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Question: A galvanometer having a resistance of 20Ω and 30 divisions on both sides has a figure of 0.005 amper...

A galvanometer having a resistance of 20Ω and 30 divisions on both sides has a figure of 0.005 ampere/division. The resistance that should be connected in series such that it can be used as a voltmeter upto 15 volt, is?
A. 80 ohms
B. 120 ohms
C. 125 ohms
D. 100 ohms

Explanation

Solution

In this question, we need to determine the value of the resistance to be connected in series with the galvanometer such that it can be used as a voltmeter measuring up to 15 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, Igmax=FOM×N{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 0.05 amperes/division and the number of divisions on either side of the galvanometer is 30. So, the maximum amount of current that can pass through the galvanometer without breaking (damaging) it is given by
Ig=FOM×N =0.005×30 =0.150Amp  {I_g} = FOM \times N \\\ = 0.005 \times 30 \\\ = 0.150Amp \\\
Now, as the resistance is connected with the galvanometer’s resistance in series, the equivalent resistance is given as Req=20+R{R_{eq}} = 20 + R.
Also, the maximum voltage drop across the galvanometer would be 15 volts.
So, following the ohms law to evaluate the value of the unknown resistance.
V=IgReq 15=0.15×(20+R) 20+R=150.15 20+R=100 R=10020 R=80Ω  V = {I_g}{R_{eq}} \\\ \Rightarrow 15 = 0.15 \times \left( {20 + R} \right) \\\ \Rightarrow 20 + R = \dfrac{{15}}{{0.15}} \\\ \Rightarrow 20 + R = 100 \\\ \Rightarrow R = 100 - 20 \\\ \therefore R = 80\Omega \\\
Hence, a resistance of 80 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 15 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.