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A multirange current meter can be constructed by using a galvanometer circuit as shown in figure. We want a current meter that can measure $10 \mathrm{~mA}, 100 \mathrm{~mA}$ and $1 \mathrm{~mA}$ using a galvanometer of resistance $10 \Omega$ and that produces maximum deflection for current of $1 \mathrm{~mA}$. Find $\mathrm{S}_1, \mathrm{~S}_2$ and $\mathrm{S}_3$ that have to be used.
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Verified Answer
A galvanometer can be converted into ammeter by connecting a very low resistance (shunt S) connected in parallel with galvanometer. The relationship is given by $I_{\mathrm{g}} \mathrm{G}=\left(\mathrm{I}-\mathrm{I}_{\mathrm{g}}\right)$ S where $\mathrm{I}_{\mathrm{g}}$ is range of galvanometer, $\mathrm{G}$ IS resistance of galvanometer.
Here,
$\mathrm{I}_{\mathrm{G}} \cdot \mathrm{G}=\left(\mathrm{I}_1-\mathrm{I}_{\mathrm{G}}\right)\left(\mathrm{S}_1+\mathrm{S}_2+\mathrm{S}_3\right)$ for $\mathrm{I}_1=10 \mathrm{~mA}$,
... (i)
$\mathrm{I}_{\mathrm{G}} \cdot\left(\mathrm{G}+\mathrm{S}_1\right)=\left(\mathrm{I}_2-\mathrm{I}_{\mathrm{G}}\right)\left(\mathrm{S}_2+\mathrm{S}_3\right)$ for $\mathrm{I}_2=100 \mathrm{~mA}$, ... (ii) and $\mathrm{I}_{\mathrm{G}} \cdot\left(\mathrm{G}+\mathrm{S}_1+\mathrm{S}_2\right)=\left(\mathrm{I}_3-\mathrm{I}_{\mathrm{G}}\right)\left(\mathrm{S}_3\right)$ for $\mathrm{I}_3=1 \mathrm{~A}$,
By solving we get,
(iii) $\begin{aligned} \mathrm{S}_1 &=1 \mathrm{~W}, \mathrm{~S}_2=0.1 \mathrm{~W} \\ \text { and } \mathrm{S}_3 &=0.001 \mathrm{~W} \end{aligned}$
Here,
$\mathrm{I}_{\mathrm{G}} \cdot \mathrm{G}=\left(\mathrm{I}_1-\mathrm{I}_{\mathrm{G}}\right)\left(\mathrm{S}_1+\mathrm{S}_2+\mathrm{S}_3\right)$ for $\mathrm{I}_1=10 \mathrm{~mA}$,
... (i)
$\mathrm{I}_{\mathrm{G}} \cdot\left(\mathrm{G}+\mathrm{S}_1\right)=\left(\mathrm{I}_2-\mathrm{I}_{\mathrm{G}}\right)\left(\mathrm{S}_2+\mathrm{S}_3\right)$ for $\mathrm{I}_2=100 \mathrm{~mA}$, ... (ii) and $\mathrm{I}_{\mathrm{G}} \cdot\left(\mathrm{G}+\mathrm{S}_1+\mathrm{S}_2\right)=\left(\mathrm{I}_3-\mathrm{I}_{\mathrm{G}}\right)\left(\mathrm{S}_3\right)$ for $\mathrm{I}_3=1 \mathrm{~A}$,
By solving we get,
(iii) $\begin{aligned} \mathrm{S}_1 &=1 \mathrm{~W}, \mathrm{~S}_2=0.1 \mathrm{~W} \\ \text { and } \mathrm{S}_3 &=0.001 \mathrm{~W} \end{aligned}$
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