A powerful circuit

On the picture above, I have assumed the directions of the currents $I_1$ , $I_2$ , and $I$ . Using Kirchoff's First Law, we can see that $I = I_1 + I_2$ . I have also marked points A and B. We can see that they have the same potentials, meaning that $U_ab = 0$ . $U_ab$ can also be expressed as $E - E - R*I_2 - R*I_1$ , meaning that I_1 = I_2 = \(\frac{I}{2} ). $U_ab$ can also be expressed as $E - R*I - R*I_1 + E$ . Switching $I$ with \(\frac{I_1}{2} ) gives us an equation from which we can express $I_1$ as \(\frac{2E}{3R} ), and that is also equal to $I_2$ . The total power is equal to the sum of all of the powers on the resistors, and is equal to 2*R*I_1^2 + R*I^2 = \(\frac{8E^2}{3R} ).