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A galvanometer shows full-scale deflection for current Ig. A resistance R1 is required to convert it into a voltmeter of range (0 - V) and a resistance R2 to convert it

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Question

A galvanometer shows full-scale deflection for current Ig. A resistance R1 is required to convert it into a voltmeter of range (0 - V) and a resistance R2 to convert it into a voltmeter of range (0 - 2V). Find the resistance of the galvanometer.

Derivation
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Solution

In the first case the potential difference between A and B,

`V = I_gR_1 + I_gG`

`I_g = V/(R_1 + G)`

Where G = Resistance of galvanometer

∴ `R_1 + G = V/I_g`

`R_1 = V/(Ig) - G`

⇒ G = `V/(Ig) - R_1`

In the second case,

`I_g = V/(R_2 + G)`

R2 is the Resistance required to change the range from 0 - 2V.

`I_g = (2V)/(R_2 + G)`

`R_2 + G = (2V)/I_g`

`R_2 = (2V)/I_g - G`

⇒ R = `(2V)/I_g - R_2`

Hence, the resistance of the galvanometer is respectively `V/(Ig) - R_1` and `(2V)/(Ig) - R_2`.

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