2. We are "jumping" a car battery because somebody left their headlights on while parked. (Jumping a battery is synonymous with the situation described in Prob. 1.12 of N&R which you solved in the previous homework.) The fresh car battery has a chemical voltage 13.7 V in series with an internal resistance of 20 m2. The dead battery has a chemical voltage of 11.9 V (not high enough to start) in series with an internal resistance of 50 m2. The jumper cables have a resistance of 3 m in the red cable and 3 m in the black cable. (a) Draw the circuit diagram for this situation and label all components. (b) What is the useful power delivered to the dead battery's chemistry?

2. We are "jumping" a car battery because somebody left their headlights on while parked. (Jumping a battery is synonymous with the situation described in Prob. 1.12 of N&R which you solved in the previous homework.) The fresh car battery has a chemical voltage 13.7 V in series with an internal resistance of 20 m2. The dead battery has a chemical voltage of 11.9 V (not high enough to start) in series with an internal resistance of 50 m2. The jumper cables have a resistance of 3 m in the red cable and 3 m in the black cable. (a) Draw the circuit diagram for this situation and label all components. (b) What is the useful power delivered to the dead battery's chemistry?