# Circuit analysis with two loops, resistors, and independent/dependent sources

I am working on this sample circuit problem, which knowledge of KCL, KVL, and how resistors in series or parallel add up in terms of resistance. However, in this circuit there is a voltage controlled current source that supplies 29 times greater current than the current I somewhere else in the circuit. I am supposed to find the voltage drop across this VCCS, labeled Vy. I think it's this element that is throwing me off. I am not sure how to go about analyzing this. Here is my work so far, hopefully the picture is clear enough. I have these equations written down from KVL. I know that there should only be two loops, I am just not sure whether which loop to analyze would be better. Although either one should give me the same result, I believe, if I do it right. But I am stuck on finding Vy, as well as the voltage drops across the other resistors, because of that VCCS in the circuit.

How should I go about analyzing this circuit?

• Apply Miller's theorem at 200 ohm resistor! After that you can simply write: $I_\beta=\frac{15.2\,\text{V}+0.8\,\text{V}}{10\,\text{k} \Omega+(29+1)\times 200\,\Omega}=1\,\text{mA}$ Oct 11 '16 at 16:37

It's a CCCS, not a VCCS.

$I_B$ is not equal to $V_T/R_T$, since the current through $R_2$ is not $I_B$.

So what is the current through $R_2$? You can get this simply by inspection.

When you've got the expression for that current, do KVL on the left mesh and you have $I_B$ (1 mA). And the rest is plain sailing.

• Ok, would the current through R2 be Ib+29Ib+(25/(200+500))? Because all of those currents flow into node N2? Or would it simply be Ib+29Ib = 30Ib? I'm confused about the current draw by the two resistors. Oct 11 '16 at 18:05
• There are two currents into the node: $I_B$ and $29I_B$ and the sum of these flows down through $R_2$ - that's KCL.
– Chu
Oct 11 '16 at 20:46
• Ok, so the current that flows through the 200 ohm resistor R2 is 30Ib. And would the current that flows through the 500 ohm resistor R3 be 29Ib? I set up KVL for the right-side loop with these values, and figured out the voltages across R2 and R3, which are 6V and 14.5V respectively. Plugging these values back into the KVL equation, I got Vy=25-(6+14.5) = 4.5V across Vy. Does this value seem right? Oct 11 '16 at 22:33
• Yes, that's correct.
– Chu
Oct 12 '16 at 0:21

This is a very quick look at it, I might have some polarities mixed up. I think the idea here is that you need two KVL and one KCL to form 3 equations to solve together.

Left loop KVL clockwise from common: 15.2V + iB*10k + 0.8V + IR2*200 = 0 (eq1)

Right loop KVL anti clockwise from common: 25 + 500*29*iB + Vy + IR2*200 = 0 (eq2)

Center VCL: iB + 29*iB = IR2 (eq3)

Now its just a matter of solving. Re-arrange eq3 to get iB as a function of IR2. Then subsituation that iB value in to eq1 to get the value of IR2. Then use IR2 in eq3 to get iB. Now you have IR2 and iB for eq2 to get Vy.