# Voltage question

Discussion in 'General Electronics Chat' started by teesee, Jan 29, 2015.

1. ### teesee Thread Starter New Member

Jan 29, 2015
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0
Basic Noobie Question
When I measure a 9 volt battery with my multimeter low and behold I get 9 volts. When I measure it with a 0-15 DCV panel volt meter (Radio Shack 270-1754) it blows right off the scale until I place a 15k ohm resistor between the negative battery terminal and the meter. I then get just short of 9 v
1) Why does the Volt meter not show 9 volts without the resistor?
2) At the same time does this also mean that the internal resistance of my multimeter is also 15k

Thanks

Jul 18, 2013
10,507
2,367
Most likely the meter is a moving coil Ma meter, and requires a series resistor to tailor to a required full scale voltage required.
http://home.cogeco.ca/~rpaisley4/Ammeter.html
It appears to need a 15k resistor in series according to the spec?
Max.

3. ### ian field Distinguished Member

Oct 27, 2012
4,413
782
All moving coil meters are mA or uA meters, if you want them to read volts they need a multiplier resistance in series - if you know the FSD current and the movement coil resistance, its easy to calculate the series resistance for a given voltage.

Usually a voltmeter will have the resistor mounted inside the case - but anyone can take the dial plate out of a mA meter and have it re-printed in volts.

4. ### MikeML AAC Fanatic!

Oct 2, 2009
5,450
1,066
Yes. Here is the spec. It specifically requires an external 15K resistor to make the meter movement into a 0-15V voltmeter.

Question to teesee: how many volts would you have to apply to meter terminals to make the meter read full scale without the external resistor. Hint, you can calculate it based on the specs I posted.

5. ### teesee Thread Starter New Member

Jan 29, 2015
12
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Hello MikeML. Thanks for the challenge .... so here goes.
I solved for I=V/R = 9volt battery/15085 (15k ohm resister + 85 ohm internal resistance) = 0.0006 amps THEN .....
I solved for V=IR = .0006 Amps * 85 (85 ohm internal resistance) = .0513 volts

Critique always welcome. Thanks

Jan 15, 2015
963
232

The details are hidden in the specifications Mike posted. Note the meter sensitivity specification. That is a hint combined with the approximate coil resistance.

Ron

7. ### ian field Distinguished Member

Oct 27, 2012
4,413
782
From the dim and distant past, I vaguely remember 2 different voltage standards for off the shelf shunts - where possible I used the lower voltage meter with the higher voltage shunt and add a 10 turn preset between the meter & shunt for calibration.

For multiplier resistors on voltmeters, its possible to use a lower resistor to accommodate the addition of a 10 turn pot for calibration - also handy if the nearest preferred value resistor isn't quite what you were hoping for.

8. ### teesee Thread Starter New Member

Jan 29, 2015
12
0
Second Attempt. If I understand what I'm reading about this type of meter I have a 1mA (current) meter movement available (Not to be exceeded) and an 85 ohm resistance which I assume has to do with the coil so the volts required as asked would be 0.085 DCV to deflect to full scale without using the multiplier resistor..
Fingers are crossed.

Thanks again

Last edited: Jan 30, 2015

Jan 15, 2015
963
232
You got it.

So you take 15 Volts - .085 Volts = 14.915 Volts / .001 Amp = 14.915 K Ohm. Thus a 15 K resistor.

Ron

10. ### teesee Thread Starter New Member

Jan 29, 2015
12
0
Ron .. Thank you very much for the smiley accolade. I'd really like to thank ALL who had input and created thought for me with this issue. Having said that, while I was googling to understand about these meters I came across this little schematic which I hopefully include.
Where my gauge had a 1 mA sensitivity and a 85 ohm internal resistance this image which I have attached shows a 1 mA and 200 ohm equivalent and they were trying to solve for the multiplier resistance. I have two questions relating to this same thread

1) In the schematic should there be a reference to a 1 mA current between the +Vin and the Rmult (49,800 ohms) or should it not be between the Rmult and the + side of the gauge?
2) Having asked that previous question ...With this very simple circuit in mind it has the multiplier resistor between the +Vin and the + terminal of the gauge. Here comes another noobie for you. I would have thought that the resistor would have to be in-line before the gauge on the negative side in order to limit the current entering into the gauge. Where am I missing the boat?

I hope I am permitted to continue this within the same thread since I believe it is totally related. If my mistake pls let me know.

Many thanks.

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Jan 15, 2015
963
232
This is your thread, take it where you wish and you are still on your original topic. Just my thinking on that.

When it comes to an analog voltmeter think of things this way. We start with an analog meter movement, movements come in several flavors but that aside for now in your movement it is a current driven device. Meters can be 1 mA F/S (Full Scale) or any number of currents. The old well loved Simpson 260 and 270 series meters used a 50 uA F/S movement. Anyway, in the example you just posted the F/S current is again 1 mA. The meter movement internal resistance is 200 Ohms so .001 Amp x 200 = .2 Volts F/S. Want to measure 50 Volts? We make a 0 to 50 Volt scale and glue it to our meter movement. So now we get 50 Volts - 0.2 Volts = 49.8 Volts / .001 Amp = 49,800 ohms 49.8 k ohms. Actually the series resistor is limiting current through the meter movement. If we draw it out in a schematic it matters not which side of the meter movement it is on. Believe me the meter won't care. Long as the range resistor for volts is in series with the meter movement it will limit the current.

Ron

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12. ### teesee Thread Starter New Member

Jan 29, 2015
12
0
Thanks Ron for the time you have spent helping me on this.
Best regards