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Unformatted text preview: as positive while the ones leaving the node are negative. Thus in the previous example, this statement means: i1 i4 i5 i2 i3 0 which is of course the same thing. It is also important to notice that KCL applies not only to nodes, but to ANY CLOSED surface. e.g. If in this circuit I choose S as a closed surface I have: 15.15 total current entering the surface = I + 2 “ “ leaving “ “ = 0 Thus I + 2 = 0 or I = ‐2A. Why is KCL true? • Kirchoff’s Voltage Law(KVL) The algebraic sum of the voltage drops around any closed path is zero at every point in time t. The following notes explain how you can apply this law and why it is true. 1. Find as a reminder, remember that VAB VA VB where VAB voltage drop between nodes A and B VA voltage of node A (with respect to a reference node, typically the ground) VB voltage of node B (with respect to a reference node, typically the ground) 2. Suppose you want to calculate VAB between the nodes A and B of a circuit. Then you start from node A and you mode towards node B. You take as positive the voltage drops that you meet the (+) sign first. You take as negative the voltage drops that you meet the (‐) sign first. For the circuit above: 15.16 VA VB V1 V2 V3 or 3. VAB V1 V2 V3 or VA V1 V2 V3 VB If you start and finish at the same point A, then obviously VAA 0 . Then if you follow the rules in (2) you will be able to write KVL for any closed path. You simply start at any point you like and you move clockwise (cw) or counter‐clockwise (ccw) [you choose!] You take as positive any...
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- Spring '08