# Finding load current in a circuit with voltage dependent voltage source

This question has been rolled back to the version that doesn't invalidate the answers (edit by Andy aka)

I'm currently studying for my Circuit Analysis class, the teacher shared some questions and their answers with us, but not the steps of reaching the answers. I solved almost all of them, but whatever I try doesn't work for this question. Can you guide me, how can I reach the solution?

Teacher says the answer is $$\I_L= 1,68 mA.\$$

Edit: I tried Norton's theorem first since it asks for the load current, but I just can't get rid of the dependent source, I tried Thevenin and superposition as well but I can't seem to make it work. It's probably my bad, I'm doing something I shouldn't with the equations.

With Norton: I tried mesh analysis after shorting RL, and used supernode because there is IS in the middle of the meshes, but it doesn't work, I can't get rid of 2Va, using $$\2V_a = 36i_1\$$ doesn't work, since that leads me to $$\i_2 = 3,648\$$ which supposed to be my IL, but it's wrong, since the real answer is supposed to be 1,68 mA.

What I did was set clockwise i1 and i2. From supernode I got:

$$\3,5 - 18i_1 - 7i_1 +2V_A = 0\\V_a = 18i_1\\3,5 -11i_1 = 0\\i_1=0,318\$$

Then, IL is i2, so: $$\i_2 = i_1+3,3 = 3,648mA\$$ which is wrong.

• so, what were your steps, so we can help you get them right? Commented Feb 6, 2021 at 11:27
• @MarcusMüller sorry, I edited the question. Commented Feb 6, 2021 at 11:38
• Thanks! I'd be careful with superposition here, indeed, but the other two approaches should take you somewhere; could you add to the question which simplifications you did before trying to apply Thevenin's or Norton's equivalent? Commented Feb 6, 2021 at 11:43
• Shouldn't you need the value of RL? Commented Feb 6, 2021 at 11:45
• It seems to me that your task is to find the RL that will give you the maximum power transferred to the load resistance (RL).
– G36
Commented Feb 6, 2021 at 13:52

## 2 Answers

I cannot see how this can be solved without knowing RL (60.9 kohm)

@Andyaka teacher is persistent about not giving RL, so I'm trying to find a way to get rid of RL

Please note that my "answer" is IS NOT a homework solution so please don't think it is.

My "answer" simply attempts to show that resistor RL needs to be known in order to determine the current through it. In other words, the question set by the teacher is flawed or the interpretation given by the OP is incorrect.

Redraw the picture with what you know about the problem and the apparent "solution": -

• The resistor values R1 and R2
• The 3.3 mA current source
• The 3.5 volts voltage source
• The "correct" current through RL of 1.68 mA

Then, because you know the current must split and add up to 3.3 mA, the current through resistors R1 and R2 must be 1.62 mA: -

This then means that the voltage at the top of the current source has to be: -

3.5 volts + (18 kΩ + 7 kΩ)•1.62 mA = 44.000 volts: -

You also know the voltage across R1 (aka VA) = 1.62 mA • 18000 = 29.16 volts

And, 2•VA = 58.32 volts hence, the voltage across resistor RL must be 44 volts + 58.32 volts = 102.32 volts: -

This cannot mean anything else other than that the resistor RL must have a value of: -

102.32 volts ÷ 1.68 mA = 60904.76 Ω.

And, if that doesn't convince you, here's the result of a simulation: -

Should I change RL to (say 50 kΩ) we get a different current distribution and the wrong answer: -

A note about maximum power transfer (following comments). If the "real" question sought to find the current in RL when the maximum power was transferred to RL then RL would equal 61 kΩ. This is because the presence of the VCVS (with a gain of 2) would make an equivalent impedance of twice R1. Given that finding the Thevenin equivalent impedance means we can ignore the current source, the total Thevenin impedance is R1 + R2 +2•R1 = 61 kΩ. And when RL equals that value, we have maximum power transfer and I(RL) would be 1.67869 mA.

• Homework questions that show effort like this one are fine Commented Feb 6, 2021 at 18:31
• @Andyaka you might want to check the question now, since I found a solution and changed the question Commented Feb 6, 2021 at 20:22
• @BerkayGünaydın No, this isn’t good. You asked a question and I completely answered it and you accepted the answer. Now you are changing that question significantly and that completely makes my answer look stupid. What you should do is, roll back your question modification and, if you have a new question based on fresh information, formally ask a brand new question. I’m trusting that you see the error of your ways in what you have done and do as I ask. Let me know when you have restored this question and have asked a new question and I shall be pleased to take a look. Commented Feb 6, 2021 at 20:27
• @BerkayGünaydın OK, you are not responding to this so I'm taking the liberty of rolling back your question to the previous version so that it doesn't invalidate the given answers. Commented Feb 6, 2021 at 23:22
• @Andyaka sorry about not responding, your answer never looked stupid and I appreciate it. The reason I mentioned you because I found what was wrong and since you were interested, thought you might wanna know. I still gonna accept your answer because it was helpful, sorry about the misunderstanding. Commented Feb 7, 2021 at 10:47

Well, we are trying to analyze the following circuit:

simulate this circuit – Schematic created using CircuitLab

When we use and apply KCL, we can write the following set of equations:

$$\text{I}_2=\text{I}_\text{k}+\text{I}_1\tag1$$

When we use and apply Ohm's law, we can write the following set of equations:

$$\begin{cases} \text{I}_1=\frac{\text{V}_\text{i}-\text{V}_1}{\text{R}_1}\\ \\ \text{I}_1=\frac{\text{V}_1-\text{V}_2}{\text{R}_2}\\ \\ \text{I}_2=\frac{\text{V}_3}{\text{R}_3} \end{cases}\tag2$$

And we also know that $$\\text{V}_3-\text{V}_2=\text{n}\cdot\left(\text{V}_1-\text{V}_\text{i}\right)\$$.

Substitute $$\(2)\$$ into $$\(1)\$$, in order to get:

$$\begin{cases} \frac{\text{V}_3}{\text{R}_3}=\text{I}_\text{k}+\frac{\text{V}_\text{i}-\text{V}_1}{\text{R}_1}\\ \\ \frac{\text{V}_3}{\text{R}_3}=\text{I}_\text{k}+\frac{\text{V}_1-\text{V}_2}{\text{R}_2}\\ \\ \text{V}_3-\text{V}_2=\text{n}\cdot\left(\text{V}_1-\text{V}_\text{i}\right) \end{cases}\tag3$$

Now, we can solve for $$\\text{I}_2\$$:

$$\text{I}_2=\frac{\text{I}_\text{k}\left(\text{R}_1\left(1+\text{n}\right)+\text{R}_2\right)+\text{V}_\text{i}}{\text{R}_1\left(1+\text{n}\right)+\text{R}_2+\text{R}_3}\tag4$$

Using your values (without knowing the values of $$\\text{R}_3\$$), we find:

$$\text{I}_2=\frac{1024}{5\left(61000+\text{R}_3\right)}\tag5$$

Edit. Because this question is about maximum power transfer we can find the value of $$\\text{R}_3\$$. In order to do that we need to solve:

$$\frac{\partial\text{P}_{\text{R}_3}}{\partial\text{R}_3}=0\tag6$$

We can use $$\(4)\$$:

$$\text{P}_{\text{R}_3}=\text{R}_3\text{I}_{\text{R}_3}^2=\text{R}_3\left(\frac{\text{I}_\text{k}\left(\text{R}_1\left(1+\text{n}\right)+\text{R}_2\right)+\text{V}_\text{i}}{\text{R}_1\left(1+\text{n}\right)+\text{R}_2+\text{R}_3}\right)^2\tag7$$

Solving $$\(6)\$$ using $$\(7)\$$ we get:

$$\text{R}_3=\text{R}_1\left(1+\text{n}\right)+\text{R}_2\tag8$$

For your circuit we get:

$$\text{R}_3=18\cdot1000\left(1+2\right)+7\cdot1000=61000\space\Omega\tag9$$

And the current and power in $$\\text{R}_3\$$ will be:

• Current: $$\text{I}_{\text{R}_3}=\text{I}_2=\frac{64}{38125}\approx0.00167869\space\text{A}\tag{10}$$
• Power: $$\text{P}_{\text{R}_3}=\text{V}_{\text{R}_3}\text{I}_{\text{R}_3}=\text{V}_3\text{I}_2=\frac{32768}{190625}\approx0.171898\space\text{W}\tag{11}$$
• Thanks for your answer, teacher is not providing R3, and insisting I don't need it to find I2. So there must be a way of getting rid of R3 somehow? Commented Feb 6, 2021 at 12:03
• @BerkayGünaydın See the edit to my answer. Commented Feb 7, 2021 at 14:37