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Q.
To convert a galvanometer into voltmeter we must connect a:
AFMCAFMC 2002
Solution:
Let $G$ be the resistance of the galvanometer and $i_{g}$ the current, which on passing through the galvanometer, produces full-scale deflection.
Suppose $V$ is the maximum potential difference to be measured which exists between points $a$ and $b$. On connecting the galvanometer across $a$ and $b$ a current $i_{g}$ flows through it. Then from Ohm's law we have.
$i_{g} =\frac{V}{G+R}$
$G+R =\frac{V}{i_{g}}$
$\Rightarrow R =\frac{V}{i_{g}}-G$
Thus, on connecting a resistance $R$ of above value in series with the galvanometer the galvanometer will become a voltmeter of range $0$ to $V$ volts.
The value of resistance should be high enough so that it does not draw any current from the circuit.
Resistance of an ideal voltmeter is high.
