/sci/ - Science & Math

here

The ruccent at R_5 and R_6 is 0.

The curcuit is not closed.
no closed curcuit => no current

>>4420890
>>4420888
Fuck thank you so much, especially for the diagram.

>>4420830
need voltage.

>>4420830
here you go.

>>4420934
appreciate it

U = R * I
R_1 = 1 Ohm
R_2 = 2 Ohm
R_3 = 3 Ohm
R_4 = 4 Ohm

I_1 = I_3 = I_4 = 2 A
I_2 = 8 A

U_1 = R_1 * I_1 = 1 Ohm * 2 A = 2 V
U_2 = R_2 * I_2 = 2 Ohm * 8 A = 16 V
U_3 = R_3 * I_3 = 3 Ohm * 2 A = 6 V
U_4 = R_4 * I_4 = 4 Ohm * 2 A = 8 V

Source = U_2 = U_1 + U_3 + U_4 = 16 V

Alright this is my last question. Its a little more involved but worth a try. Also its for the same circuit as the one used in the OP question.

14V 2Ω -- current source converted to voltage source in series with 2Ω
-3.5V 5Ω -- voltage source in series with R1+R4 (R3 is shorted by the voltage source)
11.5V 7Ω -- sum of voltage and resistance
I=11.5/7 A -- loop current
Vdc=I*4Ω=46/7 V -- voltage across R4

trade school wisdom, needs scientific verification

calc error
I=10.5/7 A -- loop current
Vdc=I*4Ω=42/7 V=6V -- voltage across R4

>>4420888
How is the current for R1 and R3 the same with R2 in parallel?

>>4421737
>me

Nvm, just saw it... i need some coffee or something

>>4421737
R_1, R_3 and R_4 are in line.

In line = same current.

R_2 is parallel to the line with R_1, R_3 and R_4

R_ab = R_2 || (R_1 + R_3 + R_4)
R_ab = 2 Ohm || (1 Ohm + 3 Ohm + 4 Ohm)
R_ab = 2 Ohm || 8 Ohm
R_ab = 2 Ohm * 8 Ohm / ( 2 Ohm + 8 Ohm )
R_ab = 16 Ohm² / 10 Ohm = 1,6 Ohm

>>4421765
Thanks for explaining it. I saw that just a moment after I posted... I was like: "Oh, man... where's my coffee when I need it?"

>>4421765
Wait...

2 Ohm || 8 Ohm
(1 / 2 Ohm) + (1 / 8 Ohm)
(5 / 8 Ohm)
Total resistance = 8/5 Ohm
Right?

>>4421773
>me

I've never seen this method before:
>R_ab = 2 Ohm * 8 Ohm / ( 2 Ohm + 8 Ohm )

>>4421765
If you had 3 resistors in parallel (R1, R2, R3) would it be:

(R1 * R2 * R3) / (R1 + R2 + R3) ?

>>4421776
NO!

R_all = 1 /( 1 / R_1 + 1 / R_2 + 1 / R _3 )

With 2 resitors:
R_all = 1 /( 1 / R_1 + 1 / R_2 )
R_all = 1 /( R_2 / ( R_1*R_2 ) + R_1 / ( R_1*R_2 ) )
R_all = 1 /( ( R_1 + R_2 ) / ( R_1*R_2 ) )
R_all = ( R_1*R_2 ) /( R_1 + R_2 ) <- this is for only 2 resitors

>>4421788
Damn, okay. It's a shame there isn't a simply formula for every amount of resistors... i might play with this later today.

>We're doing circuits in my PHY II class for the next month or so

here
better to read

>>4421799
Why, thank you. That is quite clear.

Here with 3 resitors

>>4421828
You're amazing... thank you.

further questions?

>>4421862
Is there an over-all formula that can be applied to n-resistors?

>>4421897

yes. look:
>>4421799
1. line

>>4421961
Yeah... I was wondering (without actually investigating personally) if there is a multiplication-type solution for n-resistors.

Generally, when you have resistors in parallel, the [1/x] key on your calculator is your friend.

For two resistors you normally use R=R1*R2/(R1+R2)

If you have two resistors in parallel and one is a multiple of the other (R2=n*R1, n=integer), the result is R=R2/(n+1).

Examples:
R1=10Ω, R2=10Ω, R2/R1=1, R1||R2=10/2=5Ω
R1=5Ω, R2=20Ω, R2/R1=4, R1||R2=20/5=4Ω

>>4421987

Possible.
But. look >>4421828
3 Resitors -> 3 Resistors (as multiplikation) on the top of the line
and 3 * 2 resistors (as multiplikation) on the buttom.

4 Resistors -> 4 on the top and 4*3 on the buttom
5 Resistors -> 5 on the top and 5*4 on the buttom

n Resistors -> n on the top and n*(n-1) on the buttom

looks easyler ^^ >>4421799

>>4421987
>whatthefuckamireading.jpg

Just write <div class="math">\frac{1}{R*} = \frac{1}{R1} + \frac{1}{R2} + ... + \frac{1}{Rn}</div>

you do know how to calculate with fractions, do you?

>>4422041
>you do know how to calculate with fractions, do you?
Oh yeah, I just like to play around and solve things differently if I can. It's just a bit of fun.

>>4422040
>n Resistors -> n on the top and n*(n-1) on the buttom
Hmm, interesting. And yeah, it won't be as easy as just adding all the inverses. Just interesting.