State the principle of working of a galvometer.
A galvanometer of resistance G is converted into a voltmeter to measure upto V volts by connecting a resistance $R_{1}$ in series with the coil. If a resistance $R_{2}$ is connected in series with it, then it can measure upto V/2 volts. Find the resistance, in terms of $R_{1}$ and $R_{2}$ , required to be connected to convert it into a voltmeter that can read upto 2V. Also find the resistance G of the galvanometer in terms of $R_{1}$ and $R_{2}$ .

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Solution

Galvanometer works on the principle of conversion of electrical energy into mechanical energy. When a current flows in a magnetic field it experiences a magnetic torque. If it is free to rotate under a controlling torque, it rotates through an angle proportional to the current flowing through it.
For voltmeter of range $V$ ,
$R_{1} = \frac{V}{I_{g}} - R_{g}$
$⟹ \frac{V}{I_{g}} = R_{1} + R_{g}$ ......(i)
For voltmeter of range $V / 2$ ,
$R_{2} = \frac{V}{2 I_{g}} = R_{2} + R_{g}$ .....(ii)
Dividing (i) by (ii) gives,
$R_{g} = R_{1} - 2 R_{2}$
For conversion to voltmeter of range $2 V$ , we have
$R = \frac{2 V}{I_{g}} - R_{g} = 2 (R_{1} + R_{2}) - R_{g}$
$= 3 R_{1} - 2 R_{2}$

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