0
\$\begingroup\$

Just having a little trouble understanding this solution. I understand that we are trying to achieve an equivalent voltage drop across the terminals a & b to represent the circuit. However I don't get why Vth = Vx, and it's the dependent current source that's bothering me.

Is the fact that the 60 ohm resistor and the dependent source being connected in parallel and to the same node meaning the voltage drop across both elements is equal? If the two elements were switched around it would make sense to me that Vth would be the voltage through the 60 ohm. Also I should note that calculating Vth isn't the problem it's just knowing where I should be taking the equivalent voltage from, like I guess the crux of the problem is why have they chosen the 60 ohm.

enter image description here

Text from image:

Obtain the Thevenin and Norton equivalent circuits of the circuit in Fig. 4.114 with respect to terminals a and b.

Since \$V_{Th} = V_{ab} = V_x\$, we apply KCL at the node \$\mathrm{a}\$ and obtain: $$\frac{30-V_{Th}}{12} = \frac{V_{Th}}{60} + 2V_{Th} \longrightarrow V_{Th} = \frac{150}{126} = 1.19\mathrm{V}$$

\$\endgroup\$
  • \$\begingroup\$ look on wikipedia the definitions of "parallel" and "series". \$\endgroup\$ – Vladimir Cravero Aug 28 '14 at 10:24
1
\$\begingroup\$

Redraw the circuit as follows to see why \$V_X\$ is across the 60 ohm resistor, the controlled current source and the output terminals:

enter image description here

It is often helpful to remember that two points connected by an ideal wire are the same circuit node which can be seen by 'shrinking' the wire to zero length as I've done above.

\$\endgroup\$
0
\$\begingroup\$

The problem is why Vth == Vx == Vab? Well Va is the same along the wire back to the 12 ohm resistor right? So the top of the current source and the top of the 60R and the right side of the 12R are all one point. If I reshuffle randomly a little bit is it clearer why they are equal?

schematic

simulate this circuit – Schematic created using CircuitLab

(Why is that schematic so big!) Maybe I'm misunderstanding your question!

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.