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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 Φ) = 200 *I (2): 8*e^(j Φ) = (R + j*623*L)I (3): 10*e^(j Φ) = 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 student’s understanding in dealing with sinusoidal steady state analysis problems in the frequency domain....
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 Spring '08
 SU
 Alternating Current, Volt, sinusoidal steady state, Lauren Fife Parth Khade Risa Aprilria

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