Question:
A galvanometer can be converted into a voltmeter of certain range by connecting a resistance of \(990\ \Omega\) in series with it. When the resistance of \(480\ \Omega\) is connected in series, the range is halved. The resistance of the galvanometer is
A galvanometer can be converted into a voltmeter of certain range by connecting a resistance of \(990\ \Omega\) in series with it. When the resistance of \(480\ \Omega\) is connected in series, the range is halved. The resistance of the galvanometer is
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Range is the full-scale current times the total series resistance; compare the two cases.
Updated On: Nov 12, 2026
- \(20\ \Omega\)
- \(30\ \Omega\)
- \(40\ \Omega\)
- \(50\ \Omega\)
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept
Voltmeter range \(V=I_g(G+R)\), where \(G\) is the galvanometer resistance and \(I_g\) the full-scale current.
Step 2: Two cases
\(V=I_g(G+990)\)
\(\dfrac{V}{2}=I_g(G+480)\)
Step 3: Divide
\(\dfrac{G+990}{G+480}=2\Rightarrow G+990=2G+960\)
Step 4: Solve
\(G=30\ \Omega\)
Final Answer:
The resistance of the galvanometer is 30 ohm. \[ \boxed{\text{(B) }30\ \Omega} \]
Voltmeter range \(V=I_g(G+R)\), where \(G\) is the galvanometer resistance and \(I_g\) the full-scale current.
Step 2: Two cases
\(V=I_g(G+990)\)
\(\dfrac{V}{2}=I_g(G+480)\)
Step 3: Divide
\(\dfrac{G+990}{G+480}=2\Rightarrow G+990=2G+960\)
Step 4: Solve
\(G=30\ \Omega\)
Final Answer:
The resistance of the galvanometer is 30 ohm. \[ \boxed{\text{(B) }30\ \Omega} \]
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