7 x 3x 5 5x 4 b xx 22 x m c 7x 6x2

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Unformatted text preview: 2 , i3 and i4 , measured in milliamps, mA. If we think of electrons flowing through the circuit, we can think of the voltage sources as providing the ‘push’ which makes the electrons move, the resistors as obstacles for the electrons to overcome, and the mesh current as a net rate of flow of electrons around the indicated loops. R5 i4 R1 VB1 i1 R2 R3 i2 R6 R4 The system of linear equations associated with this circuit is (R1 + R3 ) i1 − R3 i2 − R1 i4 −R3 i1 + (R2 + R3 + R4 ) i2 − R4 i3 − R2 i4 −R4 i2 + (R4 + R6 ) i3 − R6 i4 −R1 i1 − R2 i2 − R6 i3 + (R1 + R2 + R5 + R6 ) i4 i3 VB2 = = = = VB1 0 −VB2 0 1. Assuming the resistances are all 1k Ω, find the mesh currents if the battery voltages are (a) VB1 = 10V and VB2 = 5V (b) VB1 = 10V and VB2 = 0V (c) VB1 = 0V and VB2 = 10V (d) VB1 = 10V and VB2 = 10V 2. Assuming VB1 = 10V and VB2 = 5V , find the possible combinations of resistances which would yield the mesh currents you found in 1(a). 7 The authors wish to thank Don Anthan of Lakeland Community College for the design of this example. 8.4 Systems of Linear Equations: Matrix Inverses 501 Solution. 1. Substituting the resistance values into our system of equations, we get...
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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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