Question 536: Current Electricity

Question

A galvanometer has resistance '$\mathrm{G}$' and range '$\mathrm{V} g$'. How much resistance is required to read voltage upto '$\mathrm{V}$' volt?

Accepted Answer

Given: Resistance of the galvanometer = 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,
$\mathrm{R}=\frac{\mathrm{V}_8}{\mathrm{I}_8}-\mathrm{G}$
Also,
$I_g=\frac{V_8}{G}$
To increase the measuring range to $\mathrm{V}$, the new resistance value
$\begin{aligned}
\mathrm{R}^{\prime} & =\frac{\mathrm{V}}{\left(\frac{\mathrm{V}_{\mathrm{g}}}{\mathrm{G}}ight)}-\mathrm{G} \
& =\frac{\mathrm{VG}}{\mathrm{V}_{\mathrm{g}}}-\mathrm{G}=\mathrm{G}\left(\frac{\mathrm{V}}{\mathrm{V}_{\mathrm{g}}}-1ight)
\end{aligned}$

Answer

$\mathrm{G}\left(\frac{\mathrm{V}}{\mathrm{V}_{\mathrm{g}}}-1\right)$

Answer

$\mathrm{G}\left(\frac{\mathrm{V}+\mathrm{V}_{\mathrm{g}}}{\mathrm{V}}\right)$

Answer

$\mathrm{G}\left(\frac{\mathrm{V}-\mathrm{V}_{\mathrm{g}}}{\mathrm{V}}\right)$

Answer

$\mathrm{GV}_{\mathrm{g}}$

Current Electricity

A galvanometer has resistance '\mathrm{G}' and range '\mathrm{V} g'. How much resistance is required to read voltage upto '\mathrm{V}' volt?

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