Thank you for reporting, we will resolve it shortly
Q.
A galvanometer has resistance of $100\, \Omega$ and a current of $10 \,mA$ produces full scale deflection in it. The resistance to be connected in series, to get a voltmeter of range $50$ volt is
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, $V_{\max }=50 V$ and full deflection current $I_{n}=10 mA$
So, by applying the $K V L$ in above circuit diagram,
$V_{A B} =100I_{fI}+R I_{fI}$
$\Rightarrow V_{A B} =(100+R) I_{fI}$
$\therefore $ For a $50 V$ voltmeter range there must be,
$V_{A B}=50 V$ and $ I_{fI}=10 mA$
Now, substituting the values of $V_{A B}$ and $I_{fI}$ in Eq. (i) we get,
$50=(100+R) 10 \times 10^{-3}$
$\therefore \quad R=4900 \Omega$
Hence, the resistance to be connected in series, to get a voltmeter of range $50 V$ is $4900 \Omega$
