Moreover all of the usual hazards of non linear

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Unformatted text preview: so much as we are zeroing out VB1 and making VB2 = 10. We find i1 = −11.25 mA, i2 = −12.5 mA, i3 = −16.25 mA, and i4 = −10 mA, where the negatives indicate that the current is flowing in the opposite direction as is indicated on the diagram. The reader is encouraged to study the symmetry here, and if need be, hold up a mirror to the diagram to literally ‘see’ what is happening. (d) For VB1 = 10V and VB2 = 10V , we get i1 = 5 mA, i2 = 0 mA, i3 = −5 mA, and i4 = 0 mA. The mesh currents i2 and i4 being zero is a consequence of both batteries ‘pushing’ in equal but opposite directions, causing the net flow of electrons in these two regions to cancel out. Solution to 1(c) Solution to 1(d) 2. We now turn the tables and are given VB1 = 10V , VB2 = 5V , i1 = 10.625 mA, i2 = 6.25 mA, i3 = 3.125 mA and i4 = 5 mA and our unknowns are the resistance values. Rewriting our system of equations, we get 5.625R1 + 4.375R3 1.25R2 − 4.375R3 + 3.125R4 −3.125R4 − 1.875R6 −5.625R1 − 1.25R2 + 5R5 + 1.875R6 = 10...
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This note was uploaded on 05/03/2013 for the course MATH Algebra taught by Professor Wong during the Fall '13 term at Chicago Academy High School.

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