Circuit question
Please help :)

- rvc

Circuit question
Please help :)

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- rvc

|dw:1440434594670:dw|

- rvc

@rishavraj @IrishBoy123 @mathmate @Michele_Laino @e.mccormick please help :)

- rvc

|dw:1440434946834:dw|

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## More answers

- arindameducationusc

wow... nice one.. i just had my dinner... was about to sleep..
would you mind if I do tomorrow morning?

- IrishBoy123

i can have a try later today

- Michele_Laino

I think that we have to apply the second and first principle of Kirchhoff

- rvc

yep

- Michele_Laino

|dw:1440436198619:dw|
we have to suppose the existence of those currents, I1,...,I6

- rvc

okay

- Michele_Laino

now, we have to write the first principle of Kirchhoff at each node

- rvc

incoming ccurrents= outgoing currents

- Michele_Laino

yes! or algebraic sum of currents=0

- rvc

yep

- Michele_Laino

I label each node as below:
|dw:1440436439762:dw|

- rvc

oh okay

- Michele_Laino

for node A:
\[\Large {I_2} + 20 - {I_3} = 0\]

- Michele_Laino

for node B:
\[\Large {I_3} - {I_4} - 120 = 0\]

- Michele_Laino

for node C:
\[\Large {I_4} + 110 - {I_5} = 0\]

- Michele_Laino

for node D
\[\Large {I_5} - {I_6} - 60 = 0\]

- Michele_Laino

for node Y:
\[\Large {I_6} + 80 - {I_1} = 0\]

- rvc

can we asume current through ab as I1 and ax as 20-I1 ?

- Michele_Laino

for node X:
\[\Large {I_1} - {I_2} - 30 = 0\]

- Michele_Laino

with those equations, we have expressed the conservation of electrical charge

- Michele_Laino

now, we have to happly the second principle of Kirchhoff, namely the subsequent equation for electrostatic field E:
\[\Large \nabla \times {\mathbf{E}} = 0\]
in order to do that we have to establish a positive sense into our circuit, like this:
|dw:1440436999284:dw|

- Michele_Laino

here is the missing equation:
\[\large {V_{XY}} + 0.01{I_2} + 0.01{I_3} + 0.03{I_4} + 0.01{I_5} + 0.02{I_6} = 0\]

- Michele_Laino

so, you have to determine all currents, I1,...,I6, then substituting into last equation, you will get the requested voltage drop Vxy

- Michele_Laino

@rvc

- mathmate

Hmm,
There are 6 equations for 7 unknowns!
@Michele_Laino
I put 5 equations for the joints (the sixth is redundant) and the Kirchhoff's second law as 0.02*I1+0.01*I2+0.01*I3+0.03*I4+0.01*I5+0.02*I6=0 instead of using Vxy, and I seem to get satisfactory results, with I4 and I6 negative.
Do you get the similar results? @rvc

- Michele_Laino

if we collect all those equations above, we get the complete system as below:
\[\Large \left\{ \begin{gathered}
{I_2} + 20 - {I_3} = 0 \hfill \\
\hfill \\
{I_3} - {I_4} - 120 = 0 \hfill \\
\hfill \\
{I_4} + 110 - {I_5} = 0 \hfill \\
\hfill \\
{I_5} - {I_6} - 60 = 0 \hfill \\
\hfill \\
{I_6} + 80 - {I_1} = 0 \hfill \\
\hfill \\
{I_1} - {I_2} - 30 = 0 \hfill \\
\hfill \\
{V_{XY}}{\text{ }} + {\text{ }}0.01{I_2}{\text{ }} + {\text{ }}0.01{I_3}{\text{ }} + {\text{ }} \hfill \\
{\text{ + }}0.03{I_4}{\text{ }} + {\text{ }}0.01{I_5}{\text{ }} + {\text{ }}0.02{I_6}{\text{ }} = {\text{ }}0 \hfill \\
\end{gathered} \right.\]

- mathmate

What I was saying is that there are 5 independent equations out of the first 6, so the last one will fill the void by expression Vxy as 0.02*I1. Then we get to have 6 equations, and 6 unknowns (I1 to I6).

- IrishBoy123

yes, first 5 plus 0.02*I1+0.01*I2+0.01*I3+0.03*I4+0.01*I5+0.02*I6=0 gets there!
[ 61. 31. 51. -69. 41. -19.]

- mathmate

Yep, I got the same answers.

- mathmate

Don't forget to find Vxy=61*0.02=1.22 V... etc.

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