(1)圖(a)中，$R_{1}=\frac{U_{1}}{I}=\frac{4\ \text{V}}{0.4\ \text{A}}=10\ \Omega，$電源電壓$U=4\ \text{V}+IR_{2}=4\ \text{V}+0.4\ \text{A}\times R_{2}；$圖(b)中，$U=6\ \text{V}，$所以$6\ \text{V}=4\ \text{V}+0.4\ \text{A}\times R_{2}，$解得$R_{2}=5\ \Omega；$
(2)圖(b)中，$I_{1}=\frac{6\ \text{V}}{10\ \Omega}=0.6\ \text{A}，$$I_{2}=\frac{6\ \text{V}}{5\ \Omega}=1.2\ \text{A}，$總電流$I=0.6\ \text{A}+1.2\ \text{A}=1.8\ \text{A}。$
