What is the voltage drop (V1) across the 10 ohm resistor?

Can I use voltage division to get V1?

Like V1 = (20 V x 10 Ω) / (10 Ω + 4 Ω)

But when I use nodal analysis and mesh analysis I get a different answer (12.30 V).

Which one should I use?

Or have I understood the voltage division method incorrectly?

  • 6
    \$\begingroup\$ I would go with superposition. \$\endgroup\$
    – Eugene Sh.
    Commented Oct 30, 2020 at 16:47
  • 1
    \$\begingroup\$ Well, you haven't got the right answer for nodal analysis either. You are 0.5 volts out. So, if you are as good at solving it in superposition as you are in nodal then you are not going to get there. \$\endgroup\$
    – Andy aka
    Commented Oct 30, 2020 at 17:11
  • 1
    \$\begingroup\$ By using superposition I get 100/13 when 10V is closed and 60/13 when 20V is closed. After adding them up I get 160/13=12.3076V. Even using superposition I get the same answer. @Andyaka \$\endgroup\$ Commented Oct 30, 2020 at 17:37
  • 1
    \$\begingroup\$ OK I was wrong in my calculation. Reality check!! \$\endgroup\$
    – Andy aka
    Commented Oct 30, 2020 at 17:39
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    \$\begingroup\$ As others have pointed out (but you don't seem to have fully understood) you need to use the superposition principle to find the contributions of each source and then add them together. This is really as basic as homework questions get, so I don't feel that there is justification for writing up an answer on EE.SE, instead go learn this from one of the literally thousands and thousands of freely available learning resources online. \$\endgroup\$
    – user173292
    Commented Oct 30, 2020 at 17:54

5 Answers 5


Millman's theorem is a kind of multi source voltage divider theorem based on this topology: -

enter image description here enter image description here

So, re-arrange your circuit to use a 3rd source that is actually 0 volts and you're good to go: -

enter image description here

Hence \$V_{AB} = \dfrac{\frac{10}{3.33333} + \frac{0}{10} +\frac{20}{4}}{\frac{1}{3.3333}+ \frac{1}{10} + \frac{1}{4}} = \dfrac{8}{0.65} = 12.3077\$

But equally, without working out the parallel resistance of the 10 Ω and 5 Ω you could have written this (now there are two sources of 10 volts, one in series with 10 Ω and one in series with 5 Ω): -

$$V_{AB} = \dfrac{\frac{10}{10} +\frac{10}{5} + \frac{0}{10} +\frac{20}{4}}{\frac{1}{10}+\frac{1}{5}+ \frac{1}{10} + \frac{1}{4}} = \dfrac{8}{0.65} = 12.3077$$

  • 8
    \$\begingroup\$ +1 for the use of Millman's. Not enough people are aware of this simple approach \$\endgroup\$
    – user16222
    Commented Oct 30, 2020 at 18:30
  • \$\begingroup\$ Not even as simple as it could be. Working with resistance is awkward here. \$\endgroup\$ Commented Oct 31, 2020 at 16:45
  • \$\begingroup\$ @Harper-ReinstateMonica I don't follow your comment. Could it be simpler - is that what you are saying? \$\endgroup\$
    – Andy aka
    Commented Oct 31, 2020 at 17:21
  • \$\begingroup\$ @Andyaka I mean Millman's theory gets simpler when you work in conductance instead of resistance. \$\endgroup\$ Commented Oct 31, 2020 at 17:31
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    \$\begingroup\$ @Harper-ReinstateMonica isn't that what the denominator actually is? Sorry I'm not getting really what you mean here? \$\endgroup\$
    – Andy aka
    Commented Oct 31, 2020 at 17:56

Two voltage sources are contributing towards the voltage drop across V1.

So the equation \$V_1 = (20 .10)/(10+4)\$ is not correct. You ignored the 10V Source, and the 10 ohm-5 ohm resistor network in the process, as if they never exist in the circuit. Which is wrong.

Well, still 'voltage-division method' is what you use in superposition.

enter image description here

We have two voltage sources. Consider one voltage source at a time. Ideal voltage sources can be considered as a short or 0 ohm resistor, so short the other voltage source. Derive the two circuits.

Now, you can apply your voltage-divider formula on both circuits to calculate V1 on both cases. Finally, add them to get the net effect of both voltage sources. Thus, the net voltage drop V1.


You can solve this with either mesh or node analysis.

Here's my take on the problem using mesh analysis:

  • Let’s call the left mesh Mesh I and the mesh on the right, Mesh II.
  • There is a current i1 passing through Mesh I and a current i2 passing through Mesh II.
  • Notice that both i1 and i2 pass through the 10 ohm resistor in the middle.

We can build a linear system of equations of the form:

{ 3.3333333 Ω * i1 + 10 Ω * (i1 + i2) = 10 V

{ 10 Ω * (i1 + i2) + 4 Ω * i2 = 20 V

After we find values for i1 and i2, we have: V1 = 10 Ω * (i1 + i2) from Ohm's law.

Here's the MATLAB solution:

A = [13.3333333333333, 10; 10, 14];
B = [10; 20];

I = linsolve(A, B);

V = 10 * (I(1) + I(2))

Hence: V = 12.3077

  • 2
    \$\begingroup\$ It is better to have the equations all symbolic and only use the actual values at the last moment. \$\endgroup\$ Commented Nov 1, 2020 at 16:42

OK I got my initial calculation wrong but here's the source transformation method (answer now deleted because I was too hasty with my calculations). I've decided to undelete it because there is an accepted answer and the visibility of this answer makes a nice comparison.

Using source transformations: -

enter image description here

That's a total current of 8.0 amps flowing into a resistor of parallel values: -

$$R_P = \dfrac{1}{\frac{1}{10} +\frac{1}{5} + \frac{1}{10} + \frac{1}{4}} = 1.5385 \Omega$$

So, the voltage is 8.0 amps x 1.5385 Ω = 12.308 volts.

  • \$\begingroup\$ Notice that 10Ω||5Ω = 10Ω/3 and the current is I = 10V/(10Ω/3) = 3A \$\endgroup\$
    – G36
    Commented Oct 30, 2020 at 17:34
  • \$\begingroup\$ Oops yes. Fixing \$\endgroup\$
    – Andy aka
    Commented Oct 30, 2020 at 17:34

Let me give another approach, maybe simplistic.

There are several theorems or methods, but let's not forget that all those theorems come from the observation, someone found a pattern and proved its correctness.

However, when you are in front of a circuit like this, that is quite simple. Before applying theorems I think it is better to think by yourself.

Then, the first thing I see in this circuit is a parallel of two resistors that looks are there to confuse or to raise doubts, but I need to not be afraid and replace that parallel as follows.


simulate this circuit – Schematic created using CircuitLab

Then you get the following circuit where you want to find V1 and if you observe it you can see that there are three currents involved in it, Ie flowing through the equivalent parallel resistor, I2 flowing through R2 and I1 flowing through R1. You also know from Kirchoff's law (and this a law, not a theorem) that I1 = Ie + I2.


simulate this circuit

Each current's value is the following:

$$I_e=\frac{V_a-V_1}{R_e}$$ $$I_2=\frac{V_b-V_1}{R_2}$$ $$I_1=\frac{V_1}{R_1}$$

With a bit of algebra you reach your V1 = 12.3077 volt

  • \$\begingroup\$ It would be better to use subscripts for those quantities, e.g. Va. We have MathJax on this platform (and HTML <sub></sub> as the poor man's version). \$\endgroup\$ Commented Nov 1, 2020 at 16:56
  • \$\begingroup\$ @PeterMortensen, thanks for the comment, I am still learning the tools in here. Fully agree with this MathJax the readability increases \$\endgroup\$ Commented Nov 1, 2020 at 18:27

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