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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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