# How to find the currents and their directions in circuit?

I want to find the currents in blue, and also their directions.

I gave it a try and redrew the diagram like this. The arrows next to the currents represent the directions I assumed them to have at first.

My process is this:

KVL

V1 = R1 * I1;
R1 = 4ohm;
24 - V1 = 0 -> 24 = V1 -> 24 = 4 * I1;
I1 = 6A.

V3 = R3 * I3;
R3 = 10ohm;
8 - V3 = 0 -> 8 = V3 -> 8 = 10 * I3;
I3 = 0.8A.

V2 = -(R2 * I2); I set V2 as a negative value cause it's supposed to be a voltage drop from R2.
R2 = 2ohm;
V1 = R1 * I1;
8 + V2 - V1 = 0 -> (R2 * I2) = 8 - R1*I1 -> R2 * I2 = 8 - 24;
I2 = -8A.

With KCL I obtained the current I

Using the node right above R1 (in the second image) I obtained:
-I2 + I - I1 = 0 -> I = I1 + I2 = -2A

I understand that the negative sign tells that the currents actually go in the opposite direction I assumed.

The problem is that the book I have says that I = 14A and not -2A.

This difference of results happens because I'm setting V2 as a negative value (V2 = -R2 * I2 ). Cause if I set V2 as positive, then I get those 14A. But for other problems I've found I had to set the voltage drops (like V2), from resistors, as negative values to get the right answer. But in this case setting V2 as negative is giving a wrong answer.

My rule of thumb is that whenever the direction I assume, for the current when using KVL, tells me that a voltage source is causing a voltage rise (from - to +), then the voltage across any resistor must be negative. Thus if the signs of the voltage across the resistor are from + to -, (a voltage drop), then I include a negative sign next to the voltage (Vr) of the resistor, and thus the expression for Vr (resistance*current) must be positive. So that -Vr is negative.

The same for when the signs go from - to + for a resistor (according to the current direction I chose for KVL). This represents a voltage rise, and thus I include a + sign next to Vr. Thus the expression for Vr must be itself negative so that +Vr is negative.

But this is not working when I apply this to V2 from the first diagram.

Notice how in this case I have to set V0 as a negative value (V0 = -5i) because the answers are Vx = 10v and V0 = -5v. And the same happens for other problems. But I don't know why it doesn't work for the first diagram.

## 2 Answers

I didn't check your math because you made this way more complicated than it needs to be.

Every single node in the circuit has its voltage given in the problem statement.

R1 is connected between a node at 24 V and the ground node. Therefore the current through it is 24 / 4, or 6 A, and the current flows top to bottom.

R2 is connected between a node at 24 V and a node at 8 V. Therefore the voltage across it is 16 V, and the current through it is 8 A.

By KCL, $$\I\$$ is the sum of these two currents, and 8 + 6 is 14.

So $$\I\$$ is 14 A.

There was no need to set up a complete set of nodal or mesh (loop) equations to answer this question.

Edit to add

Looking at your math I see a couple problems:

1. For R3 you didn't follow the passive current convention like you did on the other components. For the passive current convention you should have the branch current going in to the positive terminal of the branch, not out.

2. When you did "8 + V2 - V1 = 0 -> (R2 * I2) = 8 - R1*I1" your KVL equation is correct, but you flipped a sign when transforming to the 2nd version. Your second equation is "V2 = 8 - V1" where it should be "V2 = -8 + V1" because you moved the 8 and V1 to the other side of the equation. (Put another way, instead of subtracting (8 - V1) from both sides of the equation, you subtracted it from the lhs and added it to the rhs)

Since the R3 current is actually irrelevant to the problem solution, the 2nd issue is likely the one that led you to the wrong answer. This has nothing to do with your choice of directions for the components, it was just an algebra error.

• Thanks a lot for your answer, really helped me understand some details I was missing. Commented Aug 25 at 2:52

### redrawn and labeled

At first glance, it's obvious that $$\V_{\text{A}}=24\:\text{V}\$$ and $$\V_{\text{B}}=8\:\text{V}\$$. All node voltages are known. So the currents can be computed, directly, using Ohm's Law:

simulate this circuit – Schematic created using CircuitLab

It seems you want to apply KCL and KVL for analysis, regardless.

### KVL

For this, you can apply all three loops. I'll go clockwise:

Using SymPy:

 loop1 = Eq( 0 + V1 - R1*(Ix-Iy), 0 )
loop2 = Eq(0 - R1*(Iy-Ix) - R2*Iy - R3*(Iy-Iz), 0 )
loop3 = Eq(0 - R3*(Iz-Iy) - V2, 0 )
solve([ loop1, loop2, loop3 ], [ Ix, Iy, Iz ])
{Ix: (-R1*V2 + R1*v1 + R2*v1)/(R1*R2),
Iy: (-V2 + v1)/R2,
Iz: (-R2*V2 - R3*V2 + R3*v1)/(R2*R3)}


From those, find: $$\I_x=14:\text{A}\$$, $$\I_y=8:\text{A}\$$, and $$\I_z=7.2:\text{A}\$$. If you take the convention that the current's arrow direction in each resistor goes from the more positive end to the less positive end, then $$\I_{R_1}=I_x-I_y=6\:\text{A}\$$, $$\I_{R_2}=I_y=8\:\text{A}\$$, and $$\I_{R_3}=I_y-I_z=800\:\text{mA}\$$.

Since you know the node voltages and Ohm's Law, you should be able to reach the same results as KVL just did.

### KCL

You need to introduce two unknown currents:

nodalA = Eq( Va/R1 + Va/R2, Vb/R2 + Iv1 )
nodalB = Eq( Vb/R2 + Vb/R3, Va/R2 + Iv2)
solve([ nodalA, nodalB ],[ Iv1, Iv2 ])
{Iv1: (R1*Va - R1*Vb + R2*Va)/(R1*R2), Iv2: (R2*Vb - R3*Va + R3*Vb)/(R2*R3)}


From here you can find: $$\I_{V_1}=14\:\text{A}\$$ and $$\I_{V_2}=-7.2\:\text{A}\$$.

*(The negative current for $$\V_2\$$ just means that I pointed the arrow wrong at the start, by assuming that $$\V_2\$$ might be providing power to the circuit, and that $$\V_2\$$ isn't providing power as I'd assumed.)