Question

RLC circuit (2) 5 of 10 > Review | Constants Learning Goal: To calculate and use complex impedance. A 25.0-9 resistor, 50 mH inductor, and 200 uF capacitor are connected as shown below to an AC voltage source with amplitude 100 V and w=200 5-1. R mat L mm Part A - Impedance of R-C Find the complex impedance of the R-C combination, in . Format: Since Z = R + Xj, your answer consists of two values: the resistive part Rand the reactance X. Enter them separated by a comma: R X View Available Hint(s) 25.0,-25.0 Submit Previous Answers Correct

Need help with B to D, Thank you!!

Answer 1

(A) The impedance of R-L circuit is expressed as follows: Z= R + joc 1 = (25.0 22)+ 10F j|200 UF (200 s-4)! 1 uF 100; = (25.0 2)+ (32)2 = (25.0 2)+(-25.0); Therefore, the complex impedance is (25.02)+(-25.0); and the answer is 25.0, -25.0 (B) The impedance of R-L circuit is expressed as follows: Z==(25.0 2)+(-25.0); The impedance of inductor is expressed as follows: Zo = jol 10 H =;(200 s-4) 50 mH 1 mH = 10j

Z eg = The total impedance for parallel combination is expressed as follows: (Z)(2) (2)+(2) ((25.0 2)+(-25.0);)(10j) ((25.02)+(-25.0);)+(10) 250j – 250( ;) (25-15j) 250j + 250 (25-15j) = = Multiply and divide by complex conjugate of (25–15j). 250j +250 (25+15j) Zeo (25-15j) (25+15j) (6250j)+(15j)(250j)+(6250)+(250)(15j) (625)-(225;) (10000j)+(3750j)+(6250) (625 +225) (6250-3750)+(10000) (625 +225) = eq Therefore, (6250–3750)+(10000j) Z (625 +225) 6250-3750) (10000) + ] (625+ 225) 625 +225) = 2.941+11.765j = 2.94+11.8j =

Therefore, the answer is 2.94 , 11.8 (C) The total current is expressed as follows: V I= Zeg (100) 20° 11.765 + 2.941 (100) 20° II (V(2.941)° +(11.765)") 275.96 = 8.246 AZ-75.96° -8.25 AZ-75.96° Therefore, the current by the AC source is 8.25 A (D) The amplitude of current through the inductor is expressed as follows: I'= |X2 (100 V) 20° V10-290° = 10 AL-90° Therefore, the current through the inductor is 10 A