Question

What is the shunt resistance \(S\) needed if the galvanometer current \(I_g\) is \(8\%\) of the total current \(I\)?

• A. \(\frac{2G}{23}\)
• B. \(\frac{8G}{23}\)
• C. \(\frac{23G}{2}\)
• D. \(\frac{G}{23}\)

Hint:

In a galvanometer–shunt combination, use \(I_gG = I_sS\). The shunt carries the majority of the current, protecting the galvanometer from large currents.

Answer 1

The Correct Option is A

Concept: A galvanometer can be converted into an ammeter by connecting a low resistance called a shunt in parallel with the galvanometer. The shunt allows most of the current to bypass the galvanometer so that only a small fraction of the total current flows through it. If \(G\) = resistance of the galvanometer, \(S\) = shunt resistance, \(I_g\) = current through the galvanometer, \(I\) = total current. Since the galvanometer and shunt are connected in parallel, the potential difference across them is the same. \[ I_g G = I_s S \] where \(I_s\) is the current through the shunt.

Step 1: Express galvanometer current as a fraction of total current.
\[ I_g = 8\% \, \text{of} \, I \] \[ I_g = 0.08I \]

Step 2: Determine the current through the shunt.
\[ I_s = I - I_g \] \[ I_s = I - 0.08I \] \[ I_s = 0.92I \]

Step 3: Use the parallel branch relation.
\[ I_g G = I_s S \] Substitute the values: \[ 0.08I \cdot G = 0.92I \cdot S \]

Step 4: Solve for shunt resistance.
Cancel \(I\) from both sides: \[ 0.08G = 0.92S \] \[ S = \frac{0.08}{0.92}G \] \[ S = \frac{8}{92}G \] \[ S = \frac{2G}{23} \]

Step 5: Final result.
\[ \boxed{S = \frac{2G}{23}} \]