Conversion Of Galvanometer To Voltmeter

Question

To convert a galvanometer of resistance $60\ \Omega$ and full-scale current $5$ mA into a voltmeter reading up to $3$ V, the series resistance that must be added is

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As set out in NCERT Class 12, Chapter 4 (Moving Charges and Magnetism), a galvanometer becomes a voltmeter when a large resistance $R$ is added in series, so that only the tiny full-scale current flows even at the maximum voltage. The condition is $V = I_g (G + R)$, giving $R = \dfrac{V}{I_g} - G$. Substituting $V = 3$ V, $I_g = 5 	imes 10^{-3}$ A, $G = 60\ \Omega$: $R = \dfrac{3}{5 	imes 10^{-3}} - 60 = 600 - 60 = 540\ \Omega$. The value $600\ \Omega$ forgets to subtract the galvanometer resistance. The value $60\ \Omega$ merely repeats $G$. The value $300\ \Omega$ uses a doubled full-scale current. Plausibility check: a voltmeter needs a large series resistance so it draws negligible current, and $540\ \Omega$ is much larger than $60\ \Omega$, consistent with that requirement; units of $	ext{V}/	ext{A} = \Omega$ confirm the dimension. As a further sanity test, the total meter resistance is $G + R = 600\ \Omega$, and at the full $3$ V this draws exactly $3/600 = 5$ mA, precisely the galvanometer's full-scale current, so the design self-consistently reaches full deflection at the intended maximum voltage.

Conversion Of Galvanometer To Voltmeter

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Conversion Of Galvanometer To Voltmeter

Select the correct option:

🔒 Solution Hidden from View

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Solution

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