When the sliding contact P of the sliding rheostat slides down, R2 becomes larger and the total resistance of the external circuit becomes larger. According to the closed ohm's law, we can know that the current I decreases, U 1=IR 1 decreases, and the terminal voltage U3=E-Ir increases.

Because U3=U 1+U2, U 1 becomes smaller, U2 becomes larger, △ U2 > △ U 1, △ U2 > △ U3.

△U 1=△I？ R 1，△U3=△I？ R, because r 1 > r, △ u 1 > △ u3. So, A is wrong and BC is right.

D, U2=E-I(R 1+r), △ U2 △ I = r 1+r > r, so D is correct.

Therefore: BCD.