A galvanometer of resistance 50 Omega is conn | vedclass

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(A) Given: Galvanometer resistance $R_G = 50 \Omega$,battery voltage $V = 3 	ext{ V}$,initial series resistance $R_1 = 2950 \Omega$,initial deflectio

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$1500$

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(A) Given: Galvanometer resistance $R_G = 50 \Omega$,battery voltage $V = 3 	ext{ V}$,initial series resistance $R_1 = 2950 \Omega$,initial deflection $	heta_1 = 30$ divisions.First,calculate the current $I_1$ for $30$ divisions:$I_1 = \frac{V}{R_G + R_1} = \frac{3}{50 + 2950} = \frac{3}{3000} = 10^{-3} 	ext{ A}$.The current per division is $k = \frac{I_1}{30} = \frac{10^{-3}}{30} 	ext{ A/division}$.For a deflection of $	heta_2 = 20$ divisions,the required current $I_2$ is:$I_2 = 20 	imes k = 20 	imes \frac{10^{-3}}{30} = \frac{2}{3} 	imes 10^{-3} 	ext{ A}$.Let the total resistance in the circuit be $R_{total} = R_G + R_{new}$.Using Ohm's law: $I_2 = \frac{V}{R_{total}} \Rightarrow \frac{2}{3} 	imes 10^{-3} = \frac{3}{50 + R_{new}}$.$50 + R_{new} = \frac{3 	imes 3}{2 	imes 10^{-3}} = 4.5 	imes 1000 = 4500 \Omega$.$R_{new} = 4500 - 50 = 4450 \Omega$.The additional resistance required is $R_{add} = R_{new} - R_1 = 4450 - 2950 = 1500 \Omega$.

$A$ galvanometer of resistance $50 \Omega$ is connected to a battery of $3 \text{ V}$ along with a resistance of $2950 \Omega$ in series,showing a full-scale deflection of $30$ divisions. The additional series resistance required to reduce the deflection to $20$ divisions is: (in $\Omega$)

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