a) Ta có mạch: \(R_1ntR_2\)
\(R_{tđ}=R_1+R_2=60+40=100\left(\Omega\right)\\ I=I_1=I_2=\frac{U}{R_{tđ}}=\frac{12}{100}=0,12\left(A\right)\)
b) Khi này, mạch là: \(R_1nt\left(R_2//R_3\right)\)
\(U_2=I_2R_2=0,075.40=3\left(V\right)\\ U_{23}=U_2=U_3=3\left(V\right)\\I=I_1=I_{23}=0,12\left(A\right)\\ R_3=\frac{U_3}{R_3}=\frac{3}{0,12}=25\left(\Omega\right)\)