A galvanometer has a resistance of 25 Ω and a maximum of 0.01 A current can be passed through it. In order to change it into an ammeter of range 10 A, the shunt resistance required is

A
$5 / 999 Ω$
B
$10 / 999 Ω$
C
$20 / 999 Ω$
D
$25 / 999 Ω$

Solution:

$\begin{array}{l} i_{g} = i \frac{S}{G + S} \Rightarrow 0 .01 = 10 \frac{S}{25 + S} \\ \Rightarrow 1000 S = 25 + S \Rightarrow S = \frac{25}{999} Ω \end{array}$

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A galvanometer has a resistance of $25 Ω$ and a maximum of 0.01 A current can be passed through it. In order to change it into an ammeter of range 10 A, the shunt resistance required is

Solution:

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