Manipal Entrance Test 2015 Physics Paper

When a galvanometer is connected to the shunt resistance, then we have
potential drop across the galvanometer = potential drop across the shunt
i.e. iG=(i0−i)S ...(i)
Here, i= current through the galvanometer
G= resistance of the galvanometer
i0= current through the shunt resistance
S= shunt resistance
From Eq. (i), we have the part of the total current that flows through the galvanometer is

i

i0

=

S

G+S

Substituting, S=2.5Ω,G=25Ω, we get

i

i0

=

1

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