Question:

We have a galvanometer of resistance \(25\Omega\). It is shunted by \(2.5\Omega\) wire. The part of the total current that flows through the galvanometer is given as

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Shunt resistance \(S = \frac{G}{n-1}\) where \(n = i_0/i\).
Updated On: Apr 20, 2026
  • \(\frac{i}{i_0} = \frac{4}{11}\)
  • \(\frac{i}{i_0} = \frac{3}{11}\)
  • \(\frac{i}{i_0} = \frac{2}{10}\)
  • \(\frac{i}{i_0} = \frac{1}{11}\)
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The Correct Option is D

Solution and Explanation

Step 1: Understanding the Concept:
In a parallel combination, voltage across galvanometer = voltage across shunt.
Step 2: Detailed Explanation:
\(iG = (i_0 - i)S\) \(\Rightarrow iG = i_0S - iS\) \(\Rightarrow i(G+S) = i_0S\)
\(\frac{i}{i_0} = \frac{S}{G+S} = \frac{2.5}{25+2.5} = \frac{2.5}{27.5} = \frac{1}{11}\).
Step 3: Final Answer:
\(\boxed{\frac{i}{i_0} = \frac{1}{11}}\)
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