Two batteries with the same EMF E are connected so that the EMF of the generated source is equal to E / 2

Two batteries with the same EMF E are connected so that the EMF of the generated source is equal to E / 2. The internal resistance of one of the batteries is r = 6 ohms. 1) What is the maximum possible resistance of the second source? Express the answer in ohms, rounded to the nearest integer 2) What is the minimum possible resistance of the second source?

From the condition of the problem, we can conclude that the power supplies are connected opposite to each other.
Suppose that the EMF is E = 180 V, then E / 2 = 90 V.
Consider the I * R voltage drop across the first power supply:
I * R = E; therefore, I = E / R = 180/6 = 30 A. This current will be the same for the entire circuit, since the sources, although they are connected oppositely, are in series, and not parallel to each other. It is known that in a series circuit, the current is the same in all its sections.
The voltage drop for connecting power supplies together can be expressed in two expressions:
1) 30 * 6 – 30 * R = 90; 2) 30 * R – 30 * 6 = 90.
In the first case, the voltage drop across the first power supply is greater than on the second, and in the second case, vice versa.
Then:
1) -30 * R = 90 – 180 = -90; R = -90 / (- 30) = 3 Ohms;
2) 30 * R = 90 +180 = 270; R = 270/30 = 9 ohms.

Answer: the maximum of the possible resistances of the second power supply is R = 9 ohms, and the minimum is R = 3 ohms.