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Unformatted text preview: MAE140 Linear Circuits Summer 1 2006 Quiz Solutions. Wednesday July 19 Question 1: Nodal Analysis Part (i): (a) Set node D to be the grounded node and refer all other nodal voltages to that. (b) By inspection, the nodal voltage v A relative to node D is v s volts. (c) Applying KCL at node B yields ( G 1 + G 2 + G 5 ) v B- G 1 v A- G 5 v C = , ( G 1 + G 2 + G 5 ) v B- G 5 v C = G 1 v s . Applying KCL at node C yields similarly,- G 5 v B + ( G 3 + G 4 + G 5 ) v C = G 3 v s . Placing these last two equations into matrix form yields the expression in the quiz. Part (ii): Inverting the matrix in Part (i), we have parenleftbigg v B v C parenrightbigg = parenleftbigg G 1 + G 2 + G 5- G 5- G 5 G 3 + G 4 + G 5 parenrightbigg- 1 parenleftbigg G 1 G 3 parenrightbigg v s . Now using the matrix formula parenleftbigg v B v C parenrightbigg = 1 det( G ) parenleftbigg G 3 + G 4 + G 5 G 5 G 5 G 1 + G 2 + G 5 parenrightbiggparenleftbigg G 1 G 3 parenrightbigg v s , where det( G ) is the determinant given by the expression corresponding to...
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