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/sci/ - Science & Math

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File: 15 KB, 783x548, coursework elec.jpg [View same] [iqdb] [saucenao] [google]
2132227 No.2132227 [Reply] [Original]

i need to work out the total resistance for this circuit, but i cant figure out which ones wud be im parallel and series cuz theres so many. sorry for the shitty diagram, all the hand writin values ae in ohms and the voltage is 10

>> No.2132233

bump

>> No.2132236

that 15 and 5 in the middle are obviously parallel, start there

>> No.2132249
File: 36 KB, 783x548, 1291132610820.jpg [View same] [iqdb] [saucenao] [google]
2132249

I believe it should be like this... not 100% sure

>> No.2132253

15 and 5 are in paralel, they become R1
R1 is in paralel with another one of 15, they become R2
R2 is in serie with 10, they become R3
R3 is in paralel with 8, they become R4
R4 is in series with 5. And that's the final one

>> No.2132264

First the 15 and 5 in the middle are parallel, find the equivalent resistance (let's call it R1)
Then R1 and the 5 (at the top right) are in parallel, find the equivalent.. R2
Then you have to resistances 10 and R2 on serie, equivalent R3=R2+10
then R3 is in parallel with the 8, find the equivalent for R3 and 8 (say R4)
Then R total is R4+5 (series resistance)

>> No.2132269

could someone give me a answer for the total reistance, im not sure im doing it right, ill post my answer up soon

>> No.2132270

>>2132269
Alright, confirmed for troll.

>> No.2132273

op here not a troll i just didnt listen in lecture now i cant do the coursework and i dont understand out lecturers accent

>> No.2132274

>>2132269
just a question the resistance at the top left is a 5 or a 15 ?

>> No.2132286

op here top left is 5

>> No.2132287

<span class="math">Rx = \frac{{15X5}/{(15+5)}} = 3.75\Omega
Ry = \frac{{3.75X8}/{(3.75+8)}} = 2.55\Omega
Total resistance = 5+5+10+2.55 = 22.55\Omega[/spoiler]

>> No.2132296

R1 = 15*5/(15+5)=3.75
R2=5*3.75/(5+3.75)=15/7
R3=10+15/7=85/7
R4=85/7*8/(85/7+8)=680/141
Rtot=R4+5=1385/141~ 9.82ohm

>> No.2132302

op here
right i got 10.32= R total
assuming R1=3.75
R2=2.14
R3=15.89
R4=5.32
R total = R4 + 5 = 10.32 is this right?

>> No.2132317

>>2132302
R2 is okay but then R3 is R2+10=12.14

>> No.2132333
File: 6 KB, 218x252, 1274290165136.jpg [View same] [iqdb] [saucenao] [google]
2132333

oo i didnt R3= 10+R2+R1 well now ive got 9.82=Rtotal so thanks for that man :)

>> No.2132541
File: 12 KB, 659x314, circuit.jpg [View same] [iqdb] [saucenao] [google]
2132541

I always redraw circuits like this