|WikiProject Electronics||(Rated Start-class, Mid-importance)|
|WikiProject Electrical engineering||(Rated Start-class, Mid-importance)|
Image of a meter
I WANT A PICTURE IN HERE!!! --Cyberman 05:29, 2 Feb 2005 (UTC)
- You asked for it, you got it! It's not a great picture, but it's better than nothing! Someone should find us or shoot us a nice photo of a D'Arsonval meter movement to go along with the diagram.
- We should probably also try to rationalize the somewhat-redundant text between here and the Galvanometer article.
- Atlant 14:09, 2 Feb 2005 (UTC)
It's good that Atlant made an illustration, but it's so different from a real-world meter movement that it really, seriously needs to be replaced; sorry! I added text, trying to be diplomatic, to the caption. Wikipedia should not have a misleading illustration.
First, the pole faces are arcs of a circle, not the curiously-curved shapes shown. Second, the front and rear hairsprings carry current to the coil, not the suggested flexible leads. Third, the coil is wound on a rectangular bobbin, not the iron core, as the image implies. Fourth, the core is stationary, supported by a non-magnetic metal plate or such. Please have a look at a Weston-type movement!
current to voltage
Could someone explain how to convert a current meter to a voltmeter using a resistor?
- Easy...you simply attach the ammeter and a huge resistor in series, and hook up the two ends of this "voltmeter" circuit to the two positions in the target circuit you want to measure the potential difference between. Then, you measure the current running through this part of the circuit (it will be small because of the huge resistor). You now have the resistance, and the current through this voltmeter circuit, and therefore can calculate the potential difference with V=IR. Ed Sanville 02:46, 2 August 2005 (UTC)
- I am trying to work out how to intergrate this into wiki, do you people think this is worth an article?mickpc
What will happen if an ammeter is connected in parallel with a resistor ? —Preceding unsigned comment added by 126.96.36.199 (talk) 18:35, 12 March 2008 (UTC)
What happens depends upon the value of the resistor. If it's a milliohm-range shunt, the ammeter measures much higher currents. If the resistor is something like a sizable fraction of an ohm, it will make the ammeter read low, the amount depending upon the relative resistance of the ammeter and the resistor. If the resistor is many ohms or greater, it will have no effect, BUT -- if that resistor has a voltage across it, the ammeter will effectively short-circuit the resistor. That's likely to cause trouble.
". Not used in practice due to non-linearity of the scale."
A superb reference, if a copy can be found
The page says that the resistance of a ammeter is kept as low as possible. How low is this? Would this affect the results when measuring?