An electric kettle has two independent heating coils. When connected individually to a 220 V supply, the first coil boils a certain mass of water in 10 minutes , while the second coil takes 40 minutes to boil the same mass of water under identical conditions. If both coils are connected in series across the same supply, the time taken to boil the water is t_S . If they are connected in parallel, the time taken is t_P

Step-by-step solution

Let the heat required to boil the water be H and the supply voltage be V . For the first coil: H = V^2 R₁ t₁ R₁ = V^2 H t₁ For the second coil: H = V^2 R₂ t₂ R₂ = V^2 H t₂ When connected in series, the equivalent resistance is R_S = R₁ + R₂ . Time taken in series t_S = H V^2 R_S = H V^2 (R₁ + R₂) = t₁ + t₂ t_S = 10 + 40 = 50 minutes When connected in parallel, the equivalent resistance is R_P = R₁ R₂ R₁ + R₂ . Time taken in parallel t_P = H V^2 R_P = H V^2 ( R₁ R₂ R₁ + R₂ ) = t₁ t₂ t₁ + t₂ t_P = 10 40 10 + 40 = 400