Question
Explain how a moving-coil galvanometer is converted into a voltmeter. Derive the necessary formula.
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Answer
A moving$-$coil galvanometer is converted into a voltmeter by increasing its effective resistance by connecting a high resistance $Rs$ in series with the galvanometer, shown in figure. The series resistance is also useful for changing the range of any given voltmeter.
Let $G$ be the resistance of the galvanometer coil and $I _{ g }$ the current required for a full$-$scale deflection.
Let $V$ be the maximum potential difference to be measured.
The value of the series resistance $R_5$ should be such that when the potential difference applied across the instrument is $V$, the current through the galvanometer is $I_g$.
In the series combination, the potential difference $V$ gets divided across the galvanometer $($resistance$, G)$ and the resistance $RS :$
$V=I_g G+I_g R_S=I_g\left(G+R_S\right)$
$\therefore R_S=\frac{V}{I_g}-G$
This is the required value of the series resistance.
The scale of the galvanometer is then calibrated so as to read the potential difference in volt or its submultiples, e.gr, $mV$, directly.
Notes:
$(1)$ A series multiplier is made of manganin wire because manganin has a very small temperature coefficient of resistivity.
$(2)$ The maximum potential difference $V_g$ that can be dropped across the galvanometer is $V_g = I_g G$.
Therefore, the above expression for the series resistance may be rewritten as
$R _{ s }=\frac{V G}{I_{ B } G}- G = \frac{V G}{V_{ g }}- G = G ( p -1)$
where $p=V / N_g$ is the range$-$multiplying factor, i.e, the voltage range of the galvanometer can be increased by a factor of $p$ by connecting a series resistance which is $(p-1)$ times the galvanometer resistance.
$\therefore p =\frac{V}{V_g}=\frac{\left(R_{ s }+G\right) I_{ g }}{G I_{ g }}=\frac{R_{ s }+G}{G}$
Since the resistance of the voltmeter is $R_V=R_S+G$,
$p =\frac{R_{ v }}{G}$
$\therefore R _{ v }= G _{ p }$
Let $G$ be the resistance of the galvanometer coil and $I _{ g }$ the current required for a full$-$scale deflection.
Let $V$ be the maximum potential difference to be measured.
The value of the series resistance $R_5$ should be such that when the potential difference applied across the instrument is $V$, the current through the galvanometer is $I_g$.
In the series combination, the potential difference $V$ gets divided across the galvanometer $($resistance$, G)$ and the resistance $RS :$
$V=I_g G+I_g R_S=I_g\left(G+R_S\right)$
$\therefore R_S=\frac{V}{I_g}-G$
This is the required value of the series resistance.
The scale of the galvanometer is then calibrated so as to read the potential difference in volt or its submultiples, e.gr, $mV$, directly.
Notes:
$(1)$ A series multiplier is made of manganin wire because manganin has a very small temperature coefficient of resistivity.
$(2)$ The maximum potential difference $V_g$ that can be dropped across the galvanometer is $V_g = I_g G$.
Therefore, the above expression for the series resistance may be rewritten as
$R _{ s }=\frac{V G}{I_{ B } G}- G = \frac{V G}{V_{ g }}- G = G ( p -1)$
where $p=V / N_g$ is the range$-$multiplying factor, i.e, the voltage range of the galvanometer can be increased by a factor of $p$ by connecting a series resistance which is $(p-1)$ times the galvanometer resistance.
$\therefore p =\frac{V}{V_g}=\frac{\left(R_{ s }+G\right) I_{ g }}{G I_{ g }}=\frac{R_{ s }+G}{G}$
Since the resistance of the voltmeter is $R_V=R_S+G$,
$p =\frac{R_{ v }}{G}$
$\therefore R _{ v }= G _{ p }$
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