# Circuit analysis with a dependent current source

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\$$.

• 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. – TimWescott Nov 29 '20 at 0:51
• True @TimWescott thanks. I will investigate in ciruitlab if you can put in loop currents. – Bucephalus Nov 29 '20 at 0:58
• Your KVL equation assumes $i_1$ is referenced (arrow direction) going to right, but your equation, $I=I_1$ assumes the opposite. – relayman357 Nov 29 '20 at 1:03
• Note that there really is only one loop here. – 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,

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

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\$$.

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

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