A galvanometer has resistance 'G' and range ' | vedclass

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(A) The galvanometer has a resistance $G$ and a full-scale deflection voltage $V_g$. The full-scale current $I_g$ is given by $I_g = \frac{V_g}{G}$.To

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

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(A) The galvanometer has a resistance $G$ and a full-scale deflection voltage $V_g$. The full-scale current $I_g$ is given by $I_g = \frac{V_g}{G}$.To convert the galvanometer into a voltmeter of range $V$,a series resistance $R$ must be connected in series with the galvanometer.The total resistance of the circuit becomes $R + G$.According to Ohm's law,for the new range $V$,the current $I_g$ remains the same:$V = I_g(R + G)$Substituting $I_g = \frac{V_g}{G}$ into the equation:$V = \left(\frac{V_g}{G}\right)(R + G)$$\frac{V}{V_g} = \frac{R+G}{G}$$\frac{V}{V_g} = \frac{R}{G} + 1$$\frac{R}{G} = \frac{V}{V_g} - 1$$R = G\left(\frac{V}{V_g} - 1\right)$

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

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