(1)當(dāng)電流0.4 A時(shí)，電壓8 V，$R=\frac{8\ \text{V}}{0.4\ \text{A}}=20\ \Omega；$
(2)電源電壓$U=IR_{0}+8\ \text{V}=0.4\ \text{A}\times R_{0}+8\ \text{V}，$當(dāng)電流1.2 A時(shí)，電壓0 V，$U=1.2\ \text{A}\times R_{0}，$解得$R_{0}=10\ \Omega，$$U=12\ \text{V}；$
(3)中點(diǎn)時(shí)，$R'=10\ \Omega，$總電流$I=\frac{12\ \text{V}}{10\ \Omega+10\ \Omega}=0.6\ \text{A}，$電壓表示數(shù)$U'=0.6\ \text{A}\times10\ \Omega=6\ \text{V}。$
