I need help with this circuit problem

Discussion in 'Homework Help' started by elecstudent, Nov 11, 2010.

1. elecstudent Thread Starter New Member

Nov 11, 2010
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I'm trying to work out the Itotal in the circuit.

I have used nodal analysis;

though unsure if this is right:

V1-25 + V1/10 + V1-V2/20 = 0

(1/10+1/20)V1 + (1/20)V2 = 1-25

0.15V1 + 0.05V2 = -24 ............. (1)

&

V2-V1/20 + V2+50/30 + V2/60 = 0

(1/20)V1 + (1/20+1/30+1/60)V2 = 50/30

0.05V1 + 0.1V2 = 1.66666 ..........(2)

I get V1 = -198.6666
V2 = 116

I used Ohms law and figured out that 25V is over R1, Ir1 = 2.5A.

I used a circuit simulation to find Itotal, which = 3.958A (yet I don't seem to be able to use Ohms Law to find Itotal. I have calculated the total 'R' in the circuit to be 8ohms (also verified by the circuit simulation); yet 75 (Vtotal) / 8 Rtotal = 9.375A (this can't be correct, the circuit simulation tells me it's definately 3.958A).

Also; by using 3.958A as the cct total, I minused 2.5A=Ir1 to find Ir2, therefore Ir2 * R2 = 29.16V over R2. so I would assume at node 2 there would be a voltage of 29.16V, yet, the simulation tells me it's -4.16V.....
So NOW I'M COMPLETLY CONFUSED

I need to know how to calculate out Itotal, and at Node 2, how to work out voltage drops and associated I's.

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2. hgmjr Moderator

Jan 28, 2005
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I can confirm for you that the simulator is correct in both the voltage at N2 and the net current flowing in the network.

You have accounted for 2.5A of the total current. The balance is flowing in the rest of the circuit.

HINT: For the purpose of calculating the voltage at N2 you can ignore R1. That simplifies the circuit quite a bit.

hgmjr

3. thatoneguy AAC Fanatic!

Feb 19, 2009
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Node 1 will have the same voltage as the battery it is in parallel with.

Pay attention to the battery polarities as well.

4. elecstudent Thread Starter New Member

Nov 11, 2010
26
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if I subtract; 75 (Vtotal) - 29.16 = 45.84 across R3, gives me Ir3 @ 1.528A.

Yet why do I get -4.16V at node 2? as I fully understand it has to be right as -4.16/ R4 = -0.0067 which is correct.

I just don't understand why I have -4.16V at node 2 when i assumed it should be 29.16V; to then have to use 75 Vtotal - 29.16V = 45.85V as a 'I' result over R3.....

and even more basic, I'm stuck at how to calculate 'I' total of the cct (without the simulation).

Please help.. i've been lookign at this problem allday (literally), I have an array of answers and ideas, yet I need direction, someone to guide which one is right and why.

Thanks

5. Jony130 AAC Fanatic!

Feb 17, 2009
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Try solve this:

( 25V - N2 )/20Ω = N2/60Ω + ( N2 - (-50V) )/30Ω

6. elecstudent Thread Starter New Member

Nov 11, 2010
26
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it should be -30.012V (as a single linear equation just for N2 with R1 removed) @ N2.. I see using nodal analysis how you did the equation.. yet why to get rid of R1?

Also how does this tie in, with the rest oif the cct?

Last edited: Nov 11, 2010
7. hgmjr Moderator

Jan 28, 2005
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Notice that the voltage across the 10 ohm resistor is 25V and will not change regardless of what other resistors and voltage sources you place in parallel with it. That I why you can remove the 10 ohm resistor for the purpose of calculating the voltage at N2.

You can prove this with your simulator. Change the value of the 10 ohm resistor and you will see that the voltage at N2 does not change.

hgmjr

8. thatoneguy AAC Fanatic!

Feb 19, 2009
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Current Analasys on each current path:
$i_1 = \frac{25-N_2}{20}$

$i_2 = \frac{N_2}{60}$

$i_3 = \frac{N_2-(-50)}{30}$

For Node 2 analysis, R1 is ignored since it is in parallel with the 25vsource, it will come into play for total current, but not for node 2 calculations.

Since all of the currents at node 2 need to sum to zero, that is how the equation posted earlier was made.

9. elecstudent Thread Starter New Member

Nov 11, 2010
26
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so if I don't use R1, do I still need to take into account the Itotal contribution it gives...

and even so, how do I calc I total, given I had to use the simulation to provide the Itotal I'm using.

I had looked at superposition for this, though it didn't work, I got different answers to the ones I'm already aware of, so assumed this to be wrong.

10. hgmjr Moderator

Jan 28, 2005
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Yes, Absolutely!!!!!

11. thatoneguy AAC Fanatic!

Feb 19, 2009
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R1 will add to the total current in the circuit, the nice thing is that it is directly across a source.

The solution should come out fine if you keep all of the polarities straight, one battery is "upside down".

12. hgmjr Moderator

Jan 28, 2005
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You will need to use thatoneguy's and jony130's equations to find N2, first.

hgmjr

13. elecstudent Thread Starter New Member

Nov 11, 2010
26
0
i understand how you got the equations you did...

I really thought it was;

N2-25/20 + N2+50/30 + N2/60

as the opposite of N2 will be the ref node=0, and everything should be inrespect of the reference, as the current sees the voltage source as it enters the branch from the ref point.

I guess I missed that completely.

I see now how the circuit is simpler without R1.

If anyone can explain to me how to rememebr this rule (+ve, -ve) how to view it in a cct.

Also, how as the circuit is to cal Itotal, am I right I can't use Ohms law, and will just have to use noddal analysis, and add/subtract the currents. to get a result.

Thanks to all that have contributed.

Thanks

14. thatoneguy AAC Fanatic!

Feb 19, 2009
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There isn't really a "rule" for it, more of a guideline.

If you put your finger (or pen/pencil) on the node, and trace the circuit out from there, when you run into one end of a voltage source, be sure to include the sign the source, so if you hit the - side, it would be -V, if you it the postiive, it would be +5v

15. elecstudent Thread Starter New Member

Nov 11, 2010
26
0
I've finally, after considering my attempt at the linear equation for N2, understand how N2 is now 4.16V.. yet the answer I got was +ve yet the simualtion indicates it's -ve.. I guess I've messed the polarities up again....

Ok I've just done it again and here's what I get;

N2+50/30 + N2/60 + N2-25/20

gives

1/30 + 50/30 + 1/60 + 1/20 + -25/20 = 0

gives

(1/30 + 1/60 + 1/20) N2 = (-50/30 + 25/20)

0.1N2 = -0.416

N2 = -4.16V

Tell me please anyone, is this the only way to work this problem out.. I had thought I could break it down to Ohms Law.. and use finding Rtotal to then use Vtotal/Rtotal for the Itotal.. but this gives a value just over 9A.. when the cct is actually 3.9A...

This value for N2 gives the right 'I's' for the cct.. but I'm unsure to get Itotal from what I have, as I have 2.5A through R1, 1.45A through R2, 1.528A through R3 and -0.069A through R4.. none add upto the Itotal of 3.9A the simulation gave.

or, I just thought... (again please tell me if I'm wrong)...

2.5+1.45+1.528+0.069 = 5.4929A (2)

Then going back to the Rtotal of 8ohms I calculated should be: 75 (Vtotal) / 8 = 9.375A (1)

then (1) - (2) = 3.8821A as a total for the circuit 'I'

am I correct???????

Last edited: Nov 11, 2010
16. elecstudent Thread Starter New Member

Nov 11, 2010
26
0
Can anyone verify if my last post was correct.

17. Jony130 AAC Fanatic!

Feb 17, 2009
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Why you don't use first Kirchhoff's law to find I_total.

Itot = I1 + I2

I1 = 25V/10Ω = 2.5A

I2 = (N1 - N2) / R2 = ( 25V - ( - 4.166V) ) / 20Ω = 1.4583A

Itot = 2.5A + 1.4583A = 3.9583A

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18. elecstudent Thread Starter New Member

Nov 11, 2010
26
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I can now, this is great news.. thanks

Can I just ask, to calc why N2 is -4.16V.

Is that V1 - Vr2 = -4.16V = N2

Then N2 - (-50) = 45.84Vr3

Is this right?

It'll be my last question.

A massive thanks to all that have contributed

19. The Electrician AAC Fanatic!

Oct 9, 2007
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You can solve it using mesh equations:

$\left[ \begin{array}{3}10&-10 &0\\-10&10+20+30&-30\\0&-30&30+60\end{array}\right]*\left[ \begin{array}{4}Itot\\I2\\I4\end{array}\right]=\left[ \begin{array}{4}25\\ 50\\-50\end{array}\right]$

which will give you the clockwise currents Itot, I2 and I4 in the three obvious meshes, left to right.

Or you can use the nodal method:

$\left[ \begin{array}{2}1 & 0 \\\frac {-1}{20} & \frac {1}{20} + \frac {1}{30}+ \frac {1}{60} \end{array}\right]*\left[ \begin{array}{1}N1\\N2\\\end{array}\right]=\left[ \begin{array}{1}25\\ \frac {-50}{30}\\\end{array}\right]$

which will give you the voltages at the nodes N1 and N2.

20. hgmjr Moderator

Jan 28, 2005
9,030
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You can calculate the voltage at N2 using Millman's Theorem:

$N2=\Large {\frac{\frac{25}{20}+\frac{-50}{30}}{\frac{1}{20}+\frac{1}{60}+\frac{1}{30}}}$

hgmjr