U3 = IA.R3 = 0,6.10 = 6V $$\to $$ U2 = U3 = 6V $$\to $$ I2 = U2/R2 = 6/10 = 0,6A $$\to $$ I = I2 + IA = 1,2A
Mà:$$I=\frac{\xi }{{{R}_{1}}+\frac{{{R}_{2}}.{{R}_{3}}}{{{R}_{2}}+{{R}_{3}}}+r}\leftrightarrow 1,2=\frac{12}{9+r}\to r=1\Omega $$