Question

A galvanometer coil has a resistance of $12\Omega$ and the meter shows full scale deflection for a current of $3\,mA$ How will you convert the metre into a voltmeter of range $0{\kern 1pt} {\kern 1pt} to{\kern 1pt} 18V?$

Answer 1

In order to solve this question, we should know that the range of voltmeter simply means the minimum and maximum value of voltage a voltmeter can read and here we will discuss the general concept of converting a galvanometer into voltmeter.

Complete step by step answer:
If $V$ is the maximum value of voltage a voltmeter range can read and G is the resistance of the coil of the galvanometer and ${I_G}$ be the current at which galvanometer shows full scale deflection then To convert galvanometer into voltmeter we simply add a resistance R in series with Galvanometer with value of resistance calculated as
$R = \dfrac{V}{{{I_G}}} - G$

Now, according to the question, we have given that maximum value in range of voltmeter as $V = 18\,Volts$
Value of resistance of galvanometer is $G = 12\Omega$
Current at which full scale deflection shows is,
${I_G} = 3\,mA = 3 \times {10^{ - 3}}\,A$
So, a resistance R we need to connect in series is given as
$R = \dfrac{V}{{{I_G}}} - G$ on putting the values we get,
$\Rightarrow R = \dfrac{{18}}{{3 \times {{10}^{ - 3}}}} - 12$
$\Rightarrow R = 6000 - 12$
$\therefore R = 5988\Omega$

Hence, to convert given galvanometer into voltmeter of range $0 - 18V$ , connect a resistance in series with it having value of $R = 5988\Omega$.

Note: It should be remembered that, galvanometer is used to detect the presence of small currents in electric circuit whereas voltmeter is uses to determine the value of potential difference across the two points in electric circuits and basic conversion used in solution as $1\,mA = {10^{ - 3}}A$ and voltmeter can read voltage up to its maximum limit which is $18V$.