Question 541: Current Electricity

Question

The range of voltmeter of resistance ' $G$ ' $\Omega$ is ' $V$ ' volt. The resistance required to be connected in series with it in order to convert it into a voltmeter of range ' $n V$ ' volt, will be

Accepted Answer

To find the resistance required to be connected in series with a voltmeter to extend its range from $V$ volts to $nV$ volts, we can use the following reasoning:
The total voltage across the voltmeter when in series with the additional resistance should equate to the new desired range. The relationship for voltmeters, which are essentially just high-resistance galvanometers, involves taking the ratio of the original and desired ranges.
Firstly, consider the following given parameters:
Original range of the voltmeter: $V$
Desired range of the voltmeter: $nV$
Original resistance of the voltmeter: $G$ ohms
When the voltmeter spans the new range $nV$, the current through the voltmeter should remain the same as when it spanned just $V$. Let $R$ be the additional resistance needed.
Using Ohm's Law, the current $I$ through the voltmeter for the range $V$ is given by:
$$ I = \frac{V}{G} $$
For the new range $nV$, the same current should flow:
$$ I = \frac{nV}{G + R} $$
Setting the two expressions for current equal gives:
$$ \frac{V}{G} = \frac{nV}{G + R} $$
Solving for $R$, multiply both sides by $(G + R)$:
$$ V(G + R) = nVG $$
Divide both sides by $V$:
$$ G + R = nG $$
Subtract $G$ from both sides:
$$ R = nG - G $$
So the required resistance $R$ is:
$$ R = (n - 1)G $$
Thus, the correct option is:
Option A
$$(\mathrm{n}-1) \mathrm{G}$$

Answer

$(\mathrm{n}-1) \mathrm{G}$

Answer

$\mathrm{G} / \mathrm{n}$

Answer

$\mathrm{nG}$

Answer

$\mathrm{\frac{G}{n}-1}$

Current Electricity

The range of voltmeter of resistance ' G ' \Omega is ' V ' volt. The resistance required to be connected in series with it in order to convert it into a voltmeter of range ' n V ' volt, will be

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