Question

A voltmeter with series resistance \(R\) has a range \((0-V_{1})\) volt. With a series resistance of \(4R\), its range becomes \((0-V_{2})\) volt. The relation between \(V_{1}\) and \(V_{2}\) is

• A. \(V_{2}=4V_{1}\)
• B. \(V_{2}\lt 4V_{1}\)
• C. \(V_{2}>4V_{1}\)
• D. \(V_{2}=\frac{V_{1}}{4}\)

Hint:

The galvanometer's own resistance is also in series with the added resistance.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
Range of a voltmeter is \(V=I_{g}(G+R)\), where G is the galvanometer resistance.

Step 2: Two ranges
\(V_{1}=I_{g}(G+R)\) and \(V_{2}=I_{g}(G+4R)\).

Step 3: Compare
\(4V_{1}=I_{g}(4G+4R)\) which is larger than \(V_{2}\) because \(4G>G\). So \(V_{2}\lt4V_{1}\).

Final Answer:
The second range is less than four times the first range. \[ \boxed{\text{(B) }V_{2}\lt 4V_{1}} \]