I'm stuck on part D on how to express Vx and Ix. I'm assuming the equation for Vx is Vx=v1 - 0 but am not entirely sure since our prof didn't provide any solutions. For Ix I don't really know what to do. Would the 1 ohm resistor play a role in here or do I write some sort of KCL equation.

Thanks alot if anyone can help!

 

\( V_x\;=\;V_1 \)

\( I_x\;=\;2i_o\;=\;2\left(\cfrac{V_1}{2\text{ Ω}}\right) \)

 

\( V_x\;=\;V_1 \)

\( I_x\;=\;2i_o\;=\;2\left(\cfrac{V_1}{2\text{ Ω}}\right) \)

Thanks so much! I was wondering, could it also be written as ix = v3−v4/ 1ohm since the current through a branch should be the same throughout? Also, I’m not quite sure how the voltage of the dependent source and the the current Ix are both 2i0 at the same time. Could you help clarify how these relationships are connected? Apologies if I’m missing something obvious - just trying to make sure I understand it fully.

Thanks.

 

There is only 1 dependent source with a current value of two times i_0. The other two sources are independent with values of 3A and +12V