Message-ID: <01C28984.9AD76150.g4jnt@thersgb.net>
From: "Andy talbot" <g4jnt@thersgb.net>
To: rsgb_lf_group@blacksheep.org
Date: Mon, 11 Nov 2002 13:16:39 -0000
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Subject: LF: RE: An answer to Monday's puzzle
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No Steve, this doesn't work.  - Although I fell into that trap initially.

For a Thevenin network,  if you connect an external 1V source, you will get 
either 2A or zero flowing depending on polarity, as both voltage sources are in 
series.

For a Norton, the external voltage source will force 1A though the internal 1R 
resistor and this will be added or subtracted, depending on polarity, from the 
internal 1A source,  which will will now be flowing completely through the 
external voltage source as this has an internal resistance of zero and will 
swamp the 1R - result zero or 2A as before.   Thev. and Norton are electrically 
identical so cannot be told apart. Remember superposition.

Congrats Hugh, you worked it out - use a thermometer, and to be really 
scientific measure the temperature when shorted, and when open circuit.

Andy


-----Original Message-----
From:	Steve Thompson [SMTP:steve@eltac.co.uk]
Sent:	2002/11/11 10:56
To:	rsgb_lf_group@blacksheep.org
Subject:	LF: An answer to Monday's puzzle

On Monday 11 November 2002 08:46, you wrote:
> You have a small box whose internal workings you can't see, but the box
> has two terminals. You measure these two terminals with a voltmeter, and
> it measures precisely 1V. You short the two terminals with an ammeter,
> and it reads precisely 1A. You then conclude the internal circuit is
> either a Thevenin source (an ideal 1V voltage source with a 1-ohm series
> resistor) or a Norton source (an ideal 1A current source with a 1-ohm
> shunt resistor). Using standard laboratory equipment, how could you
> determine which one it is.
I'm feeling pleased with myself - I honestly worked this out from scratch!

Apply a 1V external source and measure the current.

Steve


> thought the whole point of these theoretical sources is they were
>interchangeable.

>I did wonder if you could use a thermometer to measure the temperature rise,
>as the 1 ohm resistor would dissipate more heat when shortcirucited in the
>Thevenin configuraton, and none when open ciruit. Whereas the Norton config.
>would disipate more when open ciruited.
>But then, how much heat does an ideal voltage or current source dissipate ??
>It's all very theroretical. They are just mathematical tools really.
>73
>Hugh M0WYE