Question

When current \(I\) is passed through galvanometer coil, it is deflected through \(16^{\circ }\). But when it is shunted by a resistance of \(8\Omega\), the same current produces a deflection of \(4^{\circ }\). The resistance of the galvanometer is

• A. \(48\Omega\)
• B. \(16\Omega\)
• C. \(24\Omega\)
• D. \(12\Omega\)

Hint:

Deflection is proportional to the current through the coil; with a shunt S the coil current is I times S over G plus S.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
The deflection is proportional to the current through the coil. Without a shunt it is \(\theta_{1}\propto I\).

Step 2: With the shunt
The coil current becomes \(I_{g}=I\frac{S}{G+S}\).
\(\frac{\theta_{2}}{\theta_{1}}=\frac{S}{G+S}\)

Step 3: Calculation
\(\frac{4}{16}=\frac{8}{G+8}\Rightarrow G+8=32\Rightarrow G=24\,\Omega\)

Step 4: Compare the options
This is option (C).

Final Answer:
The galvanometer resistance is 24 ohm. \[ \boxed{\text{(C) }24\,\Omega} \]