A galvanometer of resistance 50 Omega is conn | vedclass

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(C) Resistance of galvanometer,$R_g = 50 \Omega$. Emf of battery,$V = 3 	ext{ V}$. Resistance connected in series,$R_s = 2950 \Omega$. Total resistan

Vedclass · Accepted Answer

$4450$

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(C) Resistance of galvanometer,$R_g = 50 \Omega$. Emf of battery,$V = 3 	ext{ V}$. Resistance connected in series,$R_s = 2950 \Omega$. Total resistance,$R' = R_g + R_s = 50 + 2950 = 3000 \Omega$. Therefore,the initial current,$I = \frac{V}{R'} = \frac{3}{3000} = 10^{-3} 	ext{ A}$. If the deflection is reduced to $20$ divisions from $30$ divisions,the new current $I'$ will be $I' = I 	imes \frac{20}{30} = 10^{-3} 	imes \frac{2}{3} = \frac{2}{3} 	imes 10^{-3} 	ext{ A}$. Let $R_E$ be the new total resistance of the circuit. Using Ohm's law,$V = I' R_E \Rightarrow R_E = \frac{V}{I'} = \frac{3}{\frac{2}{3} 	imes 10^{-3}} = \frac{9}{2} 	imes 10^3 = 4500 \Omega$. Since $R_E = R_g + R_{new}$,the new series resistance required is $R_{new} = R_E - R_g = 4500 - 50 = 4450 \Omega$.

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