An L-R-C series circuit is constructed using a 175- resistor, a 12.5-F capacitor

Question

An L-R-C series circuit is constructed using a 175- resistor, a 12.5-F capacitor, and an 8.00 mH inductor, all connected across an ac source having a variable frequency and a voltage amplitude of 25.0 V. a) At what angular frequency will the impedience be smallest. b) At the angular frequency in part (a), find the maximum current through the inductor. c) At the angular frequency in part (a), find the potential across the ac source, the resistor, the capacitor, and the inductor at the instant that the current is equal to one-half its greatest positive value. Answer to (C) below: Question: If (XL . XC)/R = 0/R =0, how is =  90° in the above solution. This solution is from the instructor's solution manual and is different than the solution posted for this problem on this site.

Explanation / Answer

a)the angular frequency at which the impedance will besmallest is w = (1/(LC)1/2) L = 8.00 mH = 8.00 * 10-3 H and C = 12.5 F =12.5 * 10-6 F b)the maximum voltage across the inductor is vL = Imax * XL * sin(wt+ /2) = Imax * XL * cos(wt) or Imax = (vL/XL *cos(wt)) vL = 25.0 V,XL = w * L and wt =0o c)the current through the resistor,the capacitor and theinductor is iR = Imax * sin(wt) iR = (Imax/2) or (Imax/2) = Imax * sin(wt) or sin(wt) = (1/2) or wt = (/6) the current through the capacitor is iC = Imax * sin(wt + /2) iC = (Imax/2) or (Imax/2) = Imax * sin(wt +/2) or wt + /2 = /6 or wt = /6 - /2 = -/3 the current through the inductor is iL = Imax * sin(wt - /2) iL = (Imax/2) or (Imax/2) = Imax * sin(wt -/2) or wt - /2 = /6 or wt = (2/3) the potential across the resistor,the capacitor,and theinductor at the instant when the current is equal to one-half itsgreatest positive value is vR = Imax * R * sin(wt) R = 175 and wt = /6, vL = Imax * XL * sin(wt+ /2) XL = wL and wt = -/3 and vC = Imax * XC *sin(wt - /2) XC = (1/wC) and wt = (2/3) c)the current through the resistor,the capacitor and theinductor is iR = Imax * sin(wt) iR = (Imax/2) or (Imax/2) = Imax * sin(wt) or sin(wt) = (1/2) or wt = (/6) the current through the capacitor is iC = Imax * sin(wt + /2) iC = (Imax/2) or (Imax/2) = Imax * sin(wt +/2) or wt + /2 = /6 or wt = /6 - /2 = -/3 the current through the inductor is iL = Imax * sin(wt - /2) iL = (Imax/2) or (Imax/2) = Imax * sin(wt -/2) or wt - /2 = /6 or wt = (2/3) the potential across the resistor,the capacitor,and theinductor at the instant when the current is equal to one-half itsgreatest positive value is vR = Imax * R * sin(wt) R = 175 and wt = /6, vL = Imax * XL * sin(wt+ /2) XL = wL and wt = -/3 and vC = Imax * XC *sin(wt - /2) XC = (1/wC) and wt = (2/3)

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