A galvanometer has resistance G and range V_g | vedclass

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(B) To convert a galvanometer into a voltmeter,a high resistance $R$ is connected in series with the galvanometer.Let $I_g$ be the full-scale deflecti

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$G(\frac{V}{V_g}-1)$

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(B) To convert a galvanometer into a voltmeter,a high resistance $R$ is connected in series with the galvanometer.Let $I_g$ be the full-scale deflection current of the galvanometer.Given,$V_g = I_g G$,so $I_g = \frac{V_g}{G}$.The total resistance of the voltmeter is $(G + R)$.For a range $V$,the total voltage is $V = I_g(G + R)$.Substituting $I_g = \frac{V_g}{G}$,we get $V = \frac{V_g}{G}(G + R)$.Rearranging for $R$: $\frac{V}{V_g} = \frac{G + R}{G} = 1 + \frac{R}{G}$.Therefore,$\frac{R}{G} = \frac{V}{V_g} - 1$.$R = G(\frac{V}{V_g} - 1)$.

$A$ galvanometer has resistance $G$ and range $V_g$. How much resistance is required to read voltage up to $V$ volt?

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