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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 shows full-scale deflection of \( 30 \) divisions. The additional series resistance
required to reduce the deflection to \( 20 \) divisions is
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\( 2950 \Omega \) in series shows full-scale deflection of \( 30 \) divisions. The additional series resistance
required to reduce the deflection to \( 20 \) divisions is
Solution:
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The correct answer is:
\( 1500 \Omega \)
Given, galvanometer resistance, \( R_{G}=50 \Omega ; \) battery, \( \mathrm{V}=3 \mathrm{~V} ; \) resistance, \( \mathrm{R}=2950 \Omega ; \) deflection \( =30 \) divisions. Now,
\( I=\frac{V}{\left(R_{G}+R\right)}=\frac{3}{(50+2950)}=0.001=1 \times 10^{-3} \mathrm{~A} \)
Thus, current when deflection is \( 30 \) divisions \( =10^{-3} \mathrm{~A} \)
\( \Rightarrow \) Current when deflection is \( 1 \) division \( =\frac{10^{-3}}{30} \)
\( \Rightarrow \) Current when deflection is \( 20 \) division \( =\frac{10^{-3}}{30} \times 20 \)
Therefore, in order to have a deflection of \( 20 \) division the resistance should be
\[
\begin{array}{l}
\frac{2}{3} \times 10^{-3}=\frac{3}{(50+R)} \\
\Rightarrow(50+R)=\frac{3 \times 3}{2 \times 10^{-3}} \\
50+R=4.5 \times 1000 \\
R=4500-50=4450
\end{array}
\]
Hence, additional resistance required \( =4450-2950=1500 \)
\( I=\frac{V}{\left(R_{G}+R\right)}=\frac{3}{(50+2950)}=0.001=1 \times 10^{-3} \mathrm{~A} \)
Thus, current when deflection is \( 30 \) divisions \( =10^{-3} \mathrm{~A} \)
\( \Rightarrow \) Current when deflection is \( 1 \) division \( =\frac{10^{-3}}{30} \)
\( \Rightarrow \) Current when deflection is \( 20 \) division \( =\frac{10^{-3}}{30} \times 20 \)
Therefore, in order to have a deflection of \( 20 \) division the resistance should be
\[
\begin{array}{l}
\frac{2}{3} \times 10^{-3}=\frac{3}{(50+R)} \\
\Rightarrow(50+R)=\frac{3 \times 3}{2 \times 10^{-3}} \\
50+R=4.5 \times 1000 \\
R=4500-50=4450
\end{array}
\]
Hence, additional resistance required \( =4450-2950=1500 \)
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