Question

\( I_g = 8\% \times I \). What is \( S \) (shunt) connected in terms of \( G \)?

• A. \( \frac{G}{11} \)
• B. \( \frac{2G}{23} \)
• C. \( \frac{3G}{25} \)
• D. \( \frac{4G}{29} \)

Hint:

The relationship \( S = \frac{I_g}{I_s} \cdot G \) is useful for calculating the shunt resistance, where \( I_g \) is the galvanometer current and \( I_s \) is the shunt current.

Answer 1

The Correct Option is B

A shunt resistance \( S \) is connected in parallel with the galvanometer resistance \( G \) to extend its range. The total current \( I \) is partitioned into two flows: \( I_g \) through the galvanometer and \( I_s \) through the shunt. Given that \( I_g = 8\% \times I = 0.08I \), the current through the shunt is calculated as \( I_s = I - I_g = I - 0.08I = 0.92I \). Step 1: Establish the relationship between \( S \) and \( G \). The voltage across \( S \) is equal to the voltage across \( G \): \(I_s \cdot S = I_g \cdot G\). Substituting the current values: \(0.92I \cdot S = 0.08I \cdot G\). Canceling \( I \) yields: \(0.92S = 0.08G\). Solving for \( S \): \(S = \frac{0.08G}{0.92}\). Step 2: Simplify the fractional expression. \(S = \frac{8G}{92} = \frac{2G}{23}\). The shunt resistance is \(S = \frac{2G}{23}\).