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rvc
 one year ago
Circuit question
Please help :)
rvc
 one year ago
Circuit question Please help :)

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rvc
 one year ago
Best ResponseYou've already chosen the best response.0@rishavraj @IrishBoy123 @mathmate @Michele_Laino @e.mccormick please help :)

arindameducationusc
 one year ago
Best ResponseYou've already chosen the best response.0wow... nice one.. i just had my dinner... was about to sleep.. would you mind if I do tomorrow morning?

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1i can have a try later today

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0I think that we have to apply the second and first principle of Kirchhoff

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0dw:1440436198619:dw we have to suppose the existence of those currents, I1,...,I6

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0now, we have to write the first principle of Kirchhoff at each node

rvc
 one year ago
Best ResponseYou've already chosen the best response.0incoming ccurrents= outgoing currents

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0yes! or algebraic sum of currents=0

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0I label each node as below: dw:1440436439762:dw

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0for node A: \[\Large {I_2} + 20  {I_3} = 0\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0for node B: \[\Large {I_3}  {I_4}  120 = 0\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0for node C: \[\Large {I_4} + 110  {I_5} = 0\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0for node D \[\Large {I_5}  {I_6}  60 = 0\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0for node Y: \[\Large {I_6} + 80  {I_1} = 0\]

rvc
 one year ago
Best ResponseYou've already chosen the best response.0can we asume current through ab as I1 and ax as 20I1 ?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0for node X: \[\Large {I_1}  {I_2}  30 = 0\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0with those equations, we have expressed the conservation of electrical charge

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0now, 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
 one year ago
Best ResponseYou've already chosen the best response.0here 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
 one year ago
Best ResponseYou've already chosen the best response.0so, you have to determine all currents, I1,...,I6, then substituting into last equation, you will get the requested voltage drop Vxy

mathmate
 one year ago
Best ResponseYou've already chosen the best response.1Hmm, 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
 one year ago
Best ResponseYou've already chosen the best response.0if 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
 one year ago
Best ResponseYou've already chosen the best response.1What 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
 one year ago
Best ResponseYou've already chosen the best response.1yes, 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
 one year ago
Best ResponseYou've already chosen the best response.1Yep, I got the same answers.

mathmate
 one year ago
Best ResponseYou've already chosen the best response.1Don't forget to find Vxy=61*0.02=1.22 V... etc.
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