Since the e.m.f of the battery is opposed to the e.m.f of the circuit, the resultant e.m.f acting in the circuit is 115 - 60 × 1⋅2 = 43V. For a current of 2⋅5A to flow through the circuit,
222 SOLUTIONS
given this e.m.f., the resistance in the circuit must equal 43 ÷ 2⋅5 = 17⋅2Ω. So the rheostat's resistance must be 17⋅2 - 60 × 0⋅02 = 16Ω.