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Calculate the self-inductance of the coil by direct method by using the following data : {(1,1.0,1.5,0.4,1.5),(2, 1.3, 2.0, 0.6, 2.0):} Frequency of AC =50 Hz. |
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Answer» Solution : Formula : `L = (sqrt(Z^(2)-R^(2)))/(2 PI F)` Using `R=V/I`, Calculating `R_(1) = 1.50 Omega` `R_(2)=1.54 Omega` & Mean `R = 1.52 Omega` Using `Z= V/I`, Calculating `Z_(1) = 3.75 Omega` `Z_(2)=3.33 Omega` Mean `Z=3.54 Omega` CALCULATION of `L=0.01019` H Deltailed Answer : For trial NUMBER (1) D.C. part, `R=V/I` `rArr R_(1)=(15)/(1) = 1.5 Omega` and for trial number (2) D.C. part `R_(2)=(2.0)/(1.3) = 1.54 Omega` Mean `R=(1.5+1.54)/(2) = 1.52 Omega` for A.C. part `Z_(1) = (1.5)/(0.4) = 3.75 Omega` `Z_(2)=(2.0)/(0.6) = 3.33 Omega` Mean `Z=3.54 Omega` Now `L= (sqrt(Z^(2)-R^(2)))/(2 pi f)` `rArr L = (sqrt((3.54)^(2)-(1.54)^(2)))/(2 xx 3.14 xx 50)` `L=0.01019 H`. |
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