Design_Studio_4__2

Design_Studio_4__2 - ω = 2 ⋅π ⋅ f = 2000π Z eq = 2 0...

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Unformatted text preview: ω = 2 ⋅π ⋅ f = 2000π Z eq = 2 0 0 + jω L + R = 2 0 0 + j 2 0 0 0 π L + R If we allow VS and V2 to be in the same phase of φ = 0, then: VS = 10 V V2 = 8 V VS = V1 + V2 V1 = VS – V2 θ θ 2 = V m 2 + θ 2 c o s ( ) + j V m 2 + θ 2 s i n ( ) …. (1) Vm Vm 4 = V m 2 + θ 2 …. (2) Taking the real part of equation (1), we get: θ 2 = 4 cos( ) Vm θ = 60° Vm From equation (2), we can solve for Vm and θ: Vm = .0667 θ=4 V1 = 4 /_60° VS 3 IS = = .0 1 + j Z eq 100 Z eq ⋅ I S = V S ( 2 0 0 + j 2 0 0 0 π L + R ) ( .0 1 + j 3 ) = 10 100 3 R = 10 100 2 + j 2 0 π L + .0 1 R + j 2 3 − 2 0 3 π L + j Take the real part: 2 + . 0 1 R + 2 0 3 π L = 1 0 .... (3) The imaginary part: 3 2 0πL − 2 3 − R = 0 …. (4) 100 From (4), we find: R = 8 0 0 − 2 0 0 0 π 3 L …. (5) Plug (5) into (3) to get: 3 H L= 8π R = 50 Ω ...
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This note was uploaded on 02/16/2009 for the course ECEN 214 taught by Professor Su during the Fall '08 term at Texas A&M.

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