\(R_N=\dfrac{R_1\cdot R_2}{R_1+R_2}=\dfrac{3\cdot6}{3+6}=2\Omega\)
\(I_N=\dfrac{\xi}{r+R_N}=\dfrac{4,5}{1+2}=1,5A\)
Xét đoạn mạch AB ở trên ta có:
\(U_{AB}=\xi-Ir=4,5-1,5\cdot1=3V\)
\(U_1=U_2=U_{AB}=3V\)
\(I_1=\dfrac{U_1}{R_1}=\dfrac{3}{3}=1A\)
\(I_2=\dfrac{U_2}{R_2}=\dfrac{3}{6}=0,5A\)


