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Nodal Circuit Analysis Using KCL ã   Most useful for when we have mostly current sources ã    Node analysis uses KCL to establish the currents Procedure (1) Choose one node as the common (or datum) node ã    Number (label) the nodes ã   Designate a voltage for each node number ã   Each node voltage is with respect to the common or datum node ã    Number of nodes used = number of nodes – 1 = n-1 ã    Note: number of nodes = branches – 1 = b-1 ã   Thus less equations with node analysis than mesh analysis (2) For each node write the KCL for current flows in each node ã   Use differences in the node voltages to calculate currents ã   Assume the current directions and write the KCL ã   Generally assume the node is a positive V relative to all others ã   Current directions different for same branch in each node ã   Often better to use conductance equations ⎥⎦⎤⎢⎣⎡+=+=+= 211211121 11  R RV  RV  RV  I  I  I   R R  (3) Solve the equations for the node voltages ã   Get currents in each branch from the voltage differences  Example Nodal Circuit Analysis ã   Consider the 2 node, 3 loop circuit below (1) Setting the base node, and node voltages ã   Set the common node to ground ã   Label voltages on the others (2) For each node write KCL for current flows ã   For node 1, ã   Defining the current directions for I 1  as into the node ã   Use differences in the node voltages for currents ã   Collect each voltage into one term ( ) 13231132111 11  I  RV  R RV  RV V  RV  =−⎥⎦⎤⎢⎣⎡+=−+   ã   Or in the conductance form ( ) ( )  13231132111  I GV GGV GV V GV   =−+=−+    Example Nodal Circuit Analysis continued ã   For node 2, ã   Defining the current directions for I 2  as into the node ã   This means current in R  3  different from node 1 ( ) 23223131222 11  I  R RV  RV  RV V  RV  =⎥⎦⎤⎢⎣⎡++−= −+   ã   Or if changing to conductance form ( ) ( )  23223131222  I GGV GV GV V GV   =++−=−+   ã    Now have two equations and unknowns (V’s) 132311 11  I  RV  R RV   =−⎥⎦⎤⎢⎣⎡+   232231  I  R1 R1V  RV  =⎥⎦⎤⎢⎣⎡++−   ã   Could solve this algebraically ã   Instead use the numerical methods  Example Nodal Circuit Analysis ã   Putting the equations in numerical form ã   for node 1: ( ) ( )  013.000025.0V 00125.0V  013.04000V 4000110001V  RV  R1 R1V  212132311 =−=−⎥⎦⎤⎢⎣⎡+=−⎥⎦⎤⎢⎣⎡+   ã   For node 2: ( ) ( ) 003.000075.000025.0 003.04000120001400011 212132231 =+−=⎥⎦⎤⎢⎣⎡++−= ⎥⎦⎤⎢⎣⎡++− V V V V  R RV  RV    ã   Using substitution method for node 1 ( ) 00125.000025.0V 013.0 V   21 +=   ã   Thus using node 2 for the solution ( )( )  003.000075.0V 00025.0 00125.000025.0V 013.0 22 =+⎥⎦⎤⎢⎣⎡ +   ( )  0026 .0003.000075.000005.0V  2  +=+−   V 8007 .00056 .0V  2  ==

Jul 23, 2017

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