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

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

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  • \$\begingroup\$ Current cannot flow via a 3k resistor because we have an open circuit on the right. \$\endgroup\$
    – G36
    Mar 24 at 15:28
  • \$\begingroup\$ 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). \$\endgroup\$
    – G36
    Mar 24 at 15:35
  • \$\begingroup\$ Trust KVL and Isc to get Rth and look here for intuitive validation with Falstad Sim tinyurl.com/ydetpfwe \$\endgroup\$
    – Tony Stewart EE75
    Mar 24 at 17:02

1 Answer 1

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Just look at the voltages and currents that can only be when there is no load: -

enter image description here

Hence: -

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

enter image description here

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