0
\$\begingroup\$

enter image description here

I want to solve the following circuit using KCL. I'm calling \$ i_1 \$ the current that flows across \$R_1 \$, \$ i_2\$ the current that flows across \$ R_2 \$ and \$ i _3 = 10A \$.

I'm assuming that the current \$ i_3 \$ is so strong that it is the only one entering the node, while \$ i_1 \$ and \$ i_2 \$ will leave the node.

\$i_1 = \left( \frac{15+V_x}{3} \right) \$ because we are going against the direction of \$ i_1 \$, so it's a voltage rise.

\$i_2 = \left( \frac{V_x}{2} \right) \$

\$i_3 = 10 A \$

So I setted KCl equation in this way:

$$ -i_1 -i_2 + i_3 = 0$$ $$ -\left( \frac{15+V_x}{3} \right) - \left( \frac{V_x}{2} \right) + 10 = 0$$ $$-2(15) -2V_x -3 V_x + 60 = 0 $$ $$V_x = \left( \frac{30}{5} \right) = 6 $$

The correct result is \$ V_x = 18V\$. And from that voltage I will calculate the currents.

I cannot find the error in my procedure. What am I doing wrong? Thank You

\$\endgroup\$
7
  • \$\begingroup\$ Yes, definitely 18 volts by another quicker method that doesn't involve writing things down (other than what I wrote here). \$\endgroup\$ – Andy aka Sep 1 '20 at 17:55
  • \$\begingroup\$ @Andyaka I'm trying to use KCL for assignment, is my method wrong? \$\endgroup\$ – Carlo Sep 1 '20 at 18:09
  • \$\begingroup\$ Why did you use a "plus" sign in (15 + Vx)/R3 ? \$\endgroup\$ – G36 Sep 1 '20 at 18:12
  • \$\begingroup\$ I've never used KCL since the day I did my final exam 40 years ago. Never needed it because there are quite frankly, just better methods. \$\endgroup\$ – Andy aka Sep 1 '20 at 18:17
  • 1
    \$\begingroup\$ Convert V1 to a current source and it's easy. KVL and KCL are generally not very helpful in the real world. But, if they are the only weapons in your arsenal then you have to use them. \$\endgroup\$ – Andy aka Sep 1 '20 at 23:31
2
\$\begingroup\$

The first problem I see is here:

$$ -\left( \frac{15+V_x}{3} \right) - \left( \frac{V_x}{2} \right) + 10 = 0$$

If you want \$i_1\$ to be the current out of the x node, then you need first term to be \$-\left(\frac{V_x - 15}{3}\right)\$, not \$-\left( \frac{15+V_x}{3} \right)\$.

Keeping track of the sign conventions you choose is one of the trickiest parts of these problems. It might help if you add current arrows indicating the directions to your schematic.

\$\endgroup\$
1
  • \$\begingroup\$ Yes it's tricky. Looking in the opposite way is the voltage drop from \$V_x\$ to 15V, so now i got it Thanks! \$\endgroup\$ – Carlo Sep 1 '20 at 21:22
0
\$\begingroup\$

From your method, it seems to me you're more like trying to solve a mystery rather than an electric circuit. It isn't and doing so is just gonna make everything much harder than it really is.

First of all, as a beginner, draw all current and voltage conventions you're employing.

schematic

simulate this circuit – Schematic created using CircuitLab

You need to understand that current flow is just an abstraction. Your statement,

I'm assuming that the current i3 is so strong that it is the only one entering the node, while i1 and i2 will leave the node

One doesn't need to need the "strength" of a current to assign its convention: it's arbitrary! So of course, there are choices more convenient than others. Sometimes it's the bipole to suggest one. You won't know whether your conventions were correct until you do all calculations - does a \$-\$ pop up? You assumed something, but it turns out to be the opposite actually.

See what I did here. You can already tell at least 2 ways (can you tell which ones?) that \$I_1\$ will have a minus sign, but I still chose that one (i.e to me, it flows into the the node, but it'll actually flow out). Then you write,
$$I_1+I_3 = I_2, \quad (15-X)/R_1 + 10 = X/R_2$$ so that \$X(1/R_2+1/R_1)=10+15/R_1\$ i.e. \$X=18V\$. From that turns out that \$I_1 = -1A\$ if you follow that convention.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.