Question

A galvanometer has resistance of $100\Omega$ and a current of $10mA$ produces full scale deflection in it. The resistance to be connected in series, to get a voltmeter of range $50$ volt is

• A. $3900\Omega$
• B. $4000\Omega$
• C. $4600\Omega$
• D. $4900\Omega$

Answer 1

Answer: (D)

A galvanometer is converted into a voltmeter by connecting a resistance in series with the galvanometer as shown in the circuit diagram,

where, $R$ is resistance of the resistor connected in series.
Given, galvanometer resistance, $R_G=100\Omega$
Voltmeter range, $V_{\max }=50V$ and full deflection current $I_n=10mA$
So, by applying the $KVL$ in above circuit diagram,
$V_{AB}=100I_{fI}+R{I}_{fI}$
$\Rightarrow V_{AB}=(100+R)I_{fI}$
$\therefore$ For a $50V$ voltmeter range there must be,
$V_{AB}=50V$ and $I_{fI}=10mA$
Now, substituting the values of $V_{AB}$ and $I_{fI}$ in Eq. (i) we get,
$50=(100+R)10\times 10^{-3}$
$\therefore R=4900\Omega$
Hence, the resistance to be connected in series, to get a voltmeter of range $50V$ is $4900\Omega$