Question:

The range of voltmeter is \(10\ \text{V}\) and its resistance is \(50\ \Omega\). To convert it to a voltmeter of range \(15\ \text{V}\), how much resistance is to be added?

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Find the full-scale current from the original range, then the total resistance needed for the new range; the difference is added in series.
Updated On: Nov 12, 2026
  • \(25\ \Omega\) resistor in parallel.
  • \(25\ \Omega\) resistor in series.
  • \(125\ \Omega\) resistor in parallel.
  • \(125\ \Omega\) resistor in series.
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The Correct Option is B

Solution and Explanation

Step 1: Understanding the Concept
A voltmeter shows full scale when the current is \(I_{g}=\frac{V}{R}\). To increase its range, a resistor is connected in series.

Step 2: Calculation
\(I_{g}=\frac{10}{50}=0.2\ \text{A}\). For \(15\ \text{V}\), the total resistance is \(\frac{15}{0.2}=75\ \Omega\). The extra resistance is \(75-50=25\ \Omega\).

Step 3: Compare the options
The resistor must be in series so that it takes the extra \(5\ \text{V}\) (\(0.2\times 25=5\ \text{V}\)). A parallel resistor would reduce the range. So the answer is \(25\ \Omega\) in series.

Final Answer:
A \(25\ \Omega\) resistor is connected in series. \[ \boxed{\text{(B) }25\ \Omega\ \text{resistor in series}} \]
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