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For the circuit shown below Vs = 6V, IS1 = 3A, IS2 = 0.08cos(5t), V2 = 2i+4i V , = V1/8 - V2/8 a)
For the circuit shown below Vs = 6V, IS1 = 3A, IS2 = 0.08cos(5t), V2 = 2i+4i V , = V1/8 - V2/8 a) Find the operating points of V1, i, V2, i2. (Take the positive root as the solution of the second degree equation. b) Make small signal analysis and approximate solutions Vn(t)= VnQ + Vn(t) and in(t) = InQ + in(t) (n=1, 2) + | Vs i1 + V1 3-T + V2 i2 IS1 4 IS2 A
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Income Tax Fundamentals 2013
Authors: Gerald E. Whittenburg, Martha Altus Buller, Steven L Gill
31st Edition
1111972516, 978-1285586618, 1285586611, 978-1285613109, 978-1111972516
