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:
To convert a galvanometer into an ammeter, a low-value "shunt" resistor is placed in parallel with it. This diverts most of the main current away from the sensitive galvanometer. Since they are in parallel, the voltage drop across both must be equal.
Step 2: Key Formula or Approach:
Parallel voltage condition: \(I_g \times G = I_s \times S\).
Current node condition: Total current \(I = I_g + I_s\).
Substitute \(I_s = I - I_g\) to get \(I_g G = (I - I_g)S\).
Step 3: Detailed Explanation:
Given values:
Galvanometer resistance \(G = 50 \, \Omega\).
Galvanometer current \(I_g = 4% \text{ of } I = 0.04I\).
Shunt current \(I_s = I - 0.04I = 0.96I\).
Equate the potential differences:
\[ V_g = V_s \] \[ I_g G = I_s S \] \[ (0.04I)(50) = (0.96I)S \] Cancel \(I\) from both sides:
\[ 0.04 \times 50 = 0.96 \times S \] \[ 2 = 0.96S \] Solve for \(S\):
\[ S = \frac{2}{0.96} = \frac{200}{96} = \frac{100}{48} = \frac{25}{12} \] \[ S \approx 2.083 \, \Omega \] Looking at the options, the value is approximately \(2 \, \Omega\).
Step 4: Final Answer:
The shunt resistance is approximately 2 \(\Omega\).