A galvanometer of resistance 50 Omega is con | vedclass

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(D) The current $I$ flowing through the galvanometer is proportional to the deflection $\phi$ in divisions,i.e.,$I \propto \phi$.Using Ohm's law,$I =

Vedclass · Accepted Answer

$7950$

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(D) The current $I$ flowing through the galvanometer is proportional to the deflection $\phi$ in divisions,i.e.,$I \propto \phi$.Using Ohm's law,$I = \frac{V}{R + G}$,where $V = 8 \ V$,$G = 50 \ \Omega$,and $R$ is the series resistance.For the first case: $I_1 = \frac{V}{R_1 + G}$ with $\phi_1 = 30$ divisions and $R_1 = 3950 \ \Omega$.For the second case: $I_2 = \frac{V}{R_2 + G}$ with $\phi_2 = 15$ divisions.Taking the ratio: $\frac{I_1}{I_2} = \frac{\phi_1}{\phi_2} \implies \frac{R_2 + G}{R_1 + G} = \frac{30}{15} = 2$.Substituting the values: $\frac{R_2 + 50}{3950 + 50} = 2$.$\frac{R_2 + 50}{4000} = 2$.$R_2 + 50 = 8000$.$R_2 = 7950 \ \Omega$.

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