# How to solve this circuit ?

Discussion in 'Homework Help' started by curry87, Jan 7, 2011.

1. ### curry87 Thread Starter Member

May 30, 2010
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0
I have a circuit with a missing voltage value at vb im trying to solve.Cant calculate the current though r1 or r2 or r3 because dont know vb.

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2. ### Jony130 AAC Fanatic!

Feb 17, 2009
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Do you know superposition?

3. ### curry87 Thread Starter Member

May 30, 2010
101
0
No i know KVL,KCL,ohms law and equivalent circuit thats about it.Other than superposition theorem what other ways are there to solve it please give an example of each thanks.

4. ### breadboardkid New Member

Jan 2, 2011
2
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I would look up a uide for vb or you will never get to know whats wrong with your circuit

-BBK

5. ### Jony130 AAC Fanatic!

Feb 17, 2009
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OK so first we start with superposition:

So first we remove V2
So circuit look like this:

And we find
Vb_1 = V1 * (8Ω||2Ω) / ( (8Ω||2Ω) + 4Ω) = 5.71428V

Second step is to remove V1

Vb_2 = V2 * (8Ω||4Ω) / ( (8Ω||4Ω) + 2Ω) = 18.28571V

And finally Vb = Vb_1+Vb_2 = 23.9 ≈ 24V

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6. ### hobbyist Distinguished Member

Aug 10, 2008
782
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Another way would be to find an equivalent, circuit.

remove R3 and find the equivalent voltage, and equivalent resistance, between the two points where R3 was at.

Then draw an equivalent circuit consisting of the equivalent voltage source (Vth), and the equivalent resistance, (Rth), in series, then in series place the R3, and solve voltage using ohms law.

There are quite a few ways to analyse linear circuits, by learning and applying as many theorems possible, gives you a better understanding of circuit analysis.

7. ### curry87 Thread Starter Member

May 30, 2010
101
0
Why is it when i apply the equation for the above circuit to this simpler circuit i get this:

vb_1 = v1 *(16ohm+16ohm)/(16ohm+16ohm)
vb_2 = v2 *(16ohm+16ohm)/(16ohm+16ohm)

vb_1 = 10v *(16)/(16) = 10v

vb_2 = 30v*(16)/(16) = 30v

vb = vb_1 +vb_2 = 40v

But on spice its says the node vb = 22.5v is above equation only good for resistors in parallel ?

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8. ### Jony130 AAC Fanatic!

Feb 17, 2009
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Becaues you have errors in your equitations
vb_1 = v1 *(16ohm+16ohm)/(16ohm+16ohm)
vb_2 = v2 *(16ohm+16ohm)/(16ohm+16ohm)

Do you understand why this is correct ?

vb_1 = v1 *(6ohm)/(16ohm) = 3.75V

vb_2 = v2 *(10ohm)/(16ohm) = 18.75V

Vb = 22.5V

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9. ### hgmjr Moderator

Jan 28, 2005
9,030
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Here is the Millman's Theorem expression for Vb:

$V_b\ =\ \Large \frac{\frac{V_2}{R_1}\ +\ \frac{V_1}{R_2}}{\frac{1}{R_1}\ +\ \frac{1}{R_2}\ +\ \frac{1}{R_3}}\ =\ \frac{R_2R_3V_2\ +\ R_1R_3V_1}{R_2R_3\ +\ R_1R_3\ +\ R_1R_2}$

Here is the Millman's Theorem expression for current in R3:

$\frac{V_b}{R_3}\ =\ \left( \Large \frac{R_2R_3V_2\ +\ R_1R_3V_1}{R_2R_3\ +\ R_1R_3\ +\ R_1R_2}\right)\ *\ \frac{1}{R_3}\ =\ \Large \frac{R_2V_2\ +\ R_1V_1}{R_2R_3\ +\ R_1R_3\ +\ R_1R_2}$

Here is the Millman's Theorem expression for current in R2:

$\frac{V_2\ -\ V_b}{R_1}\ =\ \left[ \Large V_2\ -\ \left( \frac{R_2R_3V_2\ +\ R_1R_3V_1}{R_2R_3\ +\ R_1R_3\ +\ R_1R_2}\right)\right] \ *\ \frac{1}{R_1}\ =\ \Large \frac{\frac{R_2R_3V_2}{R_1}\ +\ R_3V_1}{R_2R_3\ +\ R_1R_3\ +\ R_1R_2}$

I'll leave it as an exercise for anyone interested to form the expression for the current in R1.

hgmjr

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10. ### thatoneguy AAC Fanatic!

Feb 19, 2009
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Dang, that's a Beautiful Post!

More people need to read my sigline.

11. ### hgmjr Moderator

Jan 28, 2005
9,030
214
Thanks for the compliment.

I absolutely adore Millman's Theorem and LaTex is a really handy way to express the equations. My AAC blog is focused on Millman's.

hgmjr

12. ### hobbyist Distinguished Member

Aug 10, 2008
782
69
After reading the above post, I got my CIE books out, and found 2 pages devoted to millmans theorem.

But it wasn't in the one book called "network theorems", but rather it was in the next book called "ADVANCED network theorems".

After rereading that, this does make it much easier to analyse parrallel circuits as in the OP's post, which contains voltage sources.

It is really nice how there are a lot of network theorems to choose from to do a circuit analysis. Each theorem is a special tool in itself, to help electronic engineers analyse complicated circuitry.

13. ### hgmjr Moderator

Jan 28, 2005
9,030
214
The common misconception that Millman's Theorem is limited to voltage sources is why it has such a poor following. Millman's Theorem is not limited to circuits with voltage sources; it works equally well with circuits containing current sources.

hgmjr

14. ### hobbyist Distinguished Member

Aug 10, 2008
782
69
Yep, I know, actually current sources, eleiminates one step in the process, of calculations, because you don't have to convert the E/R to a current value.
Its already done for ya.