Series resistance for 35V voltmeter with 50Ω galvanometer

[AUAO1]

Homework Statement

A galvanometer of coil resistance 50.0 Ω deflects full scale for a current of 3.50 mA. What series resistance should be used with this galvanometer to construct a voltmeter which deflects full scale for 35.0 V

Homework Equations

/Known Variables[/B]
R1 + R2... + Rn = Rtotal
R = V/I
Vv = 35V; Rg = 50 Ω; I = 3.50 mA or 0.0035 A

The Attempt at a Solution

Since a voltmeter occurs in a series, I have used the relevant equation above (Rv + Rg = Rtotal) where Rtotal = Vv/I (since current is constant). Replacing Rtotal with Vv/I gives Rv + Rg = Vv/I, so to find the resistance of the voltmeter, it would be Rv = (Vv/I) - Rg. Substituting the values in gives (35/0.0035) - 50 so Rv = 9,950 Ω (9.95 x 10^3 Ω). This means that Rtotal = 10,000 Ω (1.00 x 10^4 Ω).

Help and feedback is very much appreciated!

 

[gneill]

Hi AUAO1,

Welcome to Physics Forums!

Your reasoning and results look good.

They ask for the required resistance to put in series with the galvanometer, which you found to be 9950 Ω. The value of Rtotal wasn't necessary, but it's nice that you calculated it, too.

 

[AUAO1]

Hello and thank you gneill!

Thank you for the quick response and feedback. It's always nice to be sure that I'm on the right track with these questions so your help is much appreciated!