Question 548: Current Electricity

Question

When moving coil galvanometer (MCG) is converted into a voltmeter, the series resistance is '$$n$$' times the resistance of galvanometer. How many times that of MCG the voltmeter is now capable of measuring voltage?

Accepted Answer

Series resistance is $$\mathrm{n}$$ times of galvanometer resistance: $$\mathrm{R}_{\mathrm{s}}=\mathrm{nR}_{\mathrm{G}}$$
Relation between voltage and current:
$$\begin{aligned}
& 	herefore \quad \mathrm{V}=\mathrm{I}\left(\mathrm{R}_{\mathrm{s}}+\mathrm{R}_{\mathrm{G}}ight) \
& 	herefore \quad \mathrm{V}=\mathrm{I}\left(\mathrm{nR}_{\mathrm{G}}+\mathrm{R}_{\mathrm{G}}ight) \
& 	herefore \quad \mathrm{V}=\mathrm{IR}_{\mathrm{G}}(\mathrm{n}+1) \
&
\end{aligned}$$

Answer

$$\mathrm{n}$$

Answer

$$\frac{\mathrm{n}+1}{\mathrm{n}}$$

Answer

$$\mathrm{n}+1$$

Answer

$$\mathrm{n}-1$$

Current Electricity

When moving coil galvanometer (MCG) is converted into a voltmeter, the series resistance is '$n$' times the resistance of galvanometer. How many times that of MCG the voltmeter is now capable of measuring voltage?

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