How we solve supernode in nodal analysis?
87871
let's consider a network like this:\n\n                           \n-node1(v1)----(+)V(-)----node 2 (v2)- \n                         \n\n\nnow this region is a supernode as voltage sourve(V) connects\nnode1 and node2 with no other element bet. node 1 and 2.\nAlso lets say that v1 and v2 are potentials of node 1 & 2 resp.\nSO how do u solve the equation? its simple.follow these 2 steps:\n\n1)write v1 = v + v2               \n2)then express v1 and v2 using KCL as seperate eqns.\nWhat is KCL?equate all currents thro the node as zero.\n\nU will get 3 equations.eqn 1 in step (1).\nand eqn 2 and 3 for KCL at node 1 and 2 respectively.\n Solve them using linear method or use Kramer's rule.(use determinants)
956604
Try it out in the lewis-dot diagram\n      ..        ..\nH-O-Ca-O-H\n      ..        ..
907364
Almost.  You would actually include the endpoints 0 and -3 (so you need to use a bracket instead of parentheses).\nD = [0, infinity)\nR = [-3, infinity)
839609
Curl, divergence, gradient.
786148
Electronic engineering - circuit analysis:\n\nSolving a circuit for the current and voltage according to Kirchhoff's Laws and these device equations:\nResistor   : v = R * i\nCapacitor: i = C * dv/dt\nInductor   : v = L * di/dt\n\nAnything complicated you generally solve by using the Laplace transform, in which case the above equations simplify to\nfor V,I = Laplace transforms of v,i:\nResistor   : V = RI\nCapacitor: I = sCV\nInductor   : V = sL\n\nThen you solve the system of equations for V,I and apply a reverse Laplace transform.
924696
This isn't the final answer, but it is a clue:\n\nhttp://www.thesheetmetalshop.com/pdf_manual/Problem1.pdf
111743
There is a great Cal Tech Video on that.  The one dealing with vectors.  It should be at most larger libraries.
816665
Look up SEMICONWELL\nIntegrated Passive Networks .
700308
Seems more like a vector analysis question to me.
525985
I think the numerical recipes book is the gold standard for solutions to these and many other kinds for problems.\n\nThis book is freely available online at\n\nhttp://www.library.cornell.edu/nr/cbookcpdf.html\n\nFor your questions take a look at chapter 10\n\nIf you can master this book you should have no shortage of programming opportunities.
725621
6Rb(s)+N2(g)--->2Rb3N(s)
469104
The algebraic sum of the currents entering a node is equal to zero. Where current entering is positive and exiting is negative (or vice versa). In other words, what comes in must come out.\nNote: this is only true for DC signals.
861221
looks like the vertex is (-5, 3)
496945
This site may help : http://www.superstringtheory.com
460850
http://en.wikipedia.org/wiki/Jakob_Steiner\nhttp://mathcircle.berkeley.edu/BMC3/constr2/node6.html
927748
You first define an array NodesList with index 0 as origin (C standard) containing all your nodes. The variable NodesNumber is the number of nodes. The key element is to define the operators S (for successor) and P (for predecessor) that will make the array behave as a circular list.\nS(NodesList[ i ]) = NodesList[ i + 1] if i < NodesNumber - 2\nand for the last node\nS(NodesList[ NodesNumber - 1 ]) = NodesList[ 0 ] \nSimilarly for P\nP(NodesList[ i ]) = NodesList[ i - 1] if i > 0\nand for the first node (origin)\nP(NodesList[ 0 ]) = NodesList[  NodesNumber - 1 ] \nThe algorithm has 2 parts\n// forward\nStartNode := NodesList[ NodesNumber - N  ]\nfor I := 1 to N\n  begin\n     StartNode := S(StartNode);\n     Node := StartNode;\n     Path := StartNode;\n     for J := 1 to N - 1\n        begin\n           Node := S(Node);\n           Add Node to Path \n        end \n      Record Path\n  end;\n// backward\nStartNode := NodesList[ N  ]\nfor I := 1 to N\n  begin\n     StartNode := P(StartNode);\n     Node := StartNode;\n     Path := StartNode;\n     for J := 1 to N - 1\n        begin\n           Node := P(Node);\n           Add Node to Path \n        end \n      Record Path\n  end;\nIn the second part the starting point is changed and S() are changed by P()
501994
http://hyperphysics.phy-astr.gsu.edu/hbase/electronic/posfe.html#c3\n\nThe link above may be of some help. Input the values you want or need.