Question

Sol. [Key Concept: Galvanometer → ammeter using a shunt] $i_{G} = 1\text{ }mA = 0.001\text{ }A$

$$i_{S} = 10 - i_{G} \simeq 10\text{ }A$$

Both shunt resistance and galvanometer are in parallel connection

$$\begin{matrix} \therefore & i_{s}r_{s} = i_{G}R_{G} \\ \Rightarrow & 10 \times r_{s} = 0.001 \times 100 \\ \Rightarrow & r_{s} = 0.01\Omega \end{matrix}$$

In a meter bridge experiment (see figure), the positions of the cell, $E$, and galvanometer, $G$, are interchanged.