I have been working through Boylestad and I am having trouble with this folling question.


simulate this circuit – Schematic created using CircuitLab

h = 50.

this one i'm finding difficult because the current source depends on another part of this network. I tried applying Kirchhoff's law around the both loops to get loop currents \$I_1\$ and \$I_2\$. $$hI = I_1 - I_2$$

So \$I = I_1\$

Then I got $$hI_1 = I_1 - I_2$$

$$0 = I_1(1-h) - I_2$$

And I used kirchhoff's around the outer loop to get the second equations. $$20 = I_1(2000) + I_2(2000)$$

So I used these two equations using determinants but I didn't get the result in the book which was \$19.62\angle 53^\circ\$

Am I on the right path here or is there a better way to try and get the correct answer?

Now correctly completed with the help of @TimWescott and @relayman357 with the correct equations: $$I_1(1+h)-I_2=0$$ $$I_1(2000) + I_2(2000) = 20\angle 53^\circ$$

Then using determinant to determine \$I_2\$ to then find \$VR2\$.

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    \$\begingroup\$ For \$I = +I_1\$ to be true, \$I_1\$ needs to be in the direction of \$I\$. For \$hI_1 = I_1 - I_2\$ to be true, \$I_1\$ needs to be opposite the direction of \$I\$. This is super easy to get wrong -- do your numbers again? It never hurts to actually draw in the loop currents. \$\endgroup\$ – TimWescott Nov 29 '20 at 0:51
  • \$\begingroup\$ True @TimWescott thanks. I will investigate in ciruitlab if you can put in loop currents. \$\endgroup\$ – Bucephalus Nov 29 '20 at 0:58
  • \$\begingroup\$ Your KVL equation assumes \$i_1\$ is referenced (arrow direction) going to right, but your equation, \$I=I_1\$ assumes the opposite. \$\endgroup\$ – relayman357 Nov 29 '20 at 1:03
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    \$\begingroup\$ Note that there really is only one loop here. \$\endgroup\$ – copper.hat Nov 29 '20 at 1:48

Your nodal equation is wrong if i'm understanding the directions you assumed for your loop currents. If my figure below is correct, then your KCL should give,

$$i_1 + hi_1 = i_2$$ $$(h+1)i_1 - i_2 = 0$$

enter image description here

You can also try writing a single KVL equation around the entire perimeter - and recognize that the current going up through R2 is \$(h+1)I\$.

  • \$\begingroup\$ Oh yeah just put in loop currents like that, yep, I get it now. Correct, the diagram was correct I forgot to take into account the opposite direction. My loop currents I had going clockwise, but I wrote them down wrong in here. I meant \$I = -I_1\$. \$\endgroup\$ – Bucephalus Nov 29 '20 at 1:04
  • \$\begingroup\$ Good deal. Easy mistake to make when you don't take time to label things well. \$\endgroup\$ – relayman357 Nov 29 '20 at 1:05
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    \$\begingroup\$ Where I went wrong in my calculations but correct in writing it down here was actually in my second equation. I had \$20\angle 53^\circ = I_1(2000) - I_2(2000)/$ which is wrong. So let me go back and use what you have shown me here and see if I can get the result. Thanks. @relayman357 \$\endgroup\$ – Bucephalus Nov 29 '20 at 1:06
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    \$\begingroup\$ Yeah that worked. I think I realise why it's such a high voltage. \$hI\$ is actually going in reverse isn't it, thus summing with the current from the voltage as it goes down \$R2 \$ branch. \$\endgroup\$ – Bucephalus Nov 29 '20 at 1:37
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    \$\begingroup\$ It will produce whatever voltage (magnitude and polarity) necessary to produce it’s controlled current, \$hI\$. \$\endgroup\$ – relayman357 Nov 29 '20 at 1:49

The current through R2 is \$(1+h)I\$, so the loop gives \$v(t) = I R_1 + (1+h) I R_2 \$ and so \$ I = {v(t) \over R_1 + (1+h) R_2 } \$, and \$v_{R_2}(t) = {(1+h)R_2\ v(t) \over R_1 + (1+h) R_2 } \$.

Substituting \$h=50, R_1=R_2 = 2k \$ gives \$v_{R_2} \approx 19.6 \angle 53^\circ\$RMS.

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    \$\begingroup\$ That's really good @copper.hat . Much more succinct than the way i did it. \$\endgroup\$ – Bucephalus Nov 29 '20 at 3:05

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