We have this circuit:

enter image description here

I make the following assumptions about the current:

enter image description here

Such that $$I_1+I_2+I_3=0$$ at the blue junction. I then draw two loops, l_1 and l_2:

enter image description here

which yields the equations $$-12 -5I_1 -3 - 10I_2=0\space(l_1)\text{ and} -5I_39-3+10I_2=0\space(l_2)$$

However, solving this equation does not yield the correct answers. Can someone please clarify what was improper about my application of KCVL and KVL?

Disclaimer: This is a homework problem (which may be for credit somewhere else, since it is from a large publisher) but it is not for credit to me. In fact, I already know: What is the current through the 10 Ω resistor in the figure? is correct with an answer of .36 A. I just can't get this without altering the KCL equation in a way I know to be wrong!

  • 1
    \$\begingroup\$ If you simulate the circuit, do you get same or different answer? \$\endgroup\$
    – Justme
    Apr 1, 2023 at 20:49
  • \$\begingroup\$ I don't readily have a way to simulate this, I'm afraid. \$\endgroup\$ Apr 2, 2023 at 0:36
  • 1
    \$\begingroup\$ @user1833028 You do now: falstad.com/circuit \$\endgroup\$
    – DKNguyen
    Apr 2, 2023 at 1:50

2 Answers 2


You can greatly simplify the schematic. This will help you to create an independent means to work out the answer. So let's do that. No simulator, just yet.


simulate this circuit – Schematic created using CircuitLab

In moving from the left schematic to the middle one, all I did was to swap the upper series pair and assign ground to one node. In moving from the middle schematic to the right one, all I did was get rid of the confusing clutter and label the node voltages, instead. (Ground is then assumed to be somewhere.)

From the right most schematic, find \$V_{_\text{X}}=\frac{V_3 R_1 R_2+V_2 R_1 R_3+V_1 R_2 R_3}{R_1 R_2+R_1 R_3+R_2 R_3}=600\:\text{mV}\$. To find the current magnitude in \$R_2\$ just compute \$\left|\frac{V_{_\text{X}}-V_2}{R_2}\right|=\left|\frac{600\:\text{mV}-\left(-3\:\text{V}\right)}{10\:\Omega}\right|=360\:\text{mA}\$.

So that makes sense.

Now let's do nodal (KCL), first. And here we'll use your original schematic on the left and ignore my simplifications. All the nodes are labeled there. Nice. I'll use your current names and directions, too.

$$\begin{align*} I_1+\frac{V_{_\text{X}}-V_{_\text{B}}}{R_2}+\frac{V_{_\text{X}}-V_{_\text{C}}}{R_3}&=0\:\text{A} \\\\ \frac{V_{_\text{A}}}{R_1} &= I_1 \\\\ \frac{V_{_\text{B}}-V_{_\text{X}}}{R_2} + I_2 &= 0\:\text{A} \\\\ \frac{V_{_\text{C}}-V_{_\text{X}}}{R_3} + I_3 &= 0\:\text{A} \end{align*}$$

That's KCL for all four nodes. In addition, you know that:

$$\begin{align*} V_{_\text{A}}+12\:\text{V}&=V_{_\text{X}} \\\\ V_{_\text{B}}&=-3\:\text{V} \\\\ V_{_\text{C}}&=-9\:\text{V} \end{align*}$$

You can either substitute into the earlier four equations in order to reduce unknown variables or else just keep all seven equations and seven unknowns (some of which are simply known already) and just solve.

Using Python and Sympy (both free) find:

var('i1 i2 i3 va vb vc vx r1 r2 r3')
ans=solve([ Eq(i1+(vx-vb)/r2+(vx-vc)/r3,0),
            Eq(vc,-9) ],
for i9 in ans:
(va, -57/5)
(vb, -3)
(vc, -9)
(vx, 3/5)
(i1, -57/25)
(i2, 9/25)
(i3, 48/25)

So, \$V_{_\text{X}}=\frac35\:\text{V}=600\:\text{mV}\$ and \$I_2=\frac9{25}\:\text{A}=360\:\text{mA}\$. Just as before.

That's KCL.

For mesh (KVL), using \$I_1\$ for your top loop and \$I_2\$ for your bottom loop, as you show them to be in your last diagram:

$$\begin{align*} 0\:\text{V}-12\:\text{V}-R_1\cdot I_1-3\:\text{V}-R_2\cdot\left(I_1+I_2\right)&=0\:\text{V} \\\\ 0\:\text{V}-R_3\cdot I_2+9\:\text{V}-3\:\text{V}-R_2\cdot\left(I_1+I_2\right)&=0\:\text{V} \end{align*}$$


ans=solve([ Eq(0-12-r1*i1-3-r2*(i1+i2),0),
for i9 in ans:
(i1, -57/25)
(i2, 48/25)

The sum of these two currents is what's passing through \$R_2\$. Since both your arrows point from right to left, we get a negative value as the result of the sum which is just another way of saying that you got the direction wrong. The magnitude is still \$360\:\text{mA}\$.

So mesh (KVL) works.

I wasn't sure about which process you needed to improve, so I provided both just to be sure I got you covered.


It looks like you are mixing the "classic KVL/KCL method" with the mesh analysis.

Your circuit looks like this:


simulate this circuit – Schematic created using CircuitLab

Notice that I marked the polarity of voltage drops across the resistors. Using assumed currents directions.

Now we can write KVL for the first loop:

$$-12V - 5\Omega \cdot I_1 - 3V + 10\Omega \cdot I_2 = 0 $$

And KVL for the secend loop:

$$-5\Omega \cdot I_3 + 9V -3V+ 10\Omega \cdot I_2 = 0 $$

And one additional equation for KCL

$$I_1 + I_2 + I_3 = 0$$

And the solution is:

\$I_1 = -2.28A\$

\$I_2 = 0.36A\$

\$I_3 = 1.92A\$

This is what the classic KVL/KCL method looks like.

  • \$\begingroup\$ So if we go - to plus we add, but if we go + to -, we subtract. Makes sense. But I can't figure out how to label the +'s and -s for resistors... why are we adding I2*R2 instead of subtracting it? Do we ALWAYS assume positive to negative in assumed current directions? \$\endgroup\$ Apr 2, 2023 at 23:57
  • \$\begingroup\$ @user1833028 Yes, we always assumed that the current is flowing from positive to negative. And based on this assumption I marked the polarity across the resistors. And we add I2*R2 because we "hit" the "-" sign first when (we go - to plus we add) we do the loop in a clockwise direction in loop number one. And counterclockwise in loop number two. \$\endgroup\$
    – G36
    Apr 3, 2023 at 14:37

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