Question 611: Current Electricity

Question

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

Accepted Answer

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, $V_{\max }=50 \mathrm{~V}$ and full deflection current $I_f=10 \mathrm{~mA}$
So, by applying the $K V L$ in above circuit diagram,
$$\begin{aligned}
& V_{A B}=100 I_n+R I_n \
& \Rightarrow \quad V_{	ext {AB }}=(100+R)_{	ext {I }}
\end{aligned}$$
$	herefore$ For a 50 V voltmeter range there must be,
$$V_{A B}=50 \mathrm{~V} 	ext { and } I_f=10 \mathrm{~mA}$$
Now, substituting the values of $V_{A B}$ and $I_{fl}$ in Eq. (i) we get,
$$\begin{array}{rlrl} 
& 50 =(100+R) 10 	imes 10^{-3} \
& 	herefore \quad R =4900 \Omega
\end{array}$$
Hence, the resistance to be connected in series, to get a voltmeter of range 50 V is $4900 \Omega$.

Answer

$3900 \Omega$

Answer

$4000 \Omega$

Answer

$4600 \Omega$

Answer

$4900 \Omega$

Current Electricity

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

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