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Q.

A galvanometer of coil resistance 100 $Ω$ is connected to a shunt of resistance 10 $Ω$ .The current through the galvanometer is $i_{1}$ and the current through the shunt is $i_{2}$ .Then the ratio $i_{1} : i_{2}$ is

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a

1:10

b

10: 1

c

11: 10

d

10: 11

answer is A.

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Detailed Solution

$S = 10 Ω, G = 100 Ω,$

$\frac{i_{g}}{i} = \frac{S}{S + G} \Rightarrow \frac{i_{g}}{i} = \frac{10}{110} = \frac{1}{11} \Rightarrow i_{g} = \frac{i}{11}$

$\begin{array}{l} \frac{i_{s}}{i} = \frac{G}{S + G} \Rightarrow \frac{i_{s}}{i} = \frac{100}{110} = \frac{10}{11} \Rightarrow i_{g} = \frac{10 i}{11}, \\ i_{1} : i_{2} = \frac{i}{11} : \frac{10 i}{11} = \frac{1}{11} : \frac{10}{11} = 1 : 10 \end{array}$

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