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Question

# When a galvanometer is shunted with a 4 Ω resistance, the deflection is reduced to one-fifth. If the galvanometer is further shunted with a 2 Ω wire, determine current in galvanometer now if initially current in galvanometer is I0. (Given: Main current remains same)

A

I013

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B

I05

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C

I08

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D

5I013

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Solution

## The correct option is A I013 In parallel, current distributes in inverse ratio of resistance In figure (ii) 4I0/5I0/5=G4 ∴G=16 Ω In figure (iii) combined resistance of 4 Ω and 2 Ω is 4/3 Ω Assuming xI0 current passes through galvanometer, then (1−x)I0 will be the current passing through 4/3 Ω. Hence we can write: xI0(1−x)I0=4/316 xI0(1−x)I0=112 12xI0=I0−xI0 13xI0=I0 x=113 So, the current in galvanometer is xI0=I013

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