Question

When a galvanometer is shunted with a \(2\ \Omega\) resistance, the deflection is reduced to one-fourth. If the galvanometer is further shunted with a \(1\ \Omega\) resistance, the further reduction in deflection is a fraction \(x\) of the deflection observed when shunted with \(2\ \Omega\). Find \(x\), assuming the main current \(I\) remains the same.

• A. \(\frac{3}{5}\)
• B. \(\frac{5}{3}\)
• C. \(\frac{2}{3}\)
• D. \(\frac{1}{3}\)

Hint:

Find the galvanometer resistance from the first shunt, then compute the current with the combined shunt.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
With a shunt \(S\) across a galvanometer of resistance \(G\), the current through the galvanometer is \(I_g=I\cdot\dfrac{S}{G+S}\). Deflection is proportional to \(I_g\).

Step 2: Find G
With \(S=2\ \Omega\), the deflection is \(\dfrac{1}{4}\) of the full value: \(\dfrac{2}{G+2}=\dfrac{1}{4}\), so \(G=6\ \Omega\).

Step 3: Add the second shunt
Placing \(1\ \Omega\) in parallel with \(2\ \Omega\) gives \(S'=\dfrac{2}{3}\ \Omega\).
Then \(\dfrac{I_g'}{I}=\dfrac{2/3}{6+2/3}=\dfrac{1}{10}\).

Step 4: Fractional reduction
The deflection falls from \(\dfrac{1}{4}\) to \(\dfrac{1}{10}\), a drop of \(\dfrac{3}{20}\).
As a fraction of \(\dfrac{1}{4}\): \(x=\dfrac{3/20}{1/4}=\dfrac{3}{5}\).

Final Answer:
The fractional further reduction is \(x=\dfrac{3}{5}\). \[ \boxed{\text{(A) }\dfrac{3}{5}} \]