Final answers: 8a: The RLC circuit has R 1500 2, L 1.5 H, C 3 Au F, erms 60 V an
ID: 1557190 • Letter: F
Question
Final answers:
8a: The RLC circuit has R 1500 2, L 1.5 H, C 3 Au F, erms 60 V and w 750 rad/s. Determine the impedance, the rms current and the phase angle. Keep the current to three significant digits (otherwise there will be rounding difficulties later in the problem). b: Write I (t) for the circuit. c: Determine the maximum voltage across each circuit element (battery, resistor, inductor and capacitor) and use that information to graph V(t) for each circuit element all on the same graph d: Determine the voltage across each circuit element at t 1 s and verify that they add up to the AC battery voltage at that time. e: Determine the power of all four circuit elements at t 1 s. Verify that energy is being conserved. Which circuit elements are gaining energy? Which are supplying energy? And which are effectively removing energy from the circuit? 00000Explanation / Answer
capacitice rectance Xc = 1/(wC) = 1/(750*3*10^-6) = 444.44 ohm
inductive reactance XL = wL = 750*1.5 = 1125 ohm
imedance Z = sqrt(R^2 + (Xc-XL)^2)
Z = sqrt(1500^2+(444.44-1125)^2)
Z = 1647.2 ohm
Irms = Erms/Z = Erms/Z = (60)/1647.2 = 0.0364 A
Imax = Irms*sqrt2 = 0.0515 A
phase angle phi = tan^-1((Xc-XL)/R)
phi = tan^-1((444.44-1125)/1500) = -0.426
===========
(b)
I(t) = Imax*sin(wt + phi)
I(t) = 0.0515*sin(750t - 0.426)
=================
(c)
VB = Erms*sqrt = 60*sqrt2 = 84.9 V
VR = Imax*R = 0.0515*1500 = 77.3 V
VL = Imax*XL = 0.0515*1125 = 57.9 V
Vc = Imax*Xc = 0.0515*444.44 = 22.9 V
=======================
at time t = 1s
I = 0.0515*(sin750*1 - 0.416) and
VB = Emax*sin(wt)
VB = 84.9*sin(750)
VB = 63.2
VR = I*R = 0.0515*sin(750-0.426)*1500 = 73.7
voltage across L
VL = VLmax*cos(wt+phi)
at t = 1
VL = 57.9*cos(750-0.426) = -17.4 V
voltage across capacitor
Vc = -Vcmax*cos(wt + phi )
at t = 1s
Vc = -22.9*cos(750 - 0.426 ) = 6.9 V
==================
(e)
PB = VB*I = 63.2*0.0515*sin((750*1) - 0.426) = 3.11 W
PR = VR*I = 73.7*0.0515*sin((750*1) - 0.426) = 3.62 W
PL = VL*I = -17.4*0.0515*sin((750*1) - 0.426) = -0.85 W
Pc = Vc*I = 6.9*0.0515*sin((750*1) - 0.426) = 0.34 W
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