# Analysis circuit using Thevenin theorem

in the book "Engineering Circuit Analysis", example 5.9, i don't understand few things, hope someone can clear up for me:

Why does the dependent current can't flow through resistor 3k? If it's because of we assume 2 terminals of vx is open circuit then isn't we can't use KVL for the outer loop as it's not a loop? And also i don't understand why vx = 8V = Voc

• Current cannot flow via a 3k resistor because we have an open circuit on the right.
– G36
Mar 24 at 15:28
• What you don't understand in the Vx = Voc = 8V. Thevenin's equivalent Vth = Vx voltage is an open-circuit voltage (we disconnect the load when we want to find Thevenin's voltage).
– G36
Mar 24 at 15:35
• Trust KVL and Isc to get Rth and look here for intuitive validation with Falstad Sim tinyurl.com/ydetpfwe Mar 24 at 17:02

$$V_x = 4 + V_x\left(\dfrac{2000}{4000}\right)\text{ or,}\hspace{1cm} V_x = 8$$