Question

A voltmeter of a certain range is constructed by connecting a resistance of 980 in series with a galvanometer. When the resistance of 470 is connected in series, the range gets halved. Find the resistance of the galvanometer.

Answer 1

Answer:40

Explanation: current same rahega potential gets half then divide the both equations and get answers

Answer 2

The resistance of the galvanometer is 40 ohm.

Explanation:

The conversion of galvanometer in voltmeter when large resistance connected to galvanometer in series.

and given as

$V=i_g(R_G+R)$

Here, $R_G$ is the resistance of galvanometer and R is the resistance and i_g is the resistance through galvanometer.

So, $V=i_g(R_G+980\Omega)$               (1)

When the resistance of 470 is connected in series, the range gets halved, then  

$\frac{V}{2} =i_g(R_G+470\Omega)$                      (2)

from both equations,

$\frac{i_g(R_G+980\Omega)}{2} =i_g(R_G+470\Omega)$

After solving, we get

$R_G=40\Omega$

Thus, the resistance of the galvanometer is 40 ohm.

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