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Unformatted text preview: From here, we calculated the Thevenin impedance. 2. Solution (1) First, since V = I Z, we solve for the equivalent impedance Zeq = 200 + R +jwL = 200 + R + j(623.19)L Since w = 2*pi*1k Using the following equations, we can solve for L and R: (1): 4*e^(j ) = 200 *I (2): 8*e^(j ) = (R + j*623*L)I (3): 10*e^(j ) = I(200 + R + j*623*L) (2) 3. Conclusions This design studio allowed students to solve real world problems involving AC circuits. Solving these problems demonstrated the students understanding in dealing with sinusoidal steady state analysis problems in the frequency domain....
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This note was uploaded on 11/03/2009 for the course ECEN 214 taught by Professor Su during the Spring '08 term at Texas A&M.
 Spring '08
 SU
 Volt

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