A galvanometer of resistance 50 Omega is conn | vedclass

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(D) The current $I$ flowing through the galvanometer circuit is given by Ohm's law: $I = \frac{V}{R_{total}}$.Initially,the total resistance is $R_{to

Vedclass · Accepted Answer

$4450$

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(D) The current $I$ flowing through the galvanometer circuit is given by Ohm's law: $I = \frac{V}{R_{total}}$.Initially,the total resistance is $R_{total,1} = 50 \Omega + 2950 \Omega = 3000 \Omega$.The current for $30$ divisions is $I_1 = \frac{3 	ext{ V}}{3000 \Omega} = 10^{-3} 	ext{ A}$.Since the deflection is proportional to the current,the current per division is $\frac{10^{-3} 	ext{ A}}{30}$.To obtain a deflection of $20$ divisions,the required current is $I_2 = 20 	imes \left( \frac{10^{-3}}{30} ight) = \frac{2}{3} 	imes 10^{-3} 	ext{ A}$.Let the new total resistance be $R_{total,2}$. Then $I_2 = \frac{3 	ext{ V}}{R_{total,2}}$.$\frac{2}{3} 	imes 10^{-3} = \frac{3}{R_{total,2}} \implies R_{total,2} = \frac{3 	imes 3}{2 	imes 10^{-3}} = 4500 \Omega$.The total resistance in the circuit is $R_{total,2} = R_G + R_{series}$.$4500 \Omega = 50 \Omega + R_{series} \implies R_{series} = 4450 \Omega$.

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