1. The problem statement, all variables and given/known data 2. Relevant equations V=ir E=V-ir 3. The attempt at a solution First part is solved. The EMF of the battery is higher than the actual output voltage due to the internal resistance of the battery. E=11.25, V measured is 10.5. The second part is boggling my mind. When both switches are closed, the circuit acts as resistors in parallel. Voltages across the branches would be the same. Current through each branch would vary. Using Kirkhoff's loop rules, starting the LOWER RIGHT corner of the diagram: Loop 1: [itex]i_b(R_b)+E -i_s=0 [/itex] Loop 2: [itex]i_b(R_b)+e-i_L=0[/itex] Equation 3: [itex]i_b = i_s + i_L[/itex] The current through the lights [itex](i_L)[/itex] is given as 14A. Therefore, we can use the idea that the voltage across the battery is the same as the voltage across the lights to solve for the resistance of the lights. V=IR [itex]10.5 = 14(R_L)[/itex] [itex]R_L = .75[/itex] At this point, I don't know how to proceed to get any closer to the solution.