Final answer:
The potential drop across an external resistor R connected to a parallel combination of two identical batteries, each with emf e and internal resistance r, is given by 2eR / (2R + r). The combined emf is e, and the effective internal resistance is r/2. We use Kirchhoff's voltage law and Ohm's law to derive the result.
Step-by-step explanation:
The question relates to finding the potential drop across a resistor R when it is connected to a parallel combination of two identical batteries, each with an electromotive force (emf) e and an internal resistance r. Considering that the batteries are identical and connected in parallel, the combined emf remains e, and the effective internal resistance of the parallel combination is r/2.
By applying Kirchhoff's voltage law to the loop containing the external resistor R and the parallel combination of batteries, the potential drop across R can be calculated as the product of the current flowing through R and the resistance R itself. The total voltage drop must equal the emf of the battery minus the voltage drop across the combined internal resistance when the current is flowing through the circuit.
To calculate the potential drop across R, first, we find the total resistance of the circuit, which is R + r/2. The total current I in the circuit is given by I = e / (R + r/2). Thus, the potential drop across R is V = IR = eR / (R + r/2). Simplifying, we get V = 2eR / (2R + r). Hence, the correct answer is option (a): 2ER/(2R+r).