Calculating Shunt Resistor for Galvanometer to Ammeter Conversion

[Magna1]

A galvonometer has a coil resitance of 460 ohms and a full scale deflection current of of 100^10*-6 A What shunt resistor is needed to convert this galv.. into an ammeter which has a full scale deflection current of 200 ^10*-3 A? What is the resitance of the ammeter when it is completed? What percentage of current flows through both Ig and Is?

I figured Rs at 230.115^10*-3 ohms. ANd Rt of 460.23 ohms. but I know my Rt should be in milli ohms??/

Also without having a voltage how can I calculate the percentage of current flow?
I used
Rs=(Ig*rg)/Im-Ig)
and
Rt=Rs+Rm

am I going in the right direction?/

 

[turin]

You need to define you're variables. I suspect that "Rt" stands for total or equivalent resistance of the modified device. In that case, your formula is incorrect. Resistances in parallel don't "add" like that; they combine like this:

R_|| = R₁R₂ / ( R₁ + R₂ )

To get the percentage of current, you can use the current divider.

 

[Averagesupernova]

You know that the meter has an internal resistance of 460 ohms. You know it will fully deflect at 100 microamperes. You NEED it to fully deflect at 200 millamperes. No problem. You know ohms law (I hope). You know how to figure what the voltage across the meter will be with the needle fully deflected. Since you know full deflection voltage and current, you know the TOTAL resistance required. The shunt is in parallel with meter coil, so it is a matter of figuring the shunt value based the 2 known resistances. One is the TOTAL resistance required, the other is the meter coil resistance. To figure parallel resistances, you add the inverse of all resistances to get the inverse of the total.

1/Rt = 1/Rshunt + 1/Rmeter

 

[Magna1]

So to get this figured out I just calculate Rt as a pair of Parallel resistors?

 

[Averagesupernova]

Originally posted by Magna1
So to get this figured out I just calculate Rt as a pair of Parallel resistors?

In my example, Rt is the total resistance. Isn't an ammeter a pair of parallel resistors? The shunt in parallel with the meter itself?