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lab8 report

# lab8 report - this 10 11)This leads to the following...

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1.) Calculation of This calculation involved the following equations: Making sure that we were using a 1x multiplier on the oscilloscope, , dA rms =2.94. Then using the first equation above the following calculation was done: , = , * = . * = . vA rms dA rms 42V10 2 94 4210 12 348V Again using the oscilloscope the value for , = . iA rms 2 01A . That led to the following calculations: , = , ( )* = . iA rms iA rms measured 2 4 02A A = , * Q vA rms , iA rms * = .* . * = . sinθ 12 4 02 sin90 49 64 , = Xm L L = Q , iA rms2 ⇒ = * * * L Q2 π f , iA rms2 = . 4 07mH = . 4 07mH 2.) Rs was given in lab to be: 3.) 4.)

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5.) In order to find the values for Lis and Lir the following three phase circuit diagram is needed: 6.) 7.) Using this diagram we make the assumption that the Lm parallel branch is an open. Doing this leads to the following equation: 8.) = + + ( + ) Z Rs Rr's jω Lis Lir 9.) From this equation we use the following equation that was given in the lab to simplify
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Unformatted text preview: this: 10.) 11.)This leads to the following simplification: 12.) = + + Z Rs Rr's jω53Lir 13.) 14.) = , , =( . , )( . ,-)= . + . Z VA rmsIA rms 12 348 ∠0 4 02 ∠ 18 2 92 j 949 15.) 16.)Equating the imaginary portion equal to jω53Lir at 120Hz gives a value for L ir =4.75mH. Using the equation above that relates L ir to L is yields a value for L is =3.16mH 17.) 18.) 19.)Phase 23.8 20.)Da .198 21.)Ia about orig val 22.) 23.) , 24.) 25.) 4.) R r ’ =-, , =-. . = . ° θpf tan 1iA rmsVA rms tan 14 02A12 348V 18 03 26.) =( , )* , * . °= . * . * . °= . Power VA rms iA rms cos18 03 12 348V 4 02A cos18 03 47 2W 27.) = , = , + = . . + Power VA rms2R VA rms2Rs Rr' 12 34820 18Ω Rr' 28.) . . -. = 12 348247 2 18 Rr' 29.) = . Rr' 3 05Ω 30.)...
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