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Meter conversion

  • Thread starter Blu3eyes
  • Start date
  • #1
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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:

Answers and Replies

  • #2
gneill
Mentor
20,797
2,777
The galvanometer and shunt resistor form a simple current divider. How do currents divide between two parallel resistive branches?
 
  • #3
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So, is the current = 200.1 mA and the V=92.046 V.
 
  • #4
gneill
Mentor
20,797
2,777
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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