potential dividers
Ok so in my book, there's a diagram that looks something like this
Then it says this (,which I don't understand at all):
This circuit is mainly used for calibrating voltmeters, which have a very high resistance. If you put something with a relatively low resistance across R1 though, you start to run into problems. You've effectively got two resistors in parallel, which will always have a total resistance less than R1. That means that Vout will be less than you've calculated, and will depend on what's connected across R1.
Isn't low resistance better? Why isn't it the case here?
Then it says this (,which I don't understand at all):
This circuit is mainly used for calibrating voltmeters, which have a very high resistance. If you put something with a relatively low resistance across R1 though, you start to run into problems. You've effectively got two resistors in parallel, which will always have a total resistance less than R1. That means that Vout will be less than you've calculated, and will depend on what's connected across R1.
Isn't low resistance better? Why isn't it the case here?
(edited 11 years ago)
Message says 'Invalid attachment link' contact administrator
Can you repost the diagram?
Can you repost the diagram?
Original post by uberteknik
I re-posted it. Does it work now?
It means that if the voltmeter you use has an internal resistance comparable to the that of the resistance across which you are measuring the voltage, you will run into problems and will then need to perform a compensation calculation to get the true value from the measured value.
i.e. if the voltmeter has an internal resistance of say 200M Ohms and you are measuring the voltage across a 100M Ohm resistor, you will get an error in the reading. i.e. the sorts of values used when calibrating the instrument.
This is because the current flowing through the circuit resistance will then be comparable to the current flowing through the voltmeter and hence the voltage dropped across the circuit resistor will be less (the original voltage across the 100M Ohm resistor is now dropped across the combined parallel resistance of 100M ll 200M = 66.6 M Ohms.) In addition, the source voltage may also now droop if it cannot provide enough current to the combined voltmeter circuit.
In an ideal world, you would want the internal resistance of the voltmeter to be infinite so that none of the circuit current os diverted through it and hence the actual voltage is unchanged.
In older AVO (analogue meters) it is common to have an internal resistance of 20K Ohms.
So it's important to know the errors when making the measurement caused by a combination of the internal voltmeter resistance and the circuit resistance across which the voltage is being measured.
i.e. if the voltmeter has an internal resistance of say 200M Ohms and you are measuring the voltage across a 100M Ohm resistor, you will get an error in the reading. i.e. the sorts of values used when calibrating the instrument.
This is because the current flowing through the circuit resistance will then be comparable to the current flowing through the voltmeter and hence the voltage dropped across the circuit resistor will be less (the original voltage across the 100M Ohm resistor is now dropped across the combined parallel resistance of 100M ll 200M = 66.6 M Ohms.) In addition, the source voltage may also now droop if it cannot provide enough current to the combined voltmeter circuit.
In an ideal world, you would want the internal resistance of the voltmeter to be infinite so that none of the circuit current os diverted through it and hence the actual voltage is unchanged.
In older AVO (analogue meters) it is common to have an internal resistance of 20K Ohms.
So it's important to know the errors when making the measurement caused by a combination of the internal voltmeter resistance and the circuit resistance across which the voltage is being measured.
(edited 11 years ago)
Original post by celina10
Ok so in my book, there's a diagram that looks something like this (,I apologize in advance for my terrible drawing skills 
)
Then it says this (,which I don't understand at all):
This circuit is mainly used for calibrating voltmeters, which have a very high resistance. If you put something with a relatively low resistance across R1 though, you start to run into problems. You've effectively got two resistors in parallel, which will always have a total resistance less than R1. That means that Vout will be less than you've calculated, and will depend on what's connected across R1.
Isn't low resistance better? Why isn't it the case here?
) Then it says this (,which I don't understand at all):
This circuit is mainly used for calibrating voltmeters, which have a very high resistance. If you put something with a relatively low resistance across R1 though, you start to run into problems. You've effectively got two resistors in parallel, which will always have a total resistance less than R1. That means that Vout will be less than you've calculated, and will depend on what's connected across R1.
Isn't low resistance better? Why isn't it the case here?
Low resistance is what you need for an ammeter. High resistance for a voltmeter.
Because an ammeter is placed in series with the component or circuit you want to measure the current in, you don't want it to add to the resistance already there. This would reduce the very current you were trying to measure. So ammeters need to have as little resistance as possible so they don't affect the circuit they are placed in.
Voltmeters are placed in parallel with the component you wish to measure the voltage across. You don't want the voltmeter to draw current out of the main circuit and through itself. This will increase the overall current in the circuit and change the voltage you were trying to measure. So voltmeters need a very high resistance so that when you connect them in parallel they don't draw any current (ideally) from that circuit.
Original post by Stonebridge
Low resistance is what you need for an ammeter. High resistance for a voltmeter.
Because an ammeter is placed in series with the component or circuit you want to measure the current in, you don't want it to add to the resistance already there. This would reduce the very current you were trying to measure. So ammeters need to have as little resistance as possible so they don't affect the circuit they are placed in.
Voltmeters are placed in parallel with the component you wish to measure the voltage across. You don't want the voltmeter to draw current out of the main circuit and through itself. This will increase the overall current in the circuit and change the voltage you were trying to measure. So voltmeters need a very high resistance so that when you connect them in parallel they don't draw any current (ideally) from that circuit.
Because an ammeter is placed in series with the component or circuit you want to measure the current in, you don't want it to add to the resistance already there. This would reduce the very current you were trying to measure. So ammeters need to have as little resistance as possible so they don't affect the circuit they are placed in.
Voltmeters are placed in parallel with the component you wish to measure the voltage across. You don't want the voltmeter to draw current out of the main circuit and through itself. This will increase the overall current in the circuit and change the voltage you were trying to measure. So voltmeters need a very high resistance so that when you connect them in parallel they don't draw any current (ideally) from that circuit.
But why does the resistance have different effects on a parallel and series circuit? And what happens to the voltage in both circuits? Does it decrease in the parallel circuit, and increase in the series or something?
Original post by steghead
i can drive.
good for you
I guess?R1 and the internal resistance of the voltmeter can be thought of as one resistor, with combined resistance 1/((1/R1) + (1/Rmeter)).
The lower the resistance of the meter, the lower the combined resistance of R1 and the resistance of the meter.
I = V/Rt, so
the current in the circuit I = Vin/(R1 + R2)
V = IR
so Vout =(R1 x Vin)/(R1 + R2)
So the smaller R1 is, The smaller percentage of the voltage input Rout is.
The lower the combined resistance of R1 and Rmeter is, the lower Rout is.
The lower the resistance of the meter, the lower the combined resistance of R1 and the resistance of the meter.
I = V/Rt, so
the current in the circuit I = Vin/(R1 + R2)
V = IR
so Vout =(R1 x Vin)/(R1 + R2)
So the smaller R1 is, The smaller percentage of the voltage input Rout is.
The lower the combined resistance of R1 and Rmeter is, the lower Rout is.
Original post by Associativity
R1 and the internal resistance of the voltmeter can be thought of as one resistor, with combined resistance 1/((1/R1) + (1/Rmeter)).
The lower the resistance of the meter, the lower the combined resistance of R1 and the resistance of the meter.
I = V/Rt, so
the current in the circuit I = Vin/(R1 + R2)
V = IR
so Vout =(R1 x Vin)/(R1 + R2)
So the smaller R1 is, The smaller percentage of the voltage input Rout is.
The lower the combined resistance of R1 and Rmeter is, the lower Rout is.
The lower the resistance of the meter, the lower the combined resistance of R1 and the resistance of the meter.
I = V/Rt, so
the current in the circuit I = Vin/(R1 + R2)
V = IR
so Vout =(R1 x Vin)/(R1 + R2)
So the smaller R1 is, The smaller percentage of the voltage input Rout is.
The lower the combined resistance of R1 and Rmeter is, the lower Rout is.
Sorry I don't get it
, I don't understand the equations.(edited 11 years ago)
Basically, the lower the resistance of the voltmeter, the lower the combined resistance of the voltmeter and R1.
The lower the combined resistance of R1 and the voltmeter, the lower Vout will be.
sorry if I have confused you.
The lower the combined resistance of R1 and the voltmeter, the lower Vout will be.
sorry if I have confused you.
Original post by celina10
Because resistances in series add as R = R1 + R1 + R3 etc (have you been taught this?)
and in parallel they add as 1/R = 1/R1 + 1/R2 + 1/R3
Do the sums.
If you add two 4 Ohm resistors in series the total increases. 4 + 4 = 8
Total 8 Ohms
If you add them in parallel the total gets less 1/R = 1/4 + 1/4 = 1/2
So R = 2 Ohms
The rule is:
Ammeters need to have low resistance and are placed in series with the part of the circuit you wish to measure.
Voltmeters are placed in parallel with the component you wish to measure the pd across (so it has the same pd across it as the other component) and need a large resistance.
The reasons I gave in my first post.
Original post by Stonebridge
Because resistances in series add as R = R1 + R1 + R3 etc (have you been taught this?)
and in parallel they add as 1/R = 1/R1 + 1/R2 + 1/R3
Do the sums.
If you add two 4 Ohm resistors in series the total increases. 4 + 4 = 8
Total 8 Ohms
If you add them in parallel the total gets less 1/R = 1/4 + 1/4 = 1/2
So R = 2 Ohms
The rule is:
Ammeters need to have low resistance and are placed in series with the part of the circuit you wish to measure.
Voltmeters are placed in parallel with the component you wish to measure the pd across (so it has the same pd across it as the other component) and need a large resistance.
The reasons I gave in my first post.
and in parallel they add as 1/R = 1/R1 + 1/R2 + 1/R3
Do the sums.
If you add two 4 Ohm resistors in series the total increases. 4 + 4 = 8
Total 8 Ohms
If you add them in parallel the total gets less 1/R = 1/4 + 1/4 = 1/2
So R = 2 Ohms
The rule is:
Ammeters need to have low resistance and are placed in series with the part of the circuit you wish to measure.
Voltmeters are placed in parallel with the component you wish to measure the pd across (so it has the same pd across it as the other component) and need a large resistance.
The reasons I gave in my first post.
Oh ok I understand, and I have been taught it but I think I just got confused. What about the voltage? When there's high resistance in a parallel circuit does it increase or decrease? And does it increase or decrease in a series circuit when the resistance is higher?
Original post by celina10
What about the voltage? When there's high resistance in a parallel circuit does it increase or decrease? And does it increase or decrease in a series circuit when the resistance is higher?
I'm sorry but the question is too vague to be answered accurately.
It depends on the circuit and what voltage you are referring to.
In ANY circuit, V=IR
This formula will tell you how the voltage will change.
If the current stays the same, then increasing the resistance will increase the voltage.
If the resistance stays the same, increasing the current will increase the voltage.
If you have a constant pd (which is often the case) then increasing the resistance decreases the current. (It's all Ohm's Law stuff.)
Without an actual circuit to discuss it's a bit difficult to go any further.
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