A galvanometer of resistance 50, Omega is con | vedclass

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(B) The initial total resistance is $R_{total,1} = R_G + R_1 = 50\, \Omega + 2950\, \Omega = 3000\, \Omega$.The current for full-scale deflection ($30

Vedclass · Accepted Answer

$4450$

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(B) The initial total resistance is $R_{total,1} = R_G + R_1 = 50\, \Omega + 2950\, \Omega = 3000\, \Omega$.The current for full-scale deflection ($30$ divisions) is $I_1 = \frac{V}{R_{total,1}} = \frac{3\, V}{3000\, \Omega} = 1 	imes 10^{-3}\, A = 1\, mA$.To reduce the deflection to $20$ divisions,the new current $I_2$ must be proportional to the divisions: $I_2 = I_1 	imes \frac{20}{30} = 1\, mA 	imes \frac{2}{3} = \frac{2}{3}\, mA$.Using Ohm's law for the new circuit,$V = I_2 	imes R_{total,2}$,where $R_{total,2} = R_G + R_2$.$3\, V = (\frac{2}{3} 	imes 10^{-3}\, A) 	imes R_{total,2}$.$R_{total,2} = \frac{3 	imes 3}{2} 	imes 10^3\, \Omega = 4500\, \Omega$.Since $R_{total,2} = R_G + R_2$,the required series resistance is $R_2 = 4500\, \Omega - 50\, \Omega = 4450\, \Omega$.

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