Resistor Circle

mpaap · Sep 17, 2009

**broken link removed** - I need help with that homework!
Check out the picture. Is there anyone who can solve it? or is there a program which solves it?

Calculate the impedance circuit is shown in the figure, the current in all resistors and resistors abeyance annually. All resistors have resistance 10oomi scheme of the limbs and voltage 44V [translated by google translator]

Supply Voltage = 44V
each resistor = 10Ω

and in Estonia It's I=U/R, not I=V/R

I know:
I,U,R, R1,R2,R3,R4,R5,R6, I1,I2,U1,U2
i need to know:
I3,I4,I5,I6,U3,U4,U5,U6

Waiting for assist, thanks

Chippie · Sep 17, 2009

I'm not going to give you the answer....but here's a little help...I've simplified the drawing but there's a little work left for you to do...

mpaap · Sep 17, 2009

hmm, i'll try
Thanks for your help.

I don't get it, how do i find from your simplified circuit a I3 or I4 or I5 or I6 ?

Chippie · Sep 17, 2009

here's a clue....read up on resistor networks...resistors in series and parallel....

mpaap · Sep 17, 2009

i do not know what to do next :/
i found all the resistors resistivity
R123456 = 11Ω
and I = 4A
What can i find next?

Chippie · Sep 17, 2009

You need to show how you arrived at your figures....

Show us your calculations.......

birdman0_o · Sep 17, 2009

How did you end up with 11ohms if it is initially given as 10?

The equations you will need : V = IR
Facts: In parallel 1/R1 + 1/R2....1/Rn = 1/Rot
In series R1 + R2 ... + Rn = Rtot

Willbe · Sep 17, 2009

I get 15 ohms in parallel with 10 and that combo in series with 5.

mpaap · Sep 18, 2009

**broken link removed**
That's what i got until

Chippie · Sep 18, 2009

ok, the combined resistance of the 5r and the two parallel ones(10 and 15) comes out at 11 ohms...
With a pd of 44 volts across the 2 resistors which are effectiviely in series, gives a current flow of 4 amps...

Calc the voltage across the first two 10 ohm resistors by V=IR...then subtract that from the 44v to give the voltage across the remaining three....

Now are we getting somewhere?

mpaap · Sep 18, 2009

Chippie said:
Now are we getting somewhere?

i wish !
it's killing me

Chippie · Sep 18, 2009

You should go back to your tutor.............

mpaap · Sep 19, 2009

and maybe you can suggest me something

Chippie · Sep 19, 2009

mpaap said:
and maybe you can suggest me something

I'm not a tutor...I've given you a way forward...You need to do a bit more thinking

ericgibbs · Sep 19, 2009

mpaap said:
and maybe you can suggest me something

hi,
Lets try it in simple steps.
You know that all the resistors are 10 ohms.

Step1:
So what is the resistance of R4 and R5 in parallel..???

Answer this first.

mpaap · Sep 19, 2009

Resistance of R4 and R5 is 5 ohms, same with R1 and R2. i calculated whole resistance, and it's 11 ohms

ericgibbs · Sep 19, 2009

mpaap said:
Resistance of R4 and R5 is 5 ohms, same with R1 and R2. i calculated whole resistance, and it's 11 ohms

So what is the current from the 44V supply?

mpaap · Sep 19, 2009

well, if the voltage is 44 and resistance is 11 then I = V / R so.. I = 44 / 11 = 4A
now what?

birdman0_o · Sep 19, 2009

Now work backwards:

V= IR

In parallel I1+ I2 + I3 = Itot

In series I is the same

In series U1 + U2 + U3 = U tot

In parallel U is the same

ericgibbs · Sep 19, 2009

mpaap said:
well, if the voltage is 44 and resistance is 11 then I = V / R so.. I = 44 / 11 = 4A
now what?

Good.

You know the value of R1 and R2 in parallel is 5R, so use ohms law to find the voltage drop across R1 and R2

Vdrop= 5ohms * 4Amps =.......

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