Please solve All parts, I will rate your answer. Thanks (5%) Problem 14: An RLC
ID: 1410069 • Letter: P
Question
Please solve All parts, I will rate your answer. Thanks
(5%) Problem 14: An RLC series circuit has a 2.5 resistor, a 85 LH inductor, and a 75 LLF capacitor. 17% Part (a) Find the circuit's impedance, in ohms, at 120 Hz 17% Part (b) Find the circuit's impedance, in ohms, at 2.5 kHz 17% Part (c) If voltage the source supplies an rms voltage of 5.58 what is the circuit's rms current, in amperes, at a frequency of 120 Hz? Grade Summary 2.232 ms 1 Deductions 18% Potential 82% 7 8 9 HOME Submissions Attempts remaining: 2 (6% per attempt atan0 acotana sinho 1 2 3 detailed view cos ho tanho cotanho 0 Degrees O Radians VO BACKSPACE DEL CLEAR Submit Hint Feedback I give up! Hints: 2 for a 0% deduction. Hints remaining: 0 Feedback: 2 for a 0% deduction Does ohm's law work in an AC circuit? The current depends on both the resistance and the reactant parts of It does ifyou use the mas current and potential with the impedance the circuit which you found in fiom part (a). A 17% Part (d) If the voltage source supplies an ms voltage of 5.58 what is the circuit's rms current, in ampere at a frequency of 2.5 s, kHz? A 17% Part (e) What is the resonant frequency in kilohertz, of the circuit A 17% Part (f What is the rms current, L in amperes, at resonanceExplanation / Answer
R = 2.5ohm
L = 120e-3H
C = 75e-6F
1) w = 2*pi*f1 = 2*3.14*120 = 753.6
Xl = wL = 753.6 * 120e-3 = 90.4
Xc = 1/wc = 1/(753.6*75e-6) = 17.69
Z = sqrt{ R^2 + (Xl-Xc)^2 } = sqrt( 2.5^2 + (90.4-17.69)^2 } = 72.75ohm
B) w = 2*pi*f2 = = 2*3.14*2.5e3 = 15.7e3
Xl = wL = 15.7e3 * 120e-3 = 1884
Xc = 1/wc = 1/(15.7e3*75e-6) = 0.85
Z = sqrt{ R^2 + (Xl-Xc)^2 } = sqrt( 2.5^2 + ((1884-0.85)^2 } = 1883.15
C) V = IZ
I = V/Z = 5.58/72.75 = 0.076A
D) V = IZ
I = V/Z = 5.58/1883.15 = 0.03mA
E)for resonance
Xl = Xc
wL = 1/wC
w^2 = 1/LC = 1/ 120e-3 * 75e-6
w = 333.33
2 * pi * f = 333.33
f = 333.33/2pi = 53.08 = 0.053kHz
F) Z = R at resonance
V = I Z
I = V/Z = 5.58/2.7 = 2.07A