a given circuit (unkown network), measurements (using the realistic instruments) of
the open-circuit (Thevenin) voltage gives Voc = 10.00[V], and the short-circuit (Norton") current gives Isc = 0.100[A].
a) Evaluate the apparent resistance, Rapparent, using the formula above (eq. 3) with
the apparent "voc" and "isc".
b) Estimate the actual equivalent resistance, Thevenin Voltage and Norton Current
taking into account the internal resistances of the sensors, rA, and rV.

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Date Posted: 7/22/2008 7:12:13 AM  Status: Live
Asker's Rating: None Provided    Moderator's Rating: N/A
Response:
Given,
(a)
(b)
Let
When the internal resistances are considered the thevenin equivalent circuit will be
 
 
By voltage division rule
When the internal resistances are considered the norton equivalent circuit will be
 
 
By current division rule
Now

Hope this helps you...



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