A series RLC circuit has R = 445 ?, L = 1.35 H, C = 3.1 µF. It is connected to a
ID: 1433117 • Letter: A
Question
A series RLC circuit has R = 445 ?, L = 1.35 H, C = 3.1 µF. It is connected to an AC source with f = 60.0 Hz and ?Vmax = 150 V.
A series RLC circuit has R-445 , L = 1.35 H, C = 3.1 F. It is connected to an AC source with f = 60.0 Hz and AVmax = 150 V (A) Determine the inductive reactance, the capacitive reactance, and the impedance of the circuit. Find the angular frequency: Use the following equation to find the inductive reactance : XL aL-(377 s. )(1.35 H)- Use the following equation to find the capacitive reactance aC (377 s3.10x 10-6F) Use the following equation to find the impedance: Z- VR(XL Xc) Z = V(445 )" + (508.94 . 855.67 ) (B) Find the maximum current in the circuit. Imax-= (C) Find the phase angle between the current and voltage. Use the the following equation to calculate the phase angle XL XC (50904 957a) 508.94 , 855.67 = tan 445 (D) Find the maximum voltage across each element. Use the following equations to calculate the maximum voltages: dVR-ImaxR-(0.27 A)(445 )- AVL = 1maxXL = (0.27 A)(508.94 )- AVC = 1maxXc = (0.27 A)(855.67 ) = E What replacement value ofL should an engineer analyzing the circuit choose such that the current leads the applied voltage by 30.0? All other values in the circuit stay the same. Solve the following equation or the inductive reactance : XL reactance and capacitive reactance into the following expression: X + R tan dj Substitute expressions or induct e Solve for L: Substitute the given values mam (377 s-1) L (377s-1)(3.10 × 10-6F) + (445 ) tan(-30.0)Explanation / Answer
A)
w=2f = 2*3.14*60 = 377 s^-1
XL = wL = (377)(1.35) = 508.95 ohm
XC= 1/wC = 1/(377*3.10*10^-6) = 855.65 ohm
Z = sqrt[R^2 + XL ^2 + XC ^2) = sqrt (445^2 + 508.95^2 +855.65^2) = 1090.5 ohm
B) Imax = V/Z = 150/1090.5 = 0.138 A
C) =tan^-1[(508.94*855.67)/(445)] = 89.94 deg
D)VR = Imax*R = (0.27)(445) = 120.15 V
VL = Imax* XL = (0.27)( 508.94) = 137.4 V
VC = Imax* XC = (0.27)( 855.67) = 231.0.3 V
E) L = (1/377)*[(1/(377*3.10*10^-6)) + (445tan^-1(89.94))] = 0.455 H
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