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 galvanometer is

• A. $25\%$
• B. $10\%$
• C. $2.5\%$
• D. $0.4\%$

Hint:

- In shunt: current divides inversely with resistance - Fraction through galvanometer = $\frac{1}{1 + (G/S)}$

Answer 1

The Correct Option is C

Concept:
When a galvanometer (resistance $G$) is converted into an ammeter by connecting a shunt $S$ in parallel: \[ R_{\text{ammeter}} = \frac{GS}{G + S} \]

Step 1: Use given condition.
\[ \frac{GS}{G + S} = \frac{G}{40} \]

Step 2: Solve for $S$.
\[ \frac{GS}{G + S} = \frac{G}{40} \Rightarrow 40GS = G(G + S) \] \[ 40GS = G^2 + GS \Rightarrow 39GS = G^2 \Rightarrow S = \frac{G}{39} \]

Step 3: Find current division.
In parallel: \[ \frac{I_g}{I_s} = \frac{S}{G} \] \[ \frac{I_g}{I_s} = \frac{G/39}{G} = \frac{1}{39} \]

Step 4: Express in terms of total current.
\[ I = I_g + I_s = I_g + 39I_g = 40I_g \] \[ \frac{I_g}{I} = \frac{1}{40} \]

Step 5: Convert to percentage.
\[ \frac{1}{40} \times 100 = 2.5\% \]