Conversion Of Galvanometer To Voltmeter

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 $R=\frac{V}{I_g}-G$. Substituting $V=3$ V, $I_g=5\times 10^{-3}$ A, $G=60\Omega$: $R=\frac{3}{5\times 10^{-3}}-60=600-60=540\Omega$. 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 $\text{V}/\text{A}=\Omega$ 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.