Question

# The galvanometer resistance is 30 Ω and it is connected to 2 V battery along with a resistance 2000 Ω in series. A full scale deflection of 25 divisions is obtained. In order to reduce this deflection to 20 divisions, the resistance in series should be

A
2507 Ω
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B
2320 Ω
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C
2180 Ω
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D
2210 Ω
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Solution

## The correct option is A 2507 ΩTotal initial resistance = 30+2000=2030 ohm. Battery voltage= 2v Current =2/2030=0.98×10−3Full scale deflection of 25 divisions was obtained .In order to reduce this deflection to 20 divisions the resistor in series should be, current i= 2025×0.98×10−3 as full scale deflection for 0.98×10−3 are 25 divisions. Let the resistance be r ohm 2=2030×0.98×10−3=r×2025×0.98×10−3r=2537ohmGalvanometer resistance 30 ohm .Resistance to be added (2537-30)=2507 ohm

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