How Do You Calculate the Series Resistance for a Voltmeter Using a Galvanometer?

[Myr73]

A galvanometer of coil resistance 50 Ω deflects full scale for a current of 3.50mA. What series resistance should be used with this galvanometer to construct a voltmeter which deflects full scale for 35.0 V ?

r=50 Ω I=3.50mA=0.0035 V=35.0v R=?


I= V/R

The Attempt at a Solution

I honestly don't know- There is very little explanation in the unit -Help me get started please?

 

[berkeman]

A galvanometer of coil resistance 50 Ω deflects full scale for a current of 3.50mA. What series resistance should be used with this galvanometer to construct a voltmeter which deflects full scale for 35.0 V ?

r=50 Ω I=3.50mA=0.0035 V=35.0v R=?


I= V/R

The Attempt at a Solution

I honestly don't know- There is very little explanation in the unit -Help me get started please?

I would recommend that you read the wikipedia article on Galvanometers. That should help you form the mental picture that you need to solve this. If you still have problems with this, post the link to the wikipedia article that you read, and ask specific questions about that material.

 

[CWatters]

A galvanometer of coil resistance 50 Ω deflects full scale for a current of 3.50mA.

What series resistance should be used with this galvanometer to construct a voltmeter which deflects full scale for 35.0 V?

First thing to do is draw and label both circuits.

It's the same galvanometer in both cases. Same deflection required (eg "full scale") so have a think about what else will need to be the same both circuits?

 

[Myr73]

Ok, So this is what I came up with, does it makes sense?
R1= 50 Ω Ig= 3.5mA=0.0035A V=35.5V R=? In series , therefore Ig=Iv=I

V=I/R V=V1+V2=IR1+IR2

V1= 0.0035/50=0.00007 V
{V-V1}/ I=R2= 10143 Ω= 10 k Ω
The voltmeter will consist of a resistance R ser = 10k Ω in series with the galvanometer.

 

[rude man]

Ok, So this is what I came up with, does it makes sense?
R1= 50 Ω Ig= 3.5mA=0.0035A V=35.5V .

No.
Where did 35.5V come from?