/sci/ - Science & Math

Kirchoff's laws and Ohm's law.

>>3027930
Right but how does each circuit separately split the current, and the resistance in each circuit. Or am I just being stupid?

>>3027940
Resistance in parallel:
<span class="math">R_t=R^1 + R^2 + R^3 ...[/spoiler]
Do it across each parallel segment, then use V=RI

4.0 + 6.0 are parallel with 5.0. That is in series with the parallel comination of the 10 ohm resisitors. That is in series with the 18 ohm resistors. You first need to calculate the total resistance of the circuit before you can do anything else.

>>3027961
Disregard that, Rt^-1=R1^-1+R2^-1....

>>3027961
<span class="math"> \displaystyle{ \frac{1}{R_T} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \cdots + \frac{1}{R_n} } [/spoiler]

FTFY

resistors in series: add.
in parallel : inverse is sum of inverses.

4+6=10
1/10+1/5=1/3.33333.....
1/10+1/10=1/5
5+3.333....+18=26.3333

so the net resistance is 26.333... ohms.
so the current in the battery is I=V/R=12.5/26.333...=75/158 amperes.

current thru 18 = I
current thru 10 = (1/2)I
current thru 5 = (2/3)I
current thru 4 = current thru 6 = (1/3)I

Voltage and current for each resistor group:
U_465 = 12.5/(R_465 + R_1010 + R_18)*R_465
U_1010 = 12.5/(R_465 + R_1010 + R_18)*R_1010
U_18 = 12.5/(R_465 + R_1010 + R_18)*R_18

I_465=I_1010=I_18=12.5/(R_465 + R_1010 + R_18)


Group 1 (465):

U_46=U_5=U_465
I_46=U_46/R_46=U_465/R_46=U_465/(R_4+R_6)

U_4=U_46/R_46*R_4
U_6=U_46/R_46*R_6
I_4=I_6=I_46

U_5=U_465
I_5=U_5/R_5


Group 2 (1010):

U_10=U_1010
I_10=U_10/R_10

Group 3 (18):
Already given in step 1.

Voltage and current for each resistor group:
U_465 = 12.5/(R_465 + R_1010 + R_18)*R_465
U_1010 = 12.5/(R_465 + R_1010 + R_18)*R_1010
U_18 = 12.5/(R_465 + R_1010 + R_18)*R_18

I_465=I_1010=I_18=12.5/(R_465 + R_1010 + R_18)

Where
R_465 = 1/(1/(R_4+R_6) + 1/R_5)
R_1010 = 1/(1/R_10 + 1/R_10)


Group 1 (465):

U_46=U_5=U_465
I_46=U_46/R_46=U_465/R_46=U_465/(R_4+R_6)

U_4=U_46/R_46*R_4
U_6=U_46/R_46*R_6
I_4=I_6=I_46

U_5=U_465
I_5=U_5/R_5


Group 2 (1010):
U_10=U_1010
I_10=U_10/R_10

Group 3 (18):
Already given in step 1.

Much simpler shortcut to do these parallel resisters:

R1 X R2/R1 + R2

Shortcut for only 2 resistors in parallel.

>>3028009
Thank you that helped me understand that alot better.

>>3028184

Also, with the 10 ohm and 10 ohm parallel resistors, that is equal to 10 ohms total in that segment, just like if it were 5 and 5 it would be 5.

>>3028184
>>3028197
WRONG!!!

>>3028196
You're welcome

>>3028184

Req=1/(1/R1+1/R2)=(R1 R2)/(R1+R2)

if R1=R2=R this gives Req=R/2 , not R

>>3028266
actually >>3028184 is correct

>>3028310

thank you good sir
l2 highschool level ohm's law

>>3028286
oh god i lolled, nice work adding there dude

>>3028310
wtf?

R1 X R2/R1 + R2 = R2+R2 = 2 R2

Also, if you combine the 2 of series resistors (in a parallel section) and treat it as one resistor you can then combine that with the other resistor that is in parallel with it, creating a single resistor with a single resistance value that includes all the resistors of that section. You can do this with all the parallel sections then add them (you should be left with a purely series circuit when you're done reducing.)

Then, you can work backwards when you know the current and figure the current for each resistor.


Don't know how clear that is but I tried :).


Also, I am
>>3028184
>>3028197
>>3028316

>>3028337
oh
R1 X R2/(R1 + R2)
better?

>>3028374

Yes, that is what I meant in my original post.

Sorry for the misunderstanding.

>>3028337

>>3028374
much

still wrong: >>3028197

1)redraw the circuit as 3 resistors in series
18 + 5 + 3.3 = 26.3 ohms total

2)calc current total
12.5/26.3=.475Amps
18ohms=.475 amps
10ohms = .237amps

3)find the voltage across the 5ohm
the resistance measured across the 5 would be 3.3 ohms from step 1
3.3 X .475 = 1.75V

4)solve everything else with simple E=IR


Note: another way to solve step 3 is V=Eapplied X (Rbranch/Rt)
or V=12.5 X (3.3/26.3)

>>3028522
oops
3.3 X .475 = 1.56V