Meter Conversion: Rs, Rm, Rt, %Ig, %Is

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Homework Help Overview

The problem involves converting a galvanometer into an ammeter by determining the necessary shunt resistance. The context includes calculating the total resistance of the ammeter and the percentage of current flowing through both the galvanometer and the shunt resistor.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to calculate the shunt resistance and the total resistance of the ammeter but notes a lack of information regarding voltage. Some participants discuss the concept of current division in parallel resistive branches and question how currents are distributed between the galvanometer and the shunt.

Discussion Status

The discussion is ongoing, with participants exploring the relationships between current, resistance, and voltage in the context of a current divider. There is no explicit consensus yet, but some guidance on the principles of current division has been provided.

Contextual Notes

The original poster indicates missing information, specifically the voltage across the galvanometer, which is necessary for completing the calculations. The problem also involves understanding the implications of the full-scale deflection current for both the galvanometer and the shunt resistor.

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Homework Statement


A galvanometer has a coil resistance of 460 Ohms and a full scale deflection current of 100 uA. What shunt resistance is needed to convert this galvanometer into an ammeter which has a full scale deflection current of 200 mA. What is the resistance of the ammeter when it is completed? What percentage of the current flows thru the galvanometer? What percentage of the current flows thru the shunt?

Homework Equations


Rs=IgRg/(I - Ig)
Rm=V/Ig - Rg
Rt=V/Ig or Rg + Rm

The Attempt at a Solution


Rs= 100 uA x 460 Ohms/ (200 mA - 100 uA) =230.115 mOhms

Rm=V/Ig - Rg
Rt=V/Ig or Rg + Rm , either case I don't have V

%Ig=??
%Is=??
 
Last edited:
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The galvanometer and shunt resistor form a simple current divider. How do currents divide between two parallel resistive branches?
 
So, is the current = 200.1 mA and the V=92.046 V.
 
Blu3eyes said:
So, is the current = 200.1 mA and the V=92.046 V.

No. It's a current divider that is dividing 200 mA through two paths. One is the galvanometer, which should have 100 μA for full scale deflection, and the rest through the shunt resistor.

Given a current I and two parallel resistances R1 and R2, what's the formula for predicting the current through, say, R1?
 

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