I require to uncover the identical Resistance across A and also B. The trouble is, i don"t recognize which one"s room in series and i beg your pardon ones are in parallel. How do I determine which ones space parallel or series?

exactly how do I identify which ones space parallel or series?

If *all* of the present leaving one resistor enters one more resistor, the 2 resistors are in series.

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The resistances of collection connected resistors deserve to be included together to discover the equivalent resistance of a solitary resistor, e.g.,

$$R_eq = R_1 + R_2 $$

If *all* the the voltage throughout one resistor is across another resistor, the two resistors space in parallel.

The *conductances* of parallel linked resistors have the right to be added together to uncover the equivalent *conductance* the a single resistor, e.g.,

$$G_eq = frac1R_eq = G_1 + G_2 = frac1R_1 + frac1R_2$$

or

$$R_eq = frac1frac1R_1 + frac1R_2 $$

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answer Oct 22 "14 at 21:31

Alfred CentauriAlfred Centauri

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Two resistors ~ above the

*same path*are

**in series**. There should be nothing else in between.Two resistors that follow

*parallel paths*room

**in parallel**. If the wire splits in to paths and gatheres into one again alter, this would be parallel paths.In your circuit, you have to piece by piece collect resistors into one

*equivalent*resistor.

Start with the $1.00;Omega$, $2.00;Omega$ and also $3.00;Omega$ ones most to the right. They space in series, so they space added:$$R_eq=1.00;Omega+ 2.00;Omega +3.00;Omega=6;Omega$$If you attracted the entirety circuit again, you could replace these 3 with *one* resistor through this value.

Now, this brand-new equivalent resistor is in parallel v the vertical $3.00;Omega$ resistor. For this reason they are included reversely:$$frac1R_eq=frac13.00;Omega+frac16.00;Omega=cdots$$These two have the right to now be replaced with one brand-new resistor of this value.

Next step would fx be to see that this brand-new equivalent resistor is in series with the $6.00;Omega$ one top top top. So include them...

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Continue in this way, till all are shortened down and there is just one indistinguishable resistor left. That will certainly the be the resulting identical resistance between A and also B.