# The measurement of current and voltage

1. Jul 6, 2009

### runfast220

1. The problem statement, all variables and given/known data
Two scales on a voltmeter measure voltages up to 20.0 and 30.0V, respectively. The resistance connected in series with the galvanometer is 1680 ohms for the 20.0V scale and 2930 ohms for the 30.0V scale. Determine the coil resistance and the full-scale current of the galvanometer that is used in the voltmeter

2. Relevant equations

V=IR

3. The attempt at a solution
First I'm confused at is this all in same circuit or in two different ones?
To find Rc I'm pretty sure I need to use the equation: V= I/R+Rc.
I'm not sure how to get this problem started.

2. Jul 6, 2009

### mgb_phys

It's two separate circuits (usually with a switch to select the range)
Both the circuits have the same unknown resistance = the of the moving coil you are trying to find.

3. Jul 6, 2009

### runfast220

How would I start to find either the current or Resistance of the coil because in the equation I would have both of those as unknowns:
V= I/R+Rc

4. Jul 6, 2009

### mgb_phys

The current in the coil at full scale is the same on both ranges.

5. Jul 6, 2009

### vk6kro

No result yet?

Two unknowns, R and I, so you need two equations.

The voltages in the meter and the series resistor add up to the supplied voltage.
So, you would have one equation like this
30 = ....... + ..........
and another like this:
20 = ........ + ............

One of the ........'s would be the same in both equations, so solving it would be easy.

If you have already done it, I can compare answers (with what I got), for you if you like.

6. Jul 7, 2009

### runfast220

I don't understand what your doing with the voltages? I thought I needed to find the current and the resistance of the coil?

7. Jul 7, 2009

### vk6kro

OK.

There is a moving coil meter movement which takes a certain amount of current to make it read full scale. There are two resistors that can be put in series with the meter so that it takes 20 volts with one and 30 volts with the other to make the meter read full scale. If the resistors were not there, the meter would read full scale with a much lower voltage across it and you wouldn't be able to measure voltages like 15 volts or 25 volts.

The Formula V = I * R applies for the voltage across the meter movement and also for the voltage across the series resistance.

We need to find two things. The resistance R of the meter movement and the current I it takes to make the meter read full scale. So, we need to get two equations to work out these things.

One thing we know is that if you have two resistors in series, the voltage across both of them equals the sum of the voltages across each of them.

So, if we put 20 volts across the resistor and meter in series, the total must equal 20 volts.

So, you can write the equations like this:
20 volts = Voltage across resistor (I * 1680) + Voltage across meter ( I * R )
30 volts = Voltage across the other resistor ( I * 2930) + Voltage across meter ( I * R )

Now, you want to find I and R. I would probably take the top equation away from the bottom one using only the parts in brackets on the right side.

Last edited: Jul 7, 2009
8. Jul 8, 2009

### runfast220

ok this is what I have so far.

20V= (I*1680)+(I*R)
30V=(I*2930)+(I*R)

(I*R)= 20 - (I*1680)
(I*R)= 30 - (I*2930)
(I*R)=(I*R)
20 - (I*1680)=30 - (I*2930)
I=125

But then when I plug 125 back into the two equations to solve for R, I get different values for R in each equations. Looking at the equations I don't understand how it would be possible for the for the current and resistance to be the same in each equation?

9. Jul 8, 2009

### vk6kro

30 - 20 = 2930 I - 1680 I = 1250 I
I = 0.008 amps or 8 mA

The current is the same because the meter is reading full scale in each case and it takes exactly 8 mA to make it read full scale.

Now you need to plug this 0.008 A back into one of the equations and get R.

10. Jul 9, 2009

### runfast220

20= (.008*1680)+(.008*R)
R=820

Thank you for all your help!

11. Jul 9, 2009

### vk6kro

Yep, that's right. Well done.

Hope you understood what was going on though.

Can you see that if you have a 20 volt supply and the two resistances (of the resistance and the meter ) add up to 2500 ohms (1680 + 820) then the current will be 8 mA?

And if the supply is 30 volts and the total resistance is 3750 ohms (2930 + 820) then the current will also be 8 mA ?

If you measure 15 volts with the 30 volt meter resistor, the current will be 4 mA and the meter will read half scale.

You may not have come across this before, but it is important stuff to know.

12. Jul 9, 2009

### runfast220

It all makes sense now. Once again thanks for the help.

13. Aug 10, 2009

### jlobo

why is a high voltage 750 miliamps fuse rated as 5 kv and not 250 volts

14. Aug 10, 2009

### vk6kro

If the fuse melts, it has to stop the current flowing. So there has to be a big enough gap that there won't be an arc formed and the current continue to flow.

Some fuses put the fuse wire under spring tension so that if the wire melts, the ends where it melted are pulled far apart. This helps to stop arcing.

A 5000 volt rating tells you this fuse could stop 5000 volts. It can also stop 250 volts, of course.