Practice question

**Heating effect** depends on I² R t, explaining why high currents cause significant heating, need for thick wires, fuses. Energy supplied by battery ε I t = I²(R+r)t, split between external and internal as per resistances. Total resistance: R = 5 + 10 = 15 Ω . Current: I = (V/R) = (10/15) = (2/3) A . Power in 10 Ω : P = I² R = ((2/3))² × 10 = (4/9) × 10 = 4.44 W . Applying I = n e A v_d, R = ρ l/A, R_t = R₀[1+αΔT], Kirchhoff's ΣI=0, ΣV=0, R_eq series/parallel, V = ε - I r and P