Question

The current through a galvanometer of resistance $50 \Omega$ is only $4\%$ of the main current when a shunt resistance is connected. The value of the shunt resistance is approximately

• A. $3 \Omega$
• B. $4 \Omega$
• C. $5 \Omega$
• D. $1 \Omega$
• E. $2 \Omega$

Hint:

If $1/n$ portion of the current passes through the galvanometer, the shunt is given by \( S = \frac{G}{n-1} \). Here \( 4\% = 1/25 \), so \( S = 50 / (25 - 1) = 50/24 \approx 2.08 \Omega \).

Answer 1

Answer: (E)

Step 1: Understanding the Concept:
A galvanometer is converted into an ammeter by connecting a low resistance, called a shunt, in parallel. The total current splits such that the potential difference across the galvanometer and the shunt is the same.
Key Formula or Approach:
\[ I_g G = (I - I_g) S \]
Where $G$ is galvanometer resistance, $S$ is shunt resistance, $I$ is main current, and $I_g$ is current through the galvanometer.

Step 2: Detailed Explanation:
Given:
\( G = 50 \Omega \)
\( I_g = 4\% \text{ of } I = 0.04I \)
Current through shunt \( I_s = I - I_g = I - 0.04I = 0.96I \)
Using the parallel potential relation:
\[ 0.04I \times 50 = 0.96I \times S \]
Divide both sides by $I$:
\[ 0.04 \times 50 = 0.96 \times S \]
\[ 2 = 0.96 S \]
\[ S = \frac{2}{0.96} = \frac{200}{96} \approx 2.08 \Omega \]
The closest approximate value is $2 \Omega$.

Step 3: Final Answer:
The shunt resistance is approximately $2 \Omega$.