A galvanometer has resistance G and range V_g | vedclass

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

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

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(C) To convert a galvanometer into a voltmeter, a high resistance $R$ must be connected in series with the galvanometer.Let $I_g$ be the full-scale deflection current of the galvanometer.Given, $V_g = I_g 	imes G$, so $I_g = \frac{V_g}{G}$.The total voltage $V$ to be measured is the sum of the voltage across the galvanometer and the voltage across the series resistance $R$.$V = I_g(G + R)$Substituting $I_g = \frac{V_g}{G}$ into the equation:$V = \frac{V_g}{G}(G + R)$$\frac{V}{V_g} = \frac{G + R}{G}$$\frac{V}{V_g} = 1 + \frac{R}{G}$$\frac{R}{G} = \frac{V}{V_g} - 1$$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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