Question

A galvanometer of resistance 'G' is converted into an ammeter of resistance $\frac{G}{40}$ by connecting a shunt 'S' to it. The part of main current passing through the shunt is

• A. 2.5%
• B. 25%
• C. 40%
• D. 97.5%

Answer 1

Answer: (D)

97.5%

Answer 2

Let the galvanometer resistance be $G$ and the shunt resistance be $S$. When connected in parallel, the effective resistance is

$$R_{\text{eff}} = \frac{G\,S}{G+S} = \frac{G}{40}$$

Solving for $S$:

$$\frac{G\,S}{G+S} = \frac{G}{40} \quad\Longrightarrow\quad \frac{S}{G+S} = \frac{1}{40} \quad\Longrightarrow\quad 40S = G+S \quad\Longrightarrow\quad 39S = G \quad\Longrightarrow\quad S = \frac{G}{39}.$$

For current division in a parallel circuit, the current through the galvanometer $I_g$ is:

$$I_g = I \cdot \frac{S}{G+S} = I \cdot \frac{\frac{G}{39}}{G+\frac{G}{39}} = I \cdot \frac{\frac{G}{39}}{\frac{40G}{39}} = \frac{I}{40} = 2.5\%\; \text{of } I.$$

Thus, the current through the shunt $I_s$ is:

$$I_s = I - I_g = I - \frac{I}{40} = \frac{39I}{40} = 97.5\%\; \text{of } I.$$