A voltmeter has a range $V$ with a series resistance $R$. With a series resistance $2R$, the range is $V'$. Th

Question

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:

Accepted Answer

1.  **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 .
2.  **Case 1 (Resistance $R$):** Let $G$ be the internal resistance of the galvanometer. The range $V$ is given by:
    $V = I_g (G + R)$
3.  **Case 2 (Resistance $2R$):** The new series resistance is $2R$. The new range $V'$ is given by:
    $V' = I_g (G + 2R)$
4.  **Comparison:** We need to compare $V'$ with $2V$.
    $2V = 2 	imes I_g (G + R) = I_g (2G + 2R)$
    $V' = I_g (G + 2R)$
    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 $V' < 2V$.
5.  **Conclusion:** The new range $V'$ is less than twice the original range $V$.

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