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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.
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 =
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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