What is the voltage drop across the 10 ohm resistor?

Question

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?

Answer 1

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

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

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): -

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

Answer 2

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.

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.

Answer 3

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

Answer 4

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: -

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.

Answer 5

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