Question

In an ammeter, 20% of the main current passes through the galvanometer. If the resistance of the galvanometer is G, then the shunt resistance in ohms is

विकल्प

• \[\frac{G}{4}\]

• \[\frac{G}{8}\]

• \[\frac{G}{9}\]

• \[\frac{G}{10}\]

Answer 1

\[\frac{G}{4}\]

Explanation:

\[\mathrm{I_{g}=\frac{20}{100}}\]\[\mathrm{I}=\frac{\mathrm{I}}{5}\]

Using S = \[\frac{\mathrm{I_{g}G}}{\mathrm{I-I_{g}}}\] we get,

\[\therefore\quad\mathrm{S=~\frac{I_gG}{I-I_g}=\frac{\left(\frac{I}{5}\right)G}{\left(I-\frac{I}{5}\right)}=\frac{\left(\frac{I}{5}\right)G}{\frac{4}{5}I}=\frac{G}{4}}\]