Question

o 23 points I 16 23.12P 105 An RLC senes orcuit has a 180 ? resistor and a 25.0 mH inducto r. At 8000 Hz, the phase angle is 45.0 (a) What is the impedance (in ohms)? 12545584 ? (b) Find the minimum possible capacitance (in nanofarads) of the circuit. 13,71 nf (c) If V 408 V is applied, what is the average power (in watts) supplied: v W ?Reading

Answer 1

Given an RLC series circuit has 180 ohm resistor and 25 mH inductor At 8000 Hz phase angle is 45^degrtee R = 180 ohm L = 25 mH t = 8000Hz theta = 45 degree c impedance z = R + J(x_L - x_c) z = squareroot R^2 + (X_L - X_C)^2 from phase angle tan phi = x_L - x_C/R X_L - X_C = tan phi times R = tan 45 times 180 X_L - X_C = 180 so z = squareroot (180)^2 + (180)^2 z = 254.55 ohm X_L - X_C = 180 X_C = X_L - 180 (2 times pi 8000 times 25 times 10^- 3) - 180 X_c = 1076 X_C = 1/wc c = 1/2 pi f X_C = 1/2 times 3.14 times 8000 times 1076 \r\n Therefore c = 18 nF avg power = r^2/R times cos theta cos theta = R/z = 180/254.5 = 0.7072 average power = (408)^2/180