Two real batteries and a resistor are connected in a single closed loop. Battery A has emf ε1 and internal resistance r1. Battery B has emf ε2 and internal resistance r2. Between the two batteries is a resistor R. All three elements are connected in series, and their orientations are such that the positive terminal of Battery A is connected to one end of resistor R, while the positive terminal of Battery B is connected to the other end of R.
Analyze the circuit in general terms, then apply your results to a specific numerical case.
(1) Part 1: Derive an expression for the current I in the loop in terms of ε1, ε2, r1, r2, and R. Clearly state your sign convention for the current direction.
(2) Part 2: Obtain the condition on ε2 that must be satisfied for the current in the loop to be zero. Express your answer as a relationship among ε1, r1, r2, and R.
(3) Part 3: Using ε1=12V, r1=0.50Ω, r2=0.30Ω, and R=4.0Ω, calculate the value of ε2 that makes the current zero.
(4) Part 4: Suppose Battery B is reversed so that its negative terminal now connects to the same end of resistor R as the positive terminal of Battery A. Without re-deriving the entire expression, explain qualitatively how the required value of ε2 to achieve zero current would change, and justify your reasoning in terms of potential changes around the loop.