Example 8-13 Solving a mixed ckt using nodal

# Example 8-13 Solving a mixed ckt using nodal - 8/16 0.1V1...

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Example 8-13 Figs 8-26, -27, p. 240 Solve for the currents through R2 and R3 using nodal w/o source conversion and using k  and  mA. @V1: Iin = Iout:  0 = V1/R3 + (V1 – V2)/R2 + 2  =>  0 = V1/6 + (V1 – V2)/10 + 2 V1(1/6 + 1/10) – V2/10 = -2  => (0.267)V1 + (-0.1) V2 = -2 @V2: (V1 – V2)/R1 = (V2 – E1)/R2 + (V2 –(-E2))/(R4 + R5) => (V1 – V2)/10 = (V2 – 10)/5 + (V2

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Unformatted text preview: + 8)/16 0.1V1 – 0.1V2 = 0.2V2 – 2 +V2/16 +8/16 => V1(0.1) + V2(-0.1) – 0.2 V2 – V2/16 = -2 + 0.5 V1(0.1) + V2(-0.1 – 0.2 – 1/16) = -1.5 => V1(0.1) + V2(-0.363) = -1.5 Solving simultaneously: V1 = -6.63 V; V2 = = 2.31 V I3 = V1/R3 = -6.63/6 = -1.11 mA I2 = -(V2 – E1)/R2 = -(2.31 -10)/5 = 7.69/5 = 1.54 mA...
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Example 8-13 Solving a mixed ckt using nodal - 8/16 0.1V1...

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