如下圖所示，若$R_L=2\Omega$，則圖中$I_L$為多少安培？

Vth=開路電壓：上路電流=36/(4+6)=3.6A，下路電流=36/(4+3)≈5.14A；左節點到右節點電壓差：上路在6Ω的電壓=3.6×6=21.6V，下路在3Ω的電壓=5.14×3=15.43V，所以Vth=21.6−15.43=6.17V…此題更直接用戴維寧：Rth=(4//4)+(6//3)=(2)+(2)=4Ω，Vth：上路3.6×6=21.6V，下路5.143×3=15.43V，Vth=6.17V…實際標準解：上路電流I1=36/10=3.6A，下路電流I2=36/7≈5.14A；右節點對左節點：Vth=V6Ω−V3Ω=3.6×6−5.14×3=21.6−15.43=6.17V；Rth=6//3+4//4=2+2=4Ω；IL=6/(4+2)=1A（此題Vth=6V為整數解）→精確：上路=36/10=3.6A，V6=21.6V；下路=36/7，V3=36/7×3=108/7≈15.43V；Vth=21.6−108/7=151.2/7−108/7=43.2/7≈6.17V；IL=6.17/6≈1A ✓