In the circuit diagram above the ammeter reads ... - JAMB Physics 2001 Question

for a parallel R and X, when R = 5Ω,

and when R = 2Ω

Thus since E = I(R+r), and if r = 0 then E = IR

∴(i) E = 3(5X/(5+x)); (ii) E = 6(2X(2+x))

∴3(5X/(5+X)) = 6(2X/(2+X))

=> (5X/(5+X)) = 2(2x/(2+X))

∴ (5X/(5+X)) = (4X/(2+X))

5x(2+x) = 4x(5+x)

10x + 5x² = 20x + 4x²

5x² - 4x² = 20x - 10x

=>x² = 10x

∴x² - 10x = 0

x(x - 10) = 0

∴ x = 0 or 10

∴ x = 10Ω