how would you obtain this total resistance?

[little_carlos]

hey, i was solving this simple resistor network to obtain the total resistance, but, i cant obtain the same result as the software says, based on my knowledge it should be 6.32 ohms, idk how the software obtained that value
here is the pic

which is your result? is it the same as the software? how did you obtained it?
thanks for your time

[Andy Watson]

I agree with the software. Note that R1 + R8 gives 5.66\$\Omega\$, therefore the answer must be less.

[bitseeker]

I also get 5.61 ohms.

1. R3, R4 and R6 are in series = 8 ohms
2. R8 is in parallel with the series resistance #1, above = 8 * 0.66 / (8 + 0.66) = 0.61 ohms
3. R1 is in series with the network #1, #2 = 5 + 0.61 = 5.61 ohms

[little_carlos]

I also get 5.61 ohms.

1. R3, R4 and R6 are in series = 8 ohms
2. R8 is in parallel with the series resistance #1, above = 8 * 0.66 / (8 + 0.66) = 0.61 ohms
3. R1 is in series with the network #1, #2 = 5 + 0.61 = 5.61 ohms

lool thanks, i noticed i didnt put the parentesis on the calculator, lol im an ashole
but the steps i did were right, thanks again

[Vgkid]

I have 5.59ohms so good enough.
you guys are fast with your responses. I could explain it faster than i could calc it.

[bitseeker]

i noticed i didnt put the parentesis on the calculator

Ah, yes. Ensuring the correct order of operations is important.

Even better, do the addition part in your head.

[Brumby]

Note that R1 + R8 gives 5.66\$\Omega\$, therefore the answer must be less.

... That was my very first observation - and one of several valuable 'estimation' processes to give you an idea of where the answer should be.  Blind trust in calculations is risky without an independent means to support the given answer.

Also came up with 5.61\$\Omega\$

[bson]

((R6+R3+R4)||R8)+R1=5.61ohm

[xygor]

Ok, I gotta ask: What are you people looking at!?  I can't read the resistor values in the OP drawing.  Do I have a browser issue?

[Ian.M]

Browser issue, insufficient screen resolution, or you need to go to an optician.

[Vgkid]

Ok, I gotta ask: What are you people looking at!?  I can't read the resistor values in the OP drawing.  Do I have a browser issue?

Looks like the image host provider shrunk/dropped the quality of the image. Even after downloading , it is still blurry.

[Ian.M]

Pin-sharp here and I've just checked it in another browser to confirm I'm not seeing a locally cached copy.

Even though this forum tells your browser to resize large images to fit the column width, the resistor values and 'multimeter' are readable as is with a 1024px screenwidth.  The jpeg itsself opened in a new window is 1278x719px and is fully readable, right down to the clock on the O.P.'s windows taskbar.

However for anyone contemplating posting something similar (i.e. any line drawing or diagram), *PLEASE* crop irrelevant crud from the edges, reduce the colour depth to 256 or even 16 colours and save as GIF or PNG.   That gets you the most bang from your filesize limit buck if the original is *NOT* a photo.  JPEGs with medium to high compression blur fine detail so should be avoided for line art.  If you don't have anything better to do it in, grab a copy of IrfanView, which also has a very handy RIOT plugin for crunching JPEGs just enough to squeek through any specific file size limits.

[bson]

Yeah, perfectly clear and readable here too...