RRB Group-D Model Paper 12 with answers for online practice

CONCEPT:

• A galvanometer can be converted into an ammeter by connecting a shunt resistance in parallel to it.
• The shunt resistance should have very low resistance. So, the ammeter (the parallel combination of galvanometer and shunt resistance) will have low resistance.
• To convert a galvanometer into an ammeter of current rating ‘I’, a small resistance ‘S’ (shunt resistance) is connected in parallel across the galvanometer.
${\mathit{V}}_{\mathit{g}}=\left(\mathit{I}-{\mathit{I}}_{\mathit{g}}\right)\mathit{S}={\mathit{I}}_{\mathit{g}}{\mathit{R}}_{\mathit{g}}$
Where ${\mathit{V}}_{\mathit{g}}$ is the voltage across the galvanometer, I is current in the circuit, ${\mathit{I}}_{\mathit{g}}$ is the current is galvanometer, ${\mathit{R}}_{\mathit{g}}$ is the resistance of galvanometer and S is the resistance of the shunt.

EXPLANATION:

• A moving coil galvanometer can be converted into a ammeter by connecting a low resistance in in parallel to the moving coil galvanometer. So option 2 is correct.