Question 565: 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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