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A galvanometer of resistance $50 \Omega$ is connected to a battery $3 \mathrm{~V}$ along with a resistance $2950 \Omega$ 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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Verified Answer
The correct answer is:
$4450 \Omega$
Resistance of galvanometer, $R_g=50 \Omega$
Emf of battery, $V=3 \mathrm{~V}$
Resistance connected in series, $R_s=2950 \Omega$
Total resistance, $R^{\prime}=R_g+R_s=50+2950=3000 \Omega$
$\therefore$ Current, $I=\frac{V}{R^{\prime}}=\frac{3}{3000}=10^{-3} \mathrm{~A}$
If the deflection has to be reduced to 20 divisions, then current, $I^{\prime}=\frac{I}{30} \times 20=\frac{2}{3} \times 10^{-3} \mathrm{~A}$
Let $R_E$ be the effective resistance of the circuit, hence
$$
\begin{aligned}
3 & =R_E I^{\prime} \\
\Rightarrow \quad R_E & =\frac{3}{I^{\prime}}=\frac{3}{\frac{2}{3} \times 10^{-3}}=4.5 \times 10^3=4500 \Omega
\end{aligned}
$$
$$
\begin{aligned}
\therefore \text { Resistance to be added } & =R_E-R_g \\
& =4500-50=4450 \Omega
\end{aligned}
$$
Emf of battery, $V=3 \mathrm{~V}$
Resistance connected in series, $R_s=2950 \Omega$
Total resistance, $R^{\prime}=R_g+R_s=50+2950=3000 \Omega$
$\therefore$ Current, $I=\frac{V}{R^{\prime}}=\frac{3}{3000}=10^{-3} \mathrm{~A}$
If the deflection has to be reduced to 20 divisions, then current, $I^{\prime}=\frac{I}{30} \times 20=\frac{2}{3} \times 10^{-3} \mathrm{~A}$
Let $R_E$ be the effective resistance of the circuit, hence
$$
\begin{aligned}
3 & =R_E I^{\prime} \\
\Rightarrow \quad R_E & =\frac{3}{I^{\prime}}=\frac{3}{\frac{2}{3} \times 10^{-3}}=4.5 \times 10^3=4500 \Omega
\end{aligned}
$$
$$
\begin{aligned}
\therefore \text { Resistance to be added } & =R_E-R_g \\
& =4500-50=4450 \Omega
\end{aligned}
$$
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