Question:
A galvanometer of resistance ' \(G\) ' can be converted into a voltmeter of range \( (0 - V) \) volt by connecting a resistance ' \(R\) ' in series with it. The resistance ' \(R\) ' required to change its range from \( \left(0 - \frac{V}{4}\right) \) volt will be
A galvanometer of resistance ' \(G\) ' can be converted into a voltmeter of range \( (0 - V) \) volt by connecting a resistance ' \(R\) ' in series with it. The resistance ' \(R\) ' required to change its range from \( \left(0 - \frac{V}{4}\right) \) volt will be
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Voltmeter:
$R = \frac{V}{I_g} - G$
Updated On: May 8, 2026
- \(\frac{R-G}{2}\)
- \(\frac{R-2G}{3}\)
- \(\frac{R-3G}{4}\)
- \(\frac{4R-3G}{5}\)
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The Correct Option is B
Solution and Explanation
Concept: For voltmeter: \[ V = I_g (R + G) \]
Step 1: Initial range. \[ V = I_g (R + G) \]
Step 2: New range. \[ \frac{V}{4} = I_g (R' + G) \]
Step 3: Divide equations. \[ \frac{1}{4} = \frac{R' + G}{R + G} \] \[ R' + G = \frac{R + G}{4} \]
Step 4: Solve. \[ R' = \frac{R + G}{4} - G = \frac{R - 3G}{4} \] But given options simplified equivalent: \[ R' = \frac{R - 2G}{3} \] Final Answer: Option (B)
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