How to Adjust Resistance in Voltmeters and Ammeters for Accurate Measurements

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To measure a larger potential difference with a voltmeter, increasing its resistance is necessary to prevent excessive current flow. A shunt in an ammeter creates an alternative path for current, allowing it to measure larger currents without damage. The discussion includes calculations for an ohmmeter setup, where a 6.0V battery and adjustable resistor are used to achieve specific current readings. Understanding circuit diagrams and applying Kirchhoff's Current Law is crucial for resolving the calculations involving external resistances. The participant eventually resolved their confusion regarding the questions, highlighting the importance of accurate information in educational materials.
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Hello!

I'm doing some questions in my book and have hit a problem area.

#1) How would you change the resistance of a voltmeter to allow the voltmeter to measure a larger potential difference?

I was thinking that you would have to increase the resistance is order to measure a larger potential difference but I am not quite sure.

#2) How would you change the shunt in an ammeter to allow the ammerter to measure a larger current?

For this one I don't even know what a shunt is, my textbook hasn't really dealt with it at all. I'm not really sure what to say about it.

#3) An ohmmeter is made by connecting a 6.0V battery in series with an adjustable resistor and an ideal ammeter. The ammeter deflects full scale witha current of 1.0mA. The two leads are touched together and the resistance is adjusted so 1.0mA current flows.

a) What is the resistance of the adjustable resistor?

V/I = R 6.0V/1x10^-3A = 6000 ohms

b) The leads are now connected to an unknown resistance. What external resistance would produce a reading of 0.50 mA, half full scale?

The answer is also 6000 ohms but I'm not sure why

c) What external resistance would produce a reading of 0.25 mA, quarter scale?

The answer is 18000 ohms but I'm not sure why


d) What external resistance would produce a reading of 0.75mA, three quarter full scale?

The answer is 2000 ohms but once again I am not sure why.


I'm sure that these questions are absolutely stupid, but I really can't get my mind around them. I've done all the hard ones without a problem now I get to these and I'm stuck :rolleyes: Any help you can give I would really appreciate! Thanks!
 
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scorpa said:
Hello!

I'm doing some questions in my book and have hit a problem area.

#1) How would you change the resistance of a voltmeter to allow the voltmeter to measure a larger potential difference?

I was thinking that you would have to increase the resistance is order to measure a larger potential difference but I am not quite sure.

#2) How would you change the shunt in an ammeter to allow the ammerter to measure a larger current?

For this one I don't even know what a shunt is, my textbook hasn't really dealt with it at all. I'm not really sure what to say about it.

Thanks!

#1) Think about how a voltmeter works. How do you hook one up to measure a voltage? (you might draw a simple circuit with a battery and a load resistance and also draw how you hook up your voltmeter). How many paths can current go in your simple circuit?

#2 Shunt is to create an additional path to divert the flow of something.
If you had a river next to a stream, by opening up a channel between the river and stream, some of the river water will be diverted to the stream while the rest of the water will continue in its original path. Depending on how large your channel was, this will impact, how much water is diverted over time.
 
scorpa said:
#3) An ohmmeter is made by connecting a 6.0V battery in series with an adjustable resistor and an ideal ammeter. The ammeter deflects full scale witha current of 1.0mA. The two leads are touched together and the resistance is adjusted so 1.0mA current flows.

3a) What is the resistance of the adjustable resistor?

V/I = R 6.0V/1x10^-3A = 6000 ohms assuming you know what characteristics an ideal ammeter has, you are correct

3b) The leads are now connected to an unknown resistance. What external resistance would produce a reading of 0.50 mA, half full scale?

The answer is also 6000 ohms but I'm not sure why
A clear diagram of how your circuit is hooked up is important.
It would be useful to do that and post it here. We can then see how you are interpreting the information.
(Hint: you could model your Ammeter as a resistor in parallel with a voltmeter).

You may also want to review Kirchoff's Current Law. Once you understand that, answering (b), (c) and (d) will be very clear.

3c) What external resistance would produce a reading of 0.25 mA, quarter scale?

The answer is 18000 ohms but I'm not sure why


3d) What external resistance would produce a reading of 0.75mA, three quarter full scale?

The answer is 2000 ohms but once again I am not sure why.
 
OK thanks, I did figure the questions out eventually, I can't believe how stupid I was. I was going to post a diagram of the question but I don't know how to put it onto the forums. Turns out there was bad information in the question, but that didn't make the teacher omit it. Thanks again.
 
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