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LF: Re: An answer to Monday's puzzle

To: [email protected]
Subject: LF: Re: An answer to Monday's puzzle
From: "g6tmk" <[email protected]>
Date: Mon, 11 Nov 2002 11:47:35 -0000
References: <[email protected]> <E18BCDB-0002zB-00@lutetium>
Reply-to: [email protected]
Sender: <[email protected]>
I 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


----- Original Message -----
From: "Steve Thompson" <[email protected]>
To: <[email protected]>
Sent: Monday, November 11, 2002 10:55 AM
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






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