Question

A shunt of resistance \[1{\text{ }}\Omega \] is connected across a galvanometer of \[120{\text{ }}\Omega \] resistance. A current of \[5.5{\text{ A }}\] gives full-scale deflection in the galvanometer. The current that will give full-scale deflection in the absence of the shunt is nearly:
A) \[5.5{\text{ A }}\]
B) \[0.5{\text{ A }}\]
C) \[0.004{\text{ A }}\]
D)\[0.045{\text{ A }}\]

Answer 1

Hint: In an ammeter, a galvanometer and a shunt resistance are connected in parallel. The current that will give full-scale deflection in the absence of the shunt is nearly equal to the current through the galvanometer when the shunt is connected.
Formula used: In this solution, we will use the following formula:
${I_g} = \dfrac{{IS}}{{G + S}}$ where $I$ is the current and $S$ is the resistance of the shunt and $G$ is the resistance of the galvanometer.

Complete step by step answer:
As we discussed in the hint, the current that will give full-scale deflection in the absence of the shunt is equal to the current through the galvanometer when the shunt is connected.
The current in the galvanometer when the shunt is connected will be equal to
${I_g} = \dfrac{{IS}}{{G + S}}$ where ${I_g}$ is the current that gives full-scale deflection in the galvanometer and $I$ is the current in the galvanometer which gives full-scale deflection in the galvanometer when the shunt is connected.
Hence the current will be
${I_g} = \dfrac{{5.5 \times 1}}{{120 + 1}}$
$ \Rightarrow {I_g} = 0.045\,A$
This will also be the current in the galvanometer when the shunt is removed that will provide full deflection.

Hence the correct choice is option (D).

Note: We should not confuse between the current in the circuit that will give full-scale deflection in the presence of the shunt and the absence of the shunt. This is because the shunt is connected in parallel with the galvanometer which will change the amount of current in the galvanometer which will then affect the direction of the deflection of the galvanometer.