A galvanometer has resistance ' $\mathrm{G}$ ' and range ' $\mathrm{V} g$ '. How much resistance is required…

A galvanometer has resistance ' $\mathrm{G}$ ' and range ' $\mathrm{V} g$ '. How much resistance is required to read voltage upto ' $\mathrm{V}$ ' volt?
  1. $\mathrm{G}\left(\frac{\mathrm{V}}{\mathrm{V}_{\mathrm{g}}}-1\right)$
  2. $\mathrm{G}\left(\frac{\mathrm{V}+\mathrm{V}_{\mathrm{g}}}{\mathrm{V}}\right)$
  3. $G\left(\frac{V-V_g}{V}\right)$
  4. $\mathrm{GV}_{\mathrm{g}}$

Solution

Given: Resistance of the galvanometer $=\mathrm{G}$ Range of the galvanometer $=\mathrm{V}_{\mathrm{g}}$ The series resistance value to be used for converting the galvanometer into a voltmeter of range 0 to $\mathrm{V}_{\mathrm{g}^{\prime}}$ is, $\mathrm{R}=\frac{\mathrm{V}_{\mathrm{z}}}{\mathrm{I}_{\mathrm{g}}}-\mathrm{G}$ Also, $\mathrm{I}_{\mathrm{g}}=\frac{\mathrm{V}_{\mathrm{g}}}{\mathrm{G}}$ To increase the measuring range to $\mathrm{V}$, the new resistance value $\mathrm{R}^{\prime}=\frac{\mathrm{V}}{\left(\frac{\mathrm{V}_{\mathrm{g}}}{\mathrm{G}}\right)}-\mathrm{G}$ $=\frac{\mathrm{VG}}{\mathrm{V}_{\mathrm{s}}}-\mathrm{G}=\mathrm{G}\left(\frac{\mathrm{V}}{\mathrm{V}_{\mathrm{g}}}-1\right)$ .

Asked in: MHT CET 2023 (10 May Shift 2)

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