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

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.

Complete step by step answer:
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.