WAEC Physics 2020 Paper

Concept:A galvanometer is converted into a voltmeter by connecting a high resistance in series with it.Explanation:Let the galvanometer resistance be $G$.Full-scale deflection current is $I_g = 0.005\ \mathrm{A}$.Maximum voltage to be measured is $V = 5\ \mathrm{V}$.The external series resistance is $R = 975\ \Omega$.In a series circuit, the total voltage is:$$V = I_g(G + R)$$Substitute the given values:$$5 = 0.005(G + 975)$$Divide both sides by $0.005$:$$G + 975 = \frac{5}{0.005} = 1000$$Therefore:$$G = 1000 - 975 = 25\ \Omega$$Answer:The resistance of the metre is $25.00\ \Omega$.