1. rvc

|dw:1440434594670:dw|

2. rvc

3. rvc

|dw:1440434946834:dw|

4. arindameducationusc

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

5. IrishBoy123

i can have a try later today

6. Michele_Laino

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

7. rvc

yep

8. Michele_Laino

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

9. rvc

okay

10. Michele_Laino

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

11. rvc

incoming ccurrents= outgoing currents

12. Michele_Laino

yes! or algebraic sum of currents=0

13. rvc

yep

14. Michele_Laino

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

15. rvc

oh okay

16. Michele_Laino

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

17. Michele_Laino

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

18. Michele_Laino

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

19. Michele_Laino

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

20. Michele_Laino

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

21. rvc

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

22. Michele_Laino

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

23. Michele_Laino

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

24. 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|

25. Michele_Laino

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

26. Michele_Laino

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

27. Michele_Laino

@rvc

28. 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

29. Michele_Laino

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

30. 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).

31. 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.]

32. mathmate

Yep, I got the same answers.

33. mathmate

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