In a parallel circuit(imagine two branches),with one DC voltage in a branch and another DC voltage in a branch, what is the method of computing the total DC voltage? The voltage through each component in a parallel circuit is the same across all components. How does this apply here?

  • 1
    $\begingroup$ You can't connect perfect voltage sources in parallel, unless they have the same voltage, and even then the distribution of currents would not be defined. If you want to do this properly, you have to insert a resistance in series with each source (except one). $\endgroup$ – CuriousOne Apr 12 '16 at 4:52
  • $\begingroup$ @CuriousOne I left out the mentioned of "resistors" thinking it was irrelevant. But yes you are spot on with the set up. How should I determine the voltage across the parallel circuit then? Is the voltage just some V across each resistor? $\endgroup$ – Physkid Apr 12 '16 at 5:01
  • $\begingroup$ Assuming there are no other currents flowing, the voltage across the resistor(s) will be the difference between the voltages of the two sources. $\endgroup$ – CuriousOne Apr 12 '16 at 5:10

Use KVL and the potential diagram for $V_1>V_2$ might help you find $V$?

enter image description here

| cite | improve this answer | |

Normally doing this with out any resistive loads is problematic, but there are ways to get by the problem if certain conditions are assumed. Though i am not sure about how valid such assumptions are. Anyway...
First you have to sum up all the currents from each parallel conductor. To do so you just use Ohm's law ans sum all the currents; $$I_{\mathrm{tot}} = \frac{U_1}{R_1} + \frac{U_2}{R_2} +\cdots+ \frac{U_n}{R_n}$$
("$n$" is the amount of parallel coupled voltage sources)

Now when you have the total current ($I_{\mathrm{tot}}$), which goes in trough the parallel coupling, you need to find the total conductance ($G_{\mathrm{tot}}$) for the parallel coupling.
(If you are unfamiliar with the concept of conductance ($G$) then $G = 1/R$)

To get "$G_{\mathrm{tot}}$" you do the following calculation: $$G_{\mathrm{tot}} = \frac{1}{R_1} + \frac{1}{R_2} +\cdots + \frac{1}{R_n}$$

To finally to get $U_{\mathrm{tot}}$ we need to divide $I_{\mathrm{tot}}$ with $G_{\mathrm{tot}}$; $$U_{\mathrm{tot}} = \frac{I_{\mathrm{tot}}}{G_{\mathrm{tot}}}$$ and there you have it.

The complete expression will look like this...
$$U_{\mathrm{tot}} = \frac{\left(\frac{U_1}{R_1} + \frac{U_2}{R_2} +\cdots + \frac{U_n}{R_n}\right)}{\frac{1}{R_1} + \frac{1}{R_2} +\cdots + \frac{1}{R_n}}$$

If you now assume that the resistance of every parallel coupled conductor; $$R = R_1 = R_2 =\cdots = R_n$$, then we can seriously simplify this expression...
$$U_{\mathrm{tot}} = \frac{\left(\frac{U_1}{R} + \frac{U_2}{R} +\cdots + \frac{U_n}{R}\right)}{\frac{1}{R} + \frac{1}{R} +\cdots + \frac{1}{R}} = \frac{\frac{U_1 + U_2 +\cdots + U_n}{R}}{\frac{n}{R}} = \frac{U_1 + U_2 +\cdots + U_n}{n}$$

Conclusion: If you assume that every parallel conductor have the same amount of resistance then $U_{\mathrm{tot}}$ becomes the arithmetic mean value of all the voltage sources; $$U_{\mathrm{tot}} = \frac{U_1 + U_2 +\cdots + U_n}{n}$$

I hope that helped :D.

| cite | improve this answer | |
  • $\begingroup$ I can also tell you that this calculation is not just trivia. It can be used when calculating the voltage across a Y coupled three face power line. |Utot| is greater than zero when there are disturbances along the power lines. My point is to say that what people often think of as an unimportant question can at the same time give useful answers to an engineer. So keep on asking ;). $\endgroup$ – Chloé Seppälä Zetterberg May 7 '17 at 23:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.