Question

When a resistance of \(100\Omega\) is connected in series with galvanometer of resistance \(R\), its range is \(V\). To double its range, another additional resistance of \(1000\Omega\) is connected in series. Find the value of \(R\).

• A. \(700\Omega\)
• B. \(800\Omega\)
• C. \(900\Omega\)
• D. \(100\Omega\)

Hint:

The range is the full-scale current times the total series resistance; doubling the range doubles the total resistance.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
With series resistance, the range is \(V=I_{g}(R+R_{s})\), where \(I_{g}\) is the full-scale current.

Step 2: Two Cases
First: \(V=I_{g}(R+100)\).
Second: \(2V=I_{g}(R+100+1000)=I_{g}(R+1100)\).

Step 3: Solve
\(\dfrac{R+1100}{R+100}=2\)
\(R+1100=2R+200\Rightarrow R=900\,\Omega\)

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

Final Answer:
The galvanometer resistance is 900 ohm. \[ \boxed{\text{(C) }900\,\Omega} \]