Question

A galvanometee has a resistane of looh
its full scale tesistance should be added so that
the ammeter can have a range of o- 10 mA?​

Answer 1

Answer:

Solution :

Given IG=50μAIG=50μA. The upper limit gives the maximum

currect to be measured, which is I=5mAI=5mA. The galvanomenter ,

resistance is G=50ΩG=50Ω. Now

S=IGGI−IG=50×10−6×505×10−3−50×10−6=50×10−6×505×10−3≈0.5ΩS=IGGI-IG=50×10-6×505×10-3-50×10-6=50×10-6×505×10-3≈0.5Ω

Explanation:

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