# Voltage that would produce full-scale deflection of a meter

• moenste
In summary: R = V / I = 1 V / 10 mA = 100 Ohm.In summary, the shunt resistance should be 0.4 Ohm and the bobbin resistance should be 100 Ohm in order to convert the galvanometer to a meter reading up to 1.0 V full-scale deflection, with the shunt inadvertently left in position. The voltage needed to produce full-scale deflection of the meter is 60.4 V.
moenste

## Homework Statement

A galvanometer of resistance 40 Ω requires a current of 10 mA to give a full-scale deflection. A shunt is put in position to convert it to a meter reading up to 1.0 A full-scale deflection. A resistance bobbin, intended to convert the galvanometer to one reading up to 1.0 V full-scale deflection, is now attached to the instrument but the shunt is inadvertently left in position. By considering the potential difference across the bobbin and the meter, calculate the voltage which would produce full-scale deflection of the meter.

2. The attempt at a solution
V = I R = 10 * 10-3 * 40 = 0.4 V and 1 * 40 = 40 V. Total V = 40.4 V.

I know that a galvanometer is an ammeter designed for more accurate measurements of A and a shunt is a low-resistance resistor. But what is a resistance bobbin?

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A resistance bobbin is an accurate resistor formed by winding resistive wire around a reel or bobbin until the right resistance is achieved. For this purpose, just think of it as an accurate resistor.

The problem statement would make more sense if "is not attached" was changed to "is attached".

moenste
Jonathan Scott said:
A resistance bobbin is an accurate resistor formed by winding resistive wire around a reel or bobbin until the right resistance is achieved. For this purpose, just think of it as an accurate resistor.

The problem statement would make more sense if "is not attached" was changed to "is attached".
Oh sorry, it's "now" instead of "not", a typo.

moenste said:
Oh sorry, it's "now" instead of "not", a typo.
That makes more sense.
Do you understand how you would add the resistor to convert the original galvanometer to measure 1V? And how leaving the ammeter shunt in will affect the result?

moenste
Jonathan Scott said:
That makes more sense.
Do you understand how you would add the resistor to convert the original galvanometer to measure 1V? And how leaving the ammeter shunt in will affect the result?
The original galvanometer measure is V = I R = 10 mA * 40 Ω = 0.4 V. Now we need to change this measure to 1 V. Right? And since it is a galvanometer (or an ammeter), we need to R = V / I = 1 V / 10 mA = 100 Ω in parallel to the galvanometer. Is this correct until this point?

Not sure about the shunt though.

Clue: the word "parallel" is not correct.

moenste
You are correct about needing a total resistance of 100 Ω in order for 1 V across it to cause 10 mA of current. So you need to increase the resistance from 40 Ω to 100 Ω.

moenste
Jonathan Scott said:
Clue: the word "parallel" is not correct.
I had a similar problem in my book and it had "in parallel". Here is it:
A moving-coil meter has a resistance of 5.0 Ω and full-scale deflection is produced by a current of 1.0 mA. How can this meter be adapted for use as: (a) a voltmeter reading up to 10 V, (b) an ammeter reading up to 2A?

Answers: (a) 9995 Ω in series, (b)2.5 * 10-3 Ω in parallel.
Why does it have in parallel then?

Jonathan Scott said:
You are correct about needing a total resistance of 100 Ω in order for 1 V across it to cause 10 mA of current. So you need to increase the resistance from 40 Ω to 100 Ω.
So we need an increase of 60 Ohm.

This is how I see the problem:

We have a galvanometer with resistance 50 Ohm and current 10 mA. A shutn of 1 A and a bobbin of 1 V. What am I missing here?

You seem to have missed a key point. For an ammeter, you need a shunt which is a low resistor in parallel to the galvanometer in order to allow most of the current to bypass it. For a volt meter you need a resistor in series with the galvanometer in order to limit the current to just the amount needed.

moenste
Jonathan Scott said:
You seem to have missed a key point. For an ammeter, you need a shunt which is a low resistor in parallel to the galvanometer in order to allow most of the current to bypass it. For a volt meter you need a resistor in series with the galvanometer in order to limit the current to just the amount needed.
This should be correct:

That's the right layout now. The shunt resistance should be such that when 1A flows in parallel through the combination of galvanometer and shunt, the current through the galvanometer is 10mA. The bobbin resistance should be such that if the shunt were not connected, the current through the galvanometer would be 10mA when there is 1V across the combination of galvanometer and resistor. You should then be able to determine the answer to the original question.

moenste
Jonathan Scott said:
That's the right layout now. The shunt resistance should be such that when 1A flows in parallel through the combination of galvanometer and shunt, the current through the galvanometer is 10mA. The bobbin resistance should be such that if the shunt were not connected, the current through the galvanometer would be 10mA when there is 1V across the combination of galvanometer and resistor. You should then be able to determine the answer to the original question.
Alright, let's see:
Jonathan Scott said:
The shunt resistance should be such that when 1A flows in parallel through the combination of galvanometer and shunt, the current through the galvanometer is 10mA.
First we find the voltage that flows through the galvanometer: V = I R = 10 * 10-3 * 40 = 0.4 V. Since shunt is parallel to the galvanometer, its voltage is also equal to 0.4 V. Therefore the resistance of the shunt is: R = V / I = 0.4 / 1 = 0.4 Ohm. Correct?

Jonathan Scott said:
The bobbin resistance should be such that if the shunt were not connected, the current through the galvanometer would be 10mA when there is 1V across the combination of galvanometer and resistor.
Hm, we have a current of I = 10 mA across the galvanometer, and both galvanometer and the bobbin (resistor) have a voltage of 1 V. I think since they are in series, therefore the current of the bobbin would also be 10 mA. And then we find the total resistance R = V / I = 1 / 10 mA = 100 Ohm. This is the total resistance. Since we know that the galvanometer has a resistance of 40 Ohm, we can find the resistance of the bobbin = 100 - 40 = 60 Ohm. And so the voltage of the bobbin is V = I R = 10 * 10-3 * 60 = 0.6 V. Correct?

moenste said:
First we find the voltage that flows through the galvanometer: V = I R = 10 * 10-3 * 40 = 0.4 V. Since shunt is parallel to the galvanometer, its voltage is also equal to 0.4 V. Therefore the resistance of the shunt is: R = V / I = 0.4 / 1 = 0.4 Ohm. Correct?
Very nearly, but if you have 1 A going through the shunt and 10 mA through the galvanometer that makes 1.01 A total. You actually want (1 A - 10 mA) going through the shunt. What's more important is the effective resistance of the combination of the galvanometer and shunt in parallel, which is actually simpler.
moenste said:
Since we know that the galvanometer has a resistance of 40 Ohm, we can find the resistance of the bobbin = 100 - 40 = 60 Ohm.
I agree.

moenste
Jonathan Scott said:
Very nearly, but if you have 1 A going through the shunt and 10 mA through the galvanometer that makes 1.01 A total. You actually want (1 A - 10 mA) going through the shunt. What's more important is the effective resistance of the combination of the galvanometer and shunt in parallel, which is actually simpler.

I agree.
Hm, I think I don't understand the problem quite well and that's why I have the difficulty with the calculations.

moenste said:
A galvanometer of resistance 40 Ω requires a current of 10 mA to give a full-scale deflection.
What does the "full-scale deflection" actually mean? I looked it up, but still not confident in the meaning. For a galvanometer of resistance 40 Ohm we need a current equal to 0.01 A to make the needle (not sure what's the correct name) go to +50 (on the right). Is this correct? FSD is like the maximum number on the apparatus? And which 50 should it go to? Left or right? Or it just means the maximum number -50 and +50 and doesn't specify the sign? Or FSD means it's always the maximum positive number?

moenste said:
A shunt is put in position to convert it to a meter reading up to 1.0 A full-scale deflection.
To convert it? What does it mean "it"?

A shunt is put to a galvanometer to make the galvanometer show the maximum number not with 0.01 A current but with 1 A current? So with a shunt the galvanometer shows FSD with I = 1 A. This is correct?

moenste said:
A resistance bobbin, intended to convert the galvanometer to one reading up to 1.0 V full-scale deflection, is now attached to the instrument
Convert the galvanometer one reading up to 1 V of FSD. What does this mean? I don't understand the logic behind it.

moenste said:
By considering the potential difference across the bobbin and the meter, calculate the voltage which would produce full-scale deflection of the meter.
OK, so we need to find the voltage of the bobbin and of the galvanometer. And then we need to find the voltage that would produce the maximum result on the galvanometer? However, what voltage do we need to find? Of the whole circuit (like of the battery)?

And we need to voltages for what? Voltage of the whole circuit = V bobbin + V galvanometer = the required answer 60.4 V. Like this?

---

I am very sorry for a large question, but maybe it's my English and these terms that confuse me. I just want to understand what we have, what we need and how to find it. At this moment it's still not 100 % clear.

You were already very near to the solution.

You don't need to know whether your galvanometer is one-way (accepting current in one direction, with zero current at the lower end of the scale) or two-way as in your picture (accepting current both ways, with zero in the middle). All you need to know is that it needs 10 mA to read the maximum value.

You can use a galvanometer to make an ammeter by allowing the majority of the current to be shunted past the galvanometer. In this case, for a 1 A result, 10 mA of the total current needs to go through the galvanometer and the other 990 mA must go through the shunt, so if you wanted to calculate the shunt resistance, it must be a fraction 10/990 of the galvanometer resistance.

A simpler approach is to consider that full scale deflection requires a voltage of 0.4V across the galvanometer (as you already calculated) so if the total current is to be 1 A in this case, the overall effective resistance of the galvanometer and shunt parallel combination must be 0.4 Ω (which means that the resistance of the shunt on its own is actually 100/99 times 0.4 Ω, and you could can use the rule for parallel resistors to calculate that combined effect works correctly).

As you previously calculated, the bobbin is 60 Ω, on the assumption that the total resistance of the voltmeter will be 100 Ω. When the shunt is in position, the resistance of the galvanometer and shunt combination has been reduced from 40 Ω to 0.4 Ω. You should now be able to calculate the voltage necessary to give a full scale deflection on that basis.

moenste
Jonathan Scott said:
You don't need to know whether your galvanometer is one-way (accepting current in one direction, with zero current at the lower end of the scale) or two-way as in your picture (accepting current both ways, with zero in the middle). All you need to know is that it needs 10 mA to read the maximum value.
OK, this part I understand. And FSD is the maximum value on the galvanometer.

Jonathan Scott said:
You can use a galvanometer to make an ammeter by allowing the majority of the current to be shunted past the galvanometer.
"Shunted past the galvanometer" means that we connect the shunt in parallel to the galvanometer (as we did in the graph) to use the galvanometer as an ammeter, right?

Jonathan Scott said:
In this case, for a 1 A result, 10 mA of the total current needs to go through the galvanometer and the other 990 mA must go through the shunt, so if you wanted to calculate the shunt resistance, it must be a fraction 10/990 of the galvanometer resistance.
So the 1 A current is the current of the whole circuit, while the current that goes throught the galvanometer is 10 mA and as a result, 1 A - 10 mA = 0.99 A -- this is the current that goes through the shunt.

Resistance of galvanometer is 40 Ohm, its current is 0.01 A, voltage is 0.4 V. Current of the shunt is 0.99 A, it's voltage is 0.4 V, its resistance is 0.4 / 0.99 = 0.4 Ohm. Correct?

Total resistance of the galvanometer and the shunt is 1 / R = 1 / 40 + 1 / 0.4 so R = 0.4 Ohm.

Jonathan Scott said:
As you previously calculated, the bobbin is 60 Ω, on the assumption that the total resistance of the voltmeter will be 100 Ω. When the shunt is in position, the resistance of the galvanometer and shunt combination has been reduced from 40 Ω to 0.4 Ω. You should now be able to calculate the voltage necessary to give a full scale deflection on that basis.
Hm, I'm now in doubt about the 60 Ohm.

We have a total current of the whole circuit equal to 1 A. We also know that the voltage of the parallel circuit is 0.4 V and the voltage of the bobbin is 1 V. Total V = 0.4 + 1 = 1.4 V. Total resistance R = V / I = 1.4 / 1 = 1.4 Ohm. Resistance of the bobbin is 1.4 Ohm - 0.4 Ohm of the parallel circuit = 1 Ohm.

Update:
Jonathan Scott said:
As you previously calculated, the bobbin is 60 Ω, on the assumption that the total resistance of the voltmeter will be 100 Ω. When the shunt is in position, the resistance of the galvanometer and shunt combination has been reduced from 40 Ω to 0.4 Ω. You should now be able to calculate the voltage necessary to give a full scale deflection on that basis.
If there is no shunt the resistance of the galvanometer would be R = V / I = 1 / 10 mA = 100 Ohm. Since it already has 40 Ohm, therefore 60 Ohm are required. They go to the bobbin. With the shunt the total resistance of the galvanometer and the shunt is 0.4 Ohm. So to find the total voltage we V = I R = 1 A across the whole circuit * (0.4 + 60) resistance of the parallel circuit and of the bobbin = 60.4 V.

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moenste said:
And FSD is the maximum value on the galvanometer.
Yes, FSD means "full scale deflection", that is deflection to the maximum reading on the scale.

moenste said:
"Shunted past the galvanometer" means that we connect the shunt in parallel to the galvanometer (as we did in the graph) to use the galvanometer as an ammeter, right?
Yes, in electric contexts a "shunt" relates to something which allows part of a current to bypass a circuit, as in a typical ammeter.

moenste said:
So the 1 A current is the current of the whole circuit, while the current that goes throught the galvanometer is 10 mA and as a result, 1 A - 10 mA = 0.99 A -- this is the current that goes through the shunt.

Resistance of galvanometer is 40 Ohm, its current is 0.01 A, voltage is 0.4 V. Current of the shunt is 0.99 A, it's voltage is 0.4 V, its resistance is 0.4 / 0.99 = 0.4 Ohm. Correct?

Total resistance of the galvanometer and the shunt is 1 / R = 1 / 40 + 1 / 0.4 so R = 0.4 Ohm.
Yes. I think that section would have been more clear if you left 0.4/0.99 as a fraction rather than rounding in two places. But as I said, it's not actually necessary to work out the exact resistance of the shunt, only of the shunt/galvanometer combination, given that the voltage across the combination must be 0.4 V when the total current is 1 A.

moenste said:
If there is no shunt the resistance of the galvanometer would be R = V / I = 1 / 10 mA = 100 Ohm. Since it already has 40 Ohm, therefore 60 Ohm are required. They go to the bobbin. With the shunt the total resistance of the galvanometer and the shunt is 0.4 Ohm. So to find the total voltage we V = I R = 1 A across the whole circuit * (0.4 + 60) resistance of the parallel circuit and of the bobbin = 60.4 V.
The wording could be tidied up a bit and the spelling is "parallel" but you now appear to have the correct result for the correct reason.

moenste
Jonathan Scott said:
it's not actually necessary to work out the exact resistance of the shunt, only of the shunt/galvanometer combination, given that the voltage across the combination must be 0.4 V when the total current is 1 A.
Yes, I see now what you mean. Indeed we have all of the numbers to find the resistance R = V of the parallel circuit 0.4 V / I of the whole circuit 1 A = 0.4 V.

Jonathan Scott said:
The wording could be tidied up a bit and the spelling is "parallel" but you now appear to have the correct result for the correct reason.
Oh, another typo, sorry for that.

What could be improved in the wording?

moenste said:
What could be improved in the wording?
That's up to you to sort out! Just read it carefully to make sure what you actually wrote, and try to write it as if you were explaining it clearly to someone else.

moenste

## 1. What is full-scale deflection of a meter?

Full-scale deflection of a meter refers to the maximum reading or value that a meter can display. This is often indicated by a "FSD" or "Full Scale" marking on the meter.

## 2. How is full-scale deflection determined for a meter?

The full-scale deflection of a meter is determined by the sensitivity and range of the meter. Sensitivity refers to the amount of voltage or current required to move the meter needle by one division, while range refers to the maximum value that the meter can measure. The product of sensitivity and range gives the full-scale deflection value.

## 3. What is the significance of knowing the voltage that produces full-scale deflection?

Knowing the voltage that produces full-scale deflection is important for accurately reading and interpreting measurements on the meter. It allows you to determine the maximum value being measured and make appropriate adjustments to the range or sensitivity of the meter if needed.

## 4. How can the voltage that produces full-scale deflection be calculated?

The voltage that produces full-scale deflection can be calculated by dividing the full-scale deflection value by the sensitivity of the meter. For example, if a meter has a full-scale deflection of 10 V and a sensitivity of 1 V/division, the voltage that produces full-scale deflection would be 10 volts.

## 5. Can the voltage that produces full-scale deflection vary for different types of meters?

Yes, the voltage that produces full-scale deflection can vary for different types of meters. It depends on the sensitivity, range, and design of the specific meter. It is important to consult the manufacturer's instructions or specifications for the specific meter being used to determine the voltage that produces full-scale deflection.

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