Question:
By connecting a resistance of \(980 \Omega\) in series with a galvanometer, it is converted into a voltmeter of a certain range. When the resistance of \(470 \Omega\) is connected in series, the range is halved. The resistance of the galvanometer is
By connecting a resistance of \(980 \Omega\) in series with a galvanometer, it is converted into a voltmeter of a certain range. When the resistance of \(470 \Omega\) is connected in series, the range is halved. The resistance of the galvanometer is
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The range V = Ig (G + R) halves when the series resistance changes from 980 to 470 ohm.
Updated On: Oct 1, 2026
- \(30 \Omega\)
- \(40 \Omega\)
- \(50 \Omega\)
- \(60 \Omega\)
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept:
A galvanometer of resistance \(G\) and full scale current \(I_g\) becomes a voltmeter of range \(V = I_g(G + R)\) when a resistance \(R\) is put in series.
Step 2: Key Formula or Approach:
First case: \(V_1 = I_g(G + 980)\). Second case: \(V_2 = I_g(G + 470)\), with \(V_2 = \frac{V_1}{2}\).
Step 3: Detailed Explanation:
\(G + 470 = \frac12(G + 980)\).
\(2G + 940 = G + 980\).
\[ G = 40\ \Omega \]
Check: \(V_1 \propto 1020\) and \(V_2 \propto 510\), which is half.
Final Answer:
The resistance of the galvanometer is \(40\ \Omega\), option (B). \[ \boxed{40\ \Omega} \]
A galvanometer of resistance \(G\) and full scale current \(I_g\) becomes a voltmeter of range \(V = I_g(G + R)\) when a resistance \(R\) is put in series.
Step 2: Key Formula or Approach:
First case: \(V_1 = I_g(G + 980)\). Second case: \(V_2 = I_g(G + 470)\), with \(V_2 = \frac{V_1}{2}\).
Step 3: Detailed Explanation:
\(G + 470 = \frac12(G + 980)\).
\(2G + 940 = G + 980\).
\[ G = 40\ \Omega \]
Check: \(V_1 \propto 1020\) and \(V_2 \propto 510\), which is half.
Final Answer:
The resistance of the galvanometer is \(40\ \Omega\), option (B). \[ \boxed{40\ \Omega} \]
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