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# An 80Ω galvanometer deflects full-scale for a potential of 20 mV. A voltmeter deflecting full-scale of 5 V is to be made using this galvanometer. We must connect(A) A resistance of 19.92 $\Omega$ parallel to the galvanometer(B) a resistance of 19.92 $\Omega$ in series with the galvanometer(C) a resistance of 20 k $\Omega$ parallel to the galvanometer(D) a resistance of 20 k $\Omega$ in series with the galvanometer

Last updated date: 11th Aug 2024
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Hint:We are given an 80 $\Omega$ galvanometer which gives a full-scale deflection for 20mV. We need to convert it into a voltmeter of given specification. We know a galvanometer can be converted into ammeter and voltameter by addition of suitable resistance either in series or parallel.

Given values are 80 $\Omega$resistance and the potential is 20mV
The current through the galvanometer producing full-scale deflection can be calculated by using the Ohm’s law.
V=IR
So, $I=\dfrac{20\times {{10}^{-3}}}{80}=2.5\times {{10}^{-4}}A$
To convert the galvanometer into a voltmeter, a high resistance is connected in series with the galvanometer.
Using the formula, $R=\dfrac{V}{{{I}_{g}}}-G$
Here G is the resistance of the galvanometer and ${{I}_{g}}$is the value of current through the galvanometer that we just calculated in the above step.
$\Rightarrow R=\dfrac{5}{2.5\times {{10}^{-4}}}-80 \\ \Rightarrow R=20000-80 \\ \therefore R=19920 \Omega \\$
Converting this into k $\Omega$ we get 19.92 k $\Omega$, so we have to connect a resistance of value 19.92k $\Omega$in series to bring the desired conversion.

Hence the correct answer is (B).

Note:If we would have to change the galvanometer into ammeter then we would have to connect a low resistance called a shunt in parallel with the given galvanometer to convert the galvanometer into ammeter.