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A galvanometer of resistance $50 \Omega$ is connected to a battery of $3 \mathrm{~V}$ along with a resistance of $2950 \Omega$ in series. A full scale deflection of 30 divisions is obtained in the galvanometer. In order to reduce this deflection to 20 divisions, the resistance in series should be
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The correct answer is:
$4450 \Omega$
$$
\begin{gathered}
30 i_0=\frac{V}{R_g+2950} ; R_g=50 \Omega \\
20 i_0=\frac{V}{R_g+R} \Rightarrow R=4450 \Omega
\end{gathered}
$$
\begin{gathered}
30 i_0=\frac{V}{R_g+2950} ; R_g=50 \Omega \\
20 i_0=\frac{V}{R_g+R} \Rightarrow R=4450 \Omega
\end{gathered}
$$
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