A galvanometer shows a reading of 0.65mA. When the galvanometer is shunted with a 4 ohm resistance, the deflection is reduced to 0.13mA. If a resistance of 2 ohm is further connected parallel to 4 ohm resistance (i.e. galvanometer is further shunted with a 2 ohm wire) the new reading will be (Assume main current remains the same)
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a
0.60mA
b
0.08mA
c
0.12mA
d
0.05mA
answer is D.
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Detailed Solution
according to shunt formula S =Gn-1here n= 0.650.13=54 =G4⇒G =16next time s=4×24+2=43again S=4 3=16n-'1⇒⇒ n= 13 i =I13=0.6513=0.05