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. Let emf = ε , internal resistance = r , external resistance = R = r . Total resistance = r + R = 2r . Current = I = ε / (2r) . Power in external resistor = I² R = (ε / 2r)² r = ε² r / (4 r²) = ε² / (4 r) . Total power = ε I = ε · ε / (2r)