Worked example: A 0–10 A ammeter uses a galvanometer (G = 90 Ω, Ig = 5 mA) with a shunt. (a) Find S. (b) If the shunt warms up and its resistance rises 2%, what fractional error appears in the reading? Step 1: S = Ig·G/(I − Ig) = (0.005 × 90)/9.995 = 0.45/9.995 ≈ 0.04502 Ω ≈ 45 mΩ. Step 2: Coil current fraction = S/(G + S); total I splits as Icoil = I·S/(G+S). If S rises 2%, Icoil rises ~2% (G ≫ S), so the meter reads ~2% high — it over-reads because more current is forced through the coil. Step 3: This is why shunts are manganin (α ≈ 10⁻⁵/°C): a 20°C rise changes S by only 0.02%, keeping the error invisible. Design moral: material choice is part of the formula.