Question

A galvanometer, whose resistance is 50ohm, has 25 divisions in it. When a current of $4 \times {10^{ - 4}}A$ passes through it, its needle (pointer) deflects by one division. To use this galvanometer as a voltmeter of range 2.5V, it should be connected to a resistance of:
A. 6250 ohms
B. 250 ohms
C. 200 ohms
D. 6200 ohms

Answer 1

In this question, we need to determine the value of the resistance to be connected with the galvanometer such that it can be used as a voltmeter measuring upto 2.5 volts of potential difference across it. For this, we will use the ohm's law along with the property of the galvanometer.

Complete step by step answer:
Let R be the resistance to be connected in series with the galvanometer.
Initially, the maximum amount of current that the galvanometer can measure without any damage is the product of the figure of merit of the galvanometer and the number of divisions on either side of the instrument (galvanometer). Mathematically, ${I_{{g_{\max }}}} = FOM \times N$ where N is the number of divisions on either side of the galvanometer.
According to the question, the figure of merit of the instrument is $4 \times {10^{ - 4}}{\text{A/division}}$ and the number of divisions on either side of the galvanometer is 25. So, the maximum amount of current that can pass through the galvanometer without breaking (damaging) it is given by
${I_g} = FOM \times N \\\ = 4 \times {10^{ - 4}} \times 25 \\\ = 100 \times {10^{ - 4}} \\\ = 0.01Amp \\\$
Now, as the resistance is connected with the galvanometer’s resistance in series, the equivalent resistance is given as ${R_{eq}} = 50 + R$.
Also, the maximum voltage drop across the galvanometer would be 2.5 volts.
So, following the ohms law to evaluate the value of the unknown resistance.
$V = {I_g}{R_{eq}} \\\ \Rightarrow 2.5 = 0.01 \times \left( {50 + R} \right) \\\ \Rightarrow 50 + R = \dfrac{{2.5}}{{0.01}} \\\ \Rightarrow 50 + R = 250 \\\ \Rightarrow R = 250 - 50 \\\ \therefore R = 200\Omega \\\$
Hence, a resistance of 200 ohms should be connected to the galvanometer with the internal resistance of 20 ohms to be used as a voltmeter for measuring a voltage drop of 2.5 volts.

So, the correct answer is “Option A”.

Note:
It is interesting to note that a galvanometer can be used as an ammeter as well as a voltmeter. For the galvanometer to act as a voltmeter, it should be connected across the element while to be used as an ammeter, it should be connected along the element.