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I'm pretty new to circuits and I only know basic rule from Ohm's law that U=I*R. I quite don't understand how was following formula for Ui created:

enter image description here

Why are there only two resistances in numerator? And why is the whole fraction multiplied by original voltage? Sorry for my dumb question, but I'm trying to figure out how does this work. I'm not sure how to take a right look at this example.

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It's not hard, but you need to have developed some very modest appreciation for voltage, current, and resistance by working through a few different examples for this to become more obvious. One of the very specific things you need to really understand better, perhaps, is the fact that when a current flows through a resistor, a voltage drop develops. Now, even my use of the word "develops" there might throw you off. It also could be just as well said that it "a difference in voltage potential must be present across it." It doesn't matter whether you think of it as the current creating the voltage, or the voltage creating the current, so long as you know that Ohm's Law tells you that there will be such a voltage if there is such a current, and visa versa.

Okay. Granting that. How can you compute the currents? Well, take your left side schematic. It's a series loop of three resistors. So the current through this loop will be:

$$I_{TOTAL} = \frac{U_{TOTAL}}{R_{TOTAL}}$$

But you know that:

$$R_{TOTAL}=R_1+R_3+R_4$$

So the current is:

$$I_{TOTAL} = \frac{U_{TOTAL}}{R_1+R_3+R_4}$$

That is where that denominator comes from, by the way.

Now, the next problem is the voltage at \$U_i\$. One side is referenced to the bottom node and the other side is in between two resistors. You could come at this from either side, approaching a computation that either adds up the voltages on the left side or else adds them up on the right. Either way, you'll get the right answer. Let's just go straight from the right side where there are two resistors there.

To compute the voltage across them, you need to know the current through them. But you know that, as we just computed it above. So let's do the sums:

$$\begin{align*} U_i &= I_{TOTAL}\cdot R_4 + I_{TOTAL}\cdot R_3 \\ \\ &= \frac{U_{TOTAL}}{R_1+R_3+R_4}\cdot R_4+\frac{U_{TOTAL}}{R_1+R_3+R_4}\cdot R_3 \\ \\ &= U_{TOTAL}\cdot\left(\frac{R_4}{R_1+R_3+R_4} +\frac{R_3}{R_1+R_3+R_4} \right) \\ \\ &= U_{TOTAL}\cdot\frac{R_3+R_4}{R_1+R_3+R_4} \end{align*}$$

Do you see that now?


Regarding \$R_i\$, I'll take a diagram approach. It wastes tons of HTML space, but a diagram is worth a thousand words they say.

schematic

simulate this circuit – Schematic created using CircuitLab

You can see the transitions there. There's only one more to go:

schematic

simulate this circuit

That's it.

So, just extracting from that schematic, the two equations are:

$$\begin{align*} U_{TH} &= \frac{U}{R_1}\cdot\frac{R_1\cdot\left(R_3+R_4\right)}{R_1+R_3+R_4} = U\cdot\frac{R_3+R_4}{R_1+R_3+R_4} \\ \\ R_{TH} &= \frac{R_1\cdot\left(R_3+R_4\right)}{R_1+R_3+R_4} \end{align*}$$

You'll notice that we re-aquired our voltage calculation from this, too. I hadn't realized that this is the approach you needed from the start, I guess. Hope that helps.

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  • \$\begingroup\$ Thank you very much for you patient answer. It's a lot clearer now. If it wouldn't bother you, could you please briefly try to explain the second part with Ri? I suppose we would advance somehow alike, but I'd like to be sure it's correct. \$\endgroup\$ – John Smith Oct 22 '16 at 20:44
  • \$\begingroup\$ @JohnSmith It's correct. But I'm not sure what tools you've acquired to gain that point. Do you understand Norton and Thevenin equivalent circuits, yet? Or would you prefer a calculus approach where \$R_i=\frac{\textrm{d}V_i}{\textrm{d}I_i}\$. What you show used a formulaic approach and not a proof, really. \$\endgroup\$ – jonk Oct 22 '16 at 20:53
  • \$\begingroup\$ It's already-solved circuit by our lecturer and I'm trying to figure out how he actually did it. I thought calculating Ri would be analogic to calculating Ui. Anyway, I'd stay away from calculus approach since that's what we're gonna do in the next semester. \$\endgroup\$ – John Smith Oct 22 '16 at 21:01
  • \$\begingroup\$ @JohnSmith Do you have Norton and Thevenin down? I'd use that approach. \$\endgroup\$ – jonk Oct 22 '16 at 21:11
  • \$\begingroup\$ I think I'll be good with that. \$\endgroup\$ – John Smith Oct 22 '16 at 21:12

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