# (a) Briefly explain how a galvanometer is converted into a voltmeter.(b) A voltmeter of a certain range is constructed by connecting a resistance of $980\; \Omega$ in series with a galvanometer. When the resistance of $470\; \Omega$ is connected in series, the range gets halved. Find the resistance of the galvanometer.

(a) A galvanometer can be converted into Voltmeter by connecting a high-value resistance R in series with the coil of the galvanometer. The value of (R) is related to the maximum voltage (V) to be measured as

$R=\frac{v}{I_{g}}-G$

(b) To determine the resistance of the galvanometer :

We have, $I_{g}=\frac{v}{R_{g}+R}$

Where, $R_{g}=$ Resistance of a galvanometer

$I_{g}=$ Current in galvanometer

Now, According to the Question,

$\frac{V}{R_{g}+980}=\frac{V}{2(R_{g}+470)}$

therefore, $2R_{g}+940=R_{g}+980$

Hence,     $R_{g}=40\; \Omega$

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