# Using KCL, KVL and Ohm’s Law find I0 in the circuit in Fig.1.1

I have spent more than 8 hours studying KCL and KVL and Ohm's law but I still can't solve this : If anyone can help me understand how to find I0, Ix and 2Ix I will really appreciate it.

• Please show the calculations you have attempted -- knowing where you are making a mistake will help us write more useful answers. – nanofarad Apr 8 at 1:46
• We don't hand out homework answers here. You need to show us that you have made a substantial effort to solve this yourself, and show all of your work. Then, if you get stuck, as a specific question. – Elliot Alderson Apr 8 at 2:03
• gentlemen i would be happy to show you my practice to finally solve this problem, how can i send pictures of my notebook ? or should i just explain what i did so you can figure out whats wrong. – zack__fj Apr 8 at 2:18
• @zack__fj You know that $I_x=2\cdot I_{6k}$ and therefore that $I_{4k}=\frac32\,I_x$. You also know that $I_o=2\cdot I_{4k}$ so that means $I_o=3\cdot I_x$. So the total in both branches must be $3\cdot I_x+\frac32\,I_x=\frac92\,I_x$. But this means that the total is $\frac{13}2\,I_x$, counting the dependent current source of $2\,I_x$. That must be equal to $6\:\text{mA}$. Unless I'm missing something. – jonk Apr 8 at 4:48
• Use superposition method. youtu.be/EX52BuZxpQM – Emre Mutlu Apr 8 at 7:47

## 1 Answer

First, I will present a method that uses Mathematica to solve this problem. When I was studying this stuff I used the method all the time (without using Mathematica of course).

Well, we are trying to analyze the following circuit: When we use and apply KCL, we can write the following set of equations:

$$\begin{cases} \text{I}_\text{a}=\text{I}_1+\text{I}_5\\ \\ \text{I}_5=\text{n}\cdot\text{I}_2+\text{I}_4\\ \\ \text{I}_1=\text{I}_2+\text{I}_3\\ \\ 0=\text{n}\cdot\text{I}_2+\text{I}_4+\text{I}_6\\ \\ \text{I}_6=\text{I}_3+\text{I}_7\\ \\ \text{I}_2=\text{I}_\text{a}+\text{I}_7 \end{cases}\tag1$$

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

$$\begin{cases} \text{I}_1=\frac{\text{V}_1-\text{V}_2}{\text{R}_1}\\ \\ \text{I}_2=\frac{\text{V}_2}{\text{R}_2}\\ \\ \text{I}_3=\frac{\text{V}_2}{\text{R}_3}\\ \\ \text{I}_4=\frac{\text{V}_1}{\text{R}_4} \end{cases}\tag2$$

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

$$\begin{cases} \text{I}_\text{a}=\frac{\text{V}_1-\text{V}_2}{\text{R}_1}+\text{I}_5\\ \\ \text{I}_5=\text{n}\cdot\frac{\text{V}_2}{\text{R}_2}+\frac{\text{V}_1}{\text{R}_4}\\ \\ \frac{\text{V}_1-\text{V}_2}{\text{R}_1}=\frac{\text{V}_2}{\text{R}_2}+\frac{\text{V}_2}{\text{R}_3}\\ \\ 0=\text{n}\cdot\frac{\text{V}_2}{\text{R}_2}+\frac{\text{V}_1}{\text{R}_4}+\text{I}_6\\ \\ \text{I}_6=\frac{\text{V}_2}{\text{R}_3}+\text{I}_7\\ \\ \frac{\text{V}_2}{\text{R}_2}=\text{I}_\text{a}+\text{I}_7 \end{cases}\tag3$$

Solving for $$\\text{I}_1\$$, gives:

$$\text{I}_4=\frac{\text{I}_\text{a}\left(\text{R}_2\text{R}_3+\text{R}_1\left(\text{R}_2+\text{R}_3\right)\right)}{\text{R}_1\left(\text{R}_2+\text{R}_3\right)+\left(1+\text{n}\right)\text{R}_3\text{R}_4+\text{R}_2\left(\text{R}_3+\text{R}_4\right)}\tag4$$

Where I used the following Mathematica-code:

In:=Clear["Global*"];
FullSimplify[
Solve[{Ia == I1 + I5, I5 == n*I2 + I4, I1 == I2 + I3,
0 == n*I2 + I4 + I6, I6 == I3 + I7, I2 == Ia + I7,
I1 == (V1 - V2)/R1, I2 == V2/R2, I3 == V2/R3, I4 == V1/R4}, {I1,
I2, I3, I4, I5, I6, I7, V1, V2}]]

Out={{I1 -> (Ia (R2 + R3) R4)/(
R1 (R2 + R3) + (1 + n) R3 R4 + R2 (R3 + R4)),
I2 -> (Ia R3 R4)/(R1 (R2 + R3) + (1 + n) R3 R4 + R2 (R3 + R4)),
I3 -> (Ia R2 R4)/(R1 (R2 + R3) + (1 + n) R3 R4 + R2 (R3 + R4)),
I4 -> (Ia (R2 R3 + R1 (R2 + R3)))/(
R1 (R2 + R3) + (1 + n) R3 R4 + R2 (R3 + R4)),
I5 -> Ia - (Ia (R2 + R3) R4)/(
R1 (R2 + R3) + (1 + n) R3 R4 + R2 (R3 + R4)),
I6 -> -((Ia (R1 (R2 + R3) + R3 (R2 + n R4)))/(
R1 (R2 + R3) + (1 + n) R3 R4 + R2 (R3 + R4))),
I7 -> Ia (-1 + (R3 R4)/(
R2 R3 + R1 (R2 + R3) + (R2 + R3 + n R3) R4)),
V1 -> (Ia (R2 R3 + R1 (R2 + R3)) R4)/(
R1 (R2 + R3) + (1 + n) R3 R4 + R2 (R3 + R4)),
V2 -> (Ia R2 R3 R4)/(R1 (R2 + R3) + (1 + n) R3 R4 + R2 (R3 + R4))}}


Using your values, we get:

$$\text{I}_4=\frac{9}{3250}\approx0.00276923\space\text{A}\tag5$$

The code gives the rest of the answers:

In:=Clear["Global*"];
Ia = 6*10^(-3);
n = 2;
R1 = 4*1000;
R2 = 3*1000;
R3 = 6*1000;
R4 = 3*1000;
FullSimplify[
Solve[{Ia == I1 + I5, I5 == n*I2 + I4, I1 == I2 + I3,
0 == n*I2 + I4 + I6, I6 == I3 + I7, I2 == Ia + I7,
I1 == (V1 - V2)/R1, I2 == V2/R2, I3 == V2/R3, I4 == V1/R4}, {I1,
I2, I3, I4, I5, I6, I7, V1, V2}]]

Out={{I1 -> 9/6500, I2 -> 3/3250, I3 -> 3/6500, I4 -> 9/3250, I5 -> 3/650,
I6 -> -(3/650), I7 -> -(33/6500), V1 -> 108/13, V2 -> 36/13}}

In:=N[%2]

Out={{I1 -> 0.00138462, I2 -> 0.000923077, I3 -> 0.000461538,
I4 -> 0.00276923, I5 -> 0.00461538, I6 -> -0.00461538,
I7 -> -0.00507692, V1 -> 8.30769, V2 -> 2.76923}}