Re: Ohm's law for AC?

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Article #2821

#2821

From: Michael A. Covington
Date: Mon, 13 Apr 1992 17:05

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What was the impedance at the end of the cable?  That is, what were you
using for a load?

The circuit looks like this:

             R1
       -----//////--------+
Signal                    / R3
source                    /
       -----//////--------+
             R2

Here R1+R2 is the resistance of the cable, and R3 is the resistance of the
load (the thing you're measuring the voltage at).

Let V represent the voltage at the source.  Then the voltage across R3
will be   V * R3 / (R1 + R2 + R3).

If R1+R2 = 5 ohms and R3 = 10 million ohms (the normal input impedance of
a digital multimeter) then R1+R2 will be negligible.
-- 
==========================================================================
Michael A. Covington, Ph.D. |  mcovingt@uga.cc.uga.edu  |  ham radio N4TMI
Artificial Intelligence Programs | U of Georgia | Athens, GA 30602  U.S.A.
==========================================================================

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Path: arc1-ca.pugleaf.net!sparky!uunet!usc!sol.ctr.columbia.edu!emory!athena.cs.uga.edu!mcovingt
From: mcovingt@athena.cs.uga.edu (Michael A. Covington)
Newsgroups: sci.electronics
Subject: Re: Ohm's law for AC?
Message-ID: <1992Apr13.170559.8310@athena.cs.uga.edu>
Date: 13 Apr 92 17:05:59 GMT
References: <oduNdSC00UhB81Orc1@andrew.cmu.edu>
Organization: University of Georgia, Athens
Lines: 25

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