A galvanometer having coil resistance $10 \ \Omega$ shows a full scale deflection for a current of $3 \mathrm{~mA}$. For it to measure a current of $8 \mathrm{~A}$, the value of the shunt should be:
JEE MAIN 2024, Diffcult
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Given $\mathrm{G}=10\ \Omega$
$ \mathrm{I}_{\mathrm{g}}=3 \mathrm{~mA} $
$ \mathrm{I}=8 \mathrm{~A}$
In case of conversion of galvanometer into ammeter.
We have $\mathrm{I}_{\mathrm{g}} \mathrm{G}=\left(\mathrm{I}-\mathrm{I}_{\mathrm{g}}\right) \mathrm{S}$
$ S=\frac{I_g G}{I-I_g} $
$ S=\frac{\left(3 \times 10^{-3}\right) 10}{8-0.003}=3.75 \times 10^{-5} \Omega$
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