Vì I1 mắc nt với mạch chính nên I1 = Im
\(R_{tđ}=14+\dfrac{8.24}{8+24}=20\left(\Omega\right)\) ; \(U=20.0,4=8\left(V\right)\)
\(U_1=0,4.14=5,6\left(V\right)\) ; \(U_2=U_3=U-U_1=8-5,6=2,4\left(V\right)\)
\(I_2=\dfrac{2,4}{8}=0,3\left(A\right)\) ; \(I_3=I_1-I_2=0,4-0,3=0,1\left(A\right)\)