Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

A voltmeter has a range $V$ with a series resistance $R$ . With a series resistance $2R$ , the range is $V'$ . The correct relation between $V$ and $V'$ is:

$V' = 2V$

$V' > 2V$

$V' >> 2V$

$V' < 2V$

Step-by-Step Solution

Principle: A voltmeter is constructed by connecting a high resistance in series with a galvanometer. The range ( $V$ ) is determined by the full-scale deflection current ( $I_g$ ) and the total resistance of the circuit .
Case 1 (Resistance $R$ ): Let $G$ be the internal resistance of the galvanometer. The range $V$ is given by: $V = I_g (G + R)$
Case 2 (Resistance $2R$ ): The new series resistance is $2R$ . The new range $V'$ is given by: $V' = I_g (G + 2R)$
Comparison: We need to compare $V'$ with $2V$ . $2V = 2 \times I_g (G + R) = I_g (2G + 2R)$ $V' = I_g (G + 2R)$ Subtracting the two expressions: $2V - V' = I_g (2G + 2R) - I_g (G + 2R) = I_g (G)$ Since the galvanometer resistance $G$ and current $I_g$ are positive, $2V - V' > 0$ , which implies $2V > V'$ or $V' < 2V$ .
Conclusion: The new range $V'$ is less than twice the original range $V$ .

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