# How can I find the voltages V1 and V2 of the circuit?

I know I can reduce the resistances and make the problem a lot easier. But my question actually is how can I solve for V1 and V2 using kirchhoff's laws? Because I tried to do it this way but I can't get the correct answer (V1 = 15 V and V2 = 20 V)

V3 is the voltage across 5 Ω resistor and Va is voltage across 1 Ω resistor. I3 is current through 5 Ω resistor, I2 for 20 kΩ resistor and I1 for the other two resistors.

For loop including 3 kΩ, 1 k&ohm and 5 k&ohm resistors: -V3 + V1 + Va = 0. V3 = -5000*I3, V1 = 3000*I1, Va = 1000*I1

Thus I get: 5*I3 + 4*I1 = 0 FIRST EQUATION

For loop including 5 k&ohm, and 20 k&ohm resistors: -V3 + V2 = 0. V2 = 20,000*I2, V3 = -5000*I3 (same as before)

Thus I get: I3 + 4*I2 = 0 SECOND EQUATION

Now using KCL I get 10 mA - I1 - I2 - I3 = 0. THIRD EQUATION

Then solving for the currents yields I1 = 25 mA and I2 = 5 mA (but that's wrong cause they should be I1 = 5 mA and I2 = 1 mA)

I read in a book that the sign you add to the voltage when using KVL is the first sign you encounter when moving along the loop (of arbitrary direction). And that if you encounter a negative sign first (of a voltage) then that voltage equals -R*I (a negative sign is added).

• There are two branches (left and middle-right and far-right) where the 10mA current source will split. All branches will have the same voltage V_2 across them. Write out the equations with three branch currents (i_1, i_2, i_3) will sum to 10mA and solve. Commented Aug 31 at 4:11
• I added my process, I'd appreciate if you could find what part of it is wrong. Commented Aug 31 at 4:30
• Asterisk is used to make text italicized. Use backslash asterisk to make and asterisk sign. I’ve taken the liberty to update it for you. Commented Aug 31 at 9:12
• Please make an updated version of the schematic with your added voltage and current designations ($v_3$, $i_3$, etc). This will make it much easier for us to follow your math than having to work out what you mean from a text description. Commented Aug 31 at 17:26

### initial thoughts

I read in a book that the sign you add to the voltage when using KVL is the first sign you encounter when moving along the loop (of arbitrary direction). And that if you encounter a negative sign first (of a voltage) then that voltage equals -R*I (a negative sign is added)

Books may bive you a lot of rules to memorize. Learn the powerful key concepts and how to apply them to any situation without aid or help from others. Learn to navigate on your own. You only have to know a few things.

KCL is powerful. KVL is powerful. Learn those well. Superposition is also powerful. Learn that, too. And stick with the passive sign convention. There are other conventions. But learning more than one, at least for now, will just get in your way.

### proper annotation and problem setup

V3 is the voltage across 5ohm resistor and Va is voltage across 1ohm resistor. I3 is current through 5ohm resistor, I2 for 20k ohm resistor and I1 for the other two resistors.

Here's with the added work I would have expected from you:

Note that all devices are labeled (it's just confusing to have numbers and to have to identify things that way.) I've labeled in the currents you designated, using the passive sign convention. To be consistent, I also labeled $$\V_{_\text{A}}\$$ with the same passive sign convention (you weren't clear about this.)

In addition, I added one possible group of mesh currents, assuming you may want to solve this using KVL/mesh. Your presentation is not KVL/mesh. It's different. So you have to be careful with it.

Also, I've labeled the nodes for the case where you may want to use nodal (KCL) analysis.

So the schematic is all set up now for a variety of ways of solving.

For loop including 3k, 1k and 5k ohm resistors: -V3 + V1 + Va = 0. V3 = -5000I3, V1 = 3000I1, Va = 1000*I1

Thus I get: 5I3 + 4I1 = 0 FIRST EQUATION

Let's stay with your currents and list out what we know:

• $$\V_2=I_1\cdot\left(R_{_\text{A}}+R_1\right)\$$
• $$\V_2=I_2\cdot R_2\$$
• $$\V_2=I_3\cdot R_3 \left[=V_3\right]\$$
• $$\I_{_\text{Z}}=I_1+I_2+I_3\$$

This is four equations and four unknowns. You know a few more things from the above, but it just duplicates information and doesn't add any. For example, $$\V_2=V_3\$$, as you have defined them. But we don't need $$\V_3\$$. It's just another name.

Let's use SymPy to solve this:

e1 = Eq( V2, I1*(Ra+R1) )
e2 = Eq( V2, I2*R2 )
e3 = Eq( V2, I3*R3 )
e4 = Eq( Iz, I1 + I2 + I3 )
solve( [ e1, e2, e3, e4 ], [ I1, I2, I3, V2 ] )
{I1: Iz*R2*R3/(R1*R2 + R1*R3 + R2*R3 + R2*Ra + R3*Ra),
I2: (Iz*R1*R3 + Iz*R3*Ra)/(R1*R2 + R1*R3 + R2*R3 + R2*Ra + R3*Ra),
I3: (Iz*R1*R2 + Iz*R2*Ra)/(R1*R2 + R1*R3 + R2*R3 + R2*Ra + R3*Ra),
Vz: (Iz*R1*R2*R3 + Iz*R2*R3*Ra)/(R1*R2 + R1*R3 + R2*R3 + R2*Ra + R3*Ra)}
for i,j in solve( [ e1, e2, e3, e4 ], [ I1, I2, I3, V2 ] ).items():
....:     i, j.subs( { R1:3e3, R2:20e3, R3:5e3, Ra:1e3, Iz:10e-3 })
....:
(I1, 0.00500000000000000)
(I2, 0.00100000000000000)
(I3, 0.00400000000000000)
(V2, 20.0000000000000)


### KVL/mesh

This could have also been set up using the mesh currents I used.

(Note here that the passive sign convention for the blue mesh currents is its own thing and ignores the passive sign convention applied to the green-labeled currents that you set up. Regardless, we use the passive sign convention as is appropriate for the approach under consideration.)

• $$\0\:\text{V}-R_1\cdot I_{_\text{B}}-R_{_\text{A}}\cdot I_{_\text{B}}-V_{_\text{Z}}=0\:\text{V}\$$
• $$\0\:\text{V}+V_{_\text{Z}}-R_3\cdot \left(I_{_\text{C}}-I_{_\text{D}}\right)=0\:\text{V}\$$
• $$\0\:\text{V}-R_3\cdot \left(I_{_\text{D}}-I_{_\text{C}}\right)-R_2\cdot I_{_\text{D}}=0\:\text{V}\$$
• $$\I_{_\text{Z}}=I_{_\text{C}}-I_{_\text{B}}\$$

Let's use SymPy to solve this:

e5 = Eq( 0 - R1*Ib - Ra*Ib - Vz, 0 )
e6 = Eq( 0 + Vz - R3*(Ic-Id), 0 )
e7 = Eq( 0 - R3*(Id-Ic) - R2*Id, 0 )
e8 = Eq( Iz, Ic-Ib )
solve( [ e5, e6, e7, e8 ], [ Ib, Ic, Id, Vz ] )
{Ib: -Iz*R2*R3/(R1*R2 + R1*R3 + R2*R3 + R2*Ra + R3*Ra),
Ic: (Iz*R1*R2 + Iz*R1*R3 + Iz*R2*Ra + Iz*R3*Ra)/(R1*R2 + R1*R3 + R2*R3 + R2*Ra + R3*Ra),
Id: (Iz*R1*R3 + Iz*R3*Ra)/(R1*R2 + R1*R3 + R2*R3 + R2*Ra + R3*Ra),
Vz: (Iz*R1*R2*R3 + Iz*R2*R3*Ra)/(R1*R2 + R1*R3 + R2*R3 + R2*Ra + R3*Ra)}
sage: for i,j in solve( [ e5, e6, e7, e8 ], [ Ib, Ic, Id, Vz ] ).items():
....:     i, j.subs( { R1:3e3, R2:20e3, R3:5e3, Ra:1e3, Iz:10e-3 })
....:
(Ib, -0.00500000000000000)
(Ic, 0.00500000000000000)
(Id, 0.00100000000000000)
(Vz, 20.0000000000000)


Note that $$\I_{_\text{Z}}=5\:\text{mA}-\left(-5\:\text{mA}\right)=10\:\text{mA}\$$. Confirming point. Also keep in mind that $$\V_{_\text{Z}}\$$, the voltage across the current source, is the same as $$\V_2\$$.

### KCL/nodal

This is pretty easy:

• $$\\frac{V_2}{R_{_\text{A}}}+\frac{V_2}{R_2}+\frac{V_2}{R_3}=\frac{V_1}{R_{_\text{A}}}+I_{_\text{Z}}\$$
• $$\\frac{V_1}{R_{_\text{A}}}+\frac{V_1}{R_1} =\frac{V_2}{R_{_\text{A}}}\$$

There are only two unknown nodes. So just two equations.

Let's use SymPy to solve this:

e9 = Eq( V2/Ra + V2/R2 + V2/R3, V1/Ra + Iz )
e10 = Eq( V1/Ra + V1/R1, V2/Ra )
solve( [ e9, e10 ], [ V1, V2 ] )
{V1: Iz*R1*R2*R3/(R1*R2 + R1*R3 + R2*R3 + R2*Ra + R3*Ra),
V2: (Iz*R1*R2*R3 + Iz*R2*R3*Ra)/(R1*R2 + R1*R3 + R2*R3 + R2*Ra + R3*Ra)}
for i,j in solve( [ e9, e10 ], [ V1, V2 ] ).items():
....:     i, j.subs( { R1:3e3, R2:20e3, R3:5e3, Ra:1e3, Iz:10e-3 })
....:
(V1, 15.0000000000000)
(V2, 20.0000000000000)


### summary

It all just works. You just need to get into the routine, is all.

If you wanted to just do it quickly without all that math involved, then you could just take $$\5\:\text{k}\Omega\,\vert\vert\, 20\:\text{k}\Omega\,\vert\vert\, \left(1\:\text{k}\Omega+3\:\text{k}\Omega\right)=2\:\text{k}\Omega\$$. Knowing the current, you know that $$\V_2=10\:\text{mA}\cdot 2\:\text{k}\Omega=20\:\text{V}\$$. And given the passive sign convention, you know that since the current arrow points into the top, then $$\V_2\$$ must be positive with respect to the bottom wire at the other end of the current source. You can work out the rest from there.

• Thanks a lot. I had a misconception about how to assign a sign to a voltage across a resistor when expressing it in terms of current/*resistance. In other words, I didn't understand when V = IR and when V = -IR (this one has a negative sign). Commented Sep 1 at 2:25
• @KevinRodriguez Signs are, perhaps, the most common mistake people make when learning. This is why a convention helps. The passive convention has it that the more positive end of a resistor is the one that the current arrow points into (or at.) Of course, you might tell me that you don't know which way the current arrow is going. But here, you just make the call. You decide. Then stick with it. You can choose either way for the current. But once you pick a direction, you are stuck with that choice. Then use the passive sign convention to set up resistor signs using your choice of direction. Commented Sep 1 at 3:46
• @KevinRodriguez Math is like that. There isn't an electronics-god out there. Nature itself has no idea what you think about the world. It just does what it does. These conventions are human-made. But nature is consistent. (Unlike humans, who are all over the place and almost never consistent.) So long as you work out a choice and apply it consistently then you will get an answer that is consistent with nature and reality. What will kill you is if you are inconsistent. That's the one thing that will give bad results. So just learn a pattern (passive convention) and stick with it. Commented Sep 1 at 3:50

how can I solve for V1 and V2 using kirchhoff's laws?

You start by simplifying the problem. If you study the circuit you should be able to realize that $$\v_1\$$ has to be 75% of $$\v_2\$$. This means that you focus your efforts into solving $$\v_2\$$. And clearly, $$\v_2\$$ has to be 20 volts based on the paralleled resistors in the circuit.

So, why are you wanting to use k-laws? Going down a mathematical route without using common-sense is always going to trip you up.

• You're right. i didn't use common sense because I was just trying to make sure that if I use k law (even if there's an easier way), I'll get the correct result. In other words, I was not understanding k laws correctly and i was trying to use this problem as a way to understand them. Commented Sep 1 at 2:28

I think this must be the definitions you describe for voltage and current.

Your first step equations look correct:

1. -V_3 + V_a + V_1 = 0, which means V_a+V_1 = V_3 (looks good)
2. -V_3 + V_2 = 0, which means V_2 = V_3 (looks good)
3. 10mA - i_1 - i_2 - i_3 = 0, which means i_1 + i_2 + i_3 = 10mA (looks good)

V_1 = 3K * i_1, V_a = 1K * i_1, and V_2 = 20K * i_2, but here is correction V_3 = +5K * i_3 (not minus, current is in same direction as defined voltage drop).

This should solve to correct answer. Maybe did you draw the i_3 current in wrong direction?

• Turns out I had a misconception about how to assign a sign to a voltage across a resistor when expressing it in terms of current * resistance. In other words, I didn't understand when V = IR and when V = -I*R (this one has a negative sign). Commented Sep 1 at 2:32