# Resistor Circle

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#### mpaap

##### New Member
**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

##### Member
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...

#### Attachments

• resistor.jpg
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#### mpaap

##### New Member
hmm, i'll try

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

Last edited:

#### Chippie

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

#### mpaap

##### New Member
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

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

#### birdman0_o

##### Active Member
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

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

#### mpaap

##### New Member
That's what i got until

#### Chippie

##### Member
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?

#### Chippie

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

#### mpaap

##### New Member
and maybe you can suggest me something

#### Chippie

##### Member
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

##### Well-Known Member
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

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

#### ericgibbs

##### Well-Known Member
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

#### mpaap

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

#### birdman0_o

##### Active Member
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

• mpaap

#### ericgibbs

##### Well-Known Member
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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