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(A) Let the full-scale deflection current of the galvanometer be $I_g$.When a resistance $R_1 = 100 \Omega$ is connected in series,the range is $V = I

Vedclass · Accepted Answer

$800$

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(A) Let the full-scale deflection current of the galvanometer be $I_g$.When a resistance $R_1 = 100 \Omega$ is connected in series,the range is $V = I_g(G + 100)$.When a resistance $R_2 = 1000 \Omega$ is connected in series,the range becomes $2V = I_g(G + 1000)$.Dividing the two equations:$\frac{2V}{V} = \frac{I_g(G + 1000)}{I_g(G + 100)}$$2 = \frac{G + 1000}{G + 100}$$2(G + 100) = G + 1000$$2G + 200 = G + 1000$$G = 1000 - 200 = 800 \Omega$.Note: Based on the calculation,the value of $G$ is $800 \Omega$. Since $800 \Omega$ is provided as option $A$,the calculation is consistent.

When a resistance of $100 \Omega$ is connected in series with a galvanometer of resistance $G$,its range is $V$. To double its range,a resistance of $1000 \Omega$ is connected in series. The value of $G$ is: (in $\Omega$)

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