Re: Wheatstone bridge problem
- From: "Chris" <cfoley1064@xxxxxxxxx>
- Date: 5 Jun 2005 09:32:16 -0700
I guess I should mention one or two other things here. The Kelvin
measurement of resistance shown above didn't show the voltmeter:
` ___ / \
` .-|___|-( A )----.
` | 10 ohm \_/ |
` | o<----.
` | | |
` 10V | .-. / \
` --- Rx | | ( V )
` - | | \_/
` | '-' |
` | | |
` | o<----'
` | |
` '----------------'
created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de
Make sure you place the voltmeter leads right on the soleniod leads
where you want to make the measurement. If you place the leads on the
ammeter and the battery, you will be measuring the resistance of the
leads between the ammeter and the solenoid, and between the solenoid
and the battery. And also, you'll need a 10 ohm, _10 watt_ resistor.
10V * 1A = 10 watts, P = V * I
Second, a lot of people have those cheapie DVMs that only measure DC
current to 200mA or so. If you can't find a DVM that can measure DC
current of 1A, use a 100 ohm 1 watt resistor in place of the 10 ohm
resistor, giving you 100mA test current. You will only be measuring
20mV for an 0.2 ohm test resistor, but you'll still be way ahead of the
game if you've got a 200mV DC range on your DVM, despite reduced
voltmeter accuracy.
By the way, don't be too surprised if your calculated ohms are quite a
bit different than what the Wheatstone bridge led you to expect.
You'll be getting a more accurate reading this way, though.
If you have any questions or additional problems, don't hesitate to
post back.
Good luck
Chris
.
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