Question

17.34 Using Fig. 17.70, design a problem to help other end students better understand circuit responses to a Fourier series. vt) (+) C + vo(t) Figure 17.70 For Prob. 17.34.

Solve for Vo, EXPLAIN EVERY MAJOR STEP

Answer 1

Page! RLC Circuit 1 Scule R - m i um ct.ct) VoCE) romana The above circuit is a series RLC ckt having la single loop. Let R- Resistance L- Inductance - c- capacitance Let u(t) be the source voltage [u(t)= I'm SimCwt) up-be the vtg acros reis too R. VL - be the nuity auos industol L Ve=Volt)- be the utg acous capauts c Het ict) = Imax bin (wt) be the current in the loop - - - - - - Noltage response achos each elementi The voltage across a pure revista CRI, NR is in phare with carrent Î UR Resistame CIn phare)

LIST page 2 rure The voltage rachols inductos, leads by ao current 90° f I Inductance . The mustare voltage pure capacitor, the current across e lags capacitance ... and Vc are 180 out of phare of opposite to each other Apply KuL to the circuit, VCE) - VR-Ve-Vc 0 VE) - IR - Ldi / azo --V(t) = IR+LA+ end

Page 3 Phasos diagram of sories RLC cirant! The result uits is obtained by Vectors UL & Vc & then the sum adding the to vectos URL NL-Vc T----- Vet) zuet tao poder From above phase diagram on right hand side the voltage vectore produced rectangulas triangle chawing ucts as hypotenure : Using Pythagoras theshem we get; VCt)² = UR ²+ CULP-ucj? Aucts = Jup² + CUL-VEJ?

1 Page 4 رها و لا = لا ،، from eqati equation o . (UL-vc)² = u(+²-wp² VL-VC = NUC£3?VR? Ve = ve- Vucts* -up? -Voltz Ve > VL- Juces?=vx? No Hage avors each component; VR = IR sun Ewtto) = IR V = IXL bis Cwt +90) = I jwL Ve=I Xcsin (wt-96) = Į عمان of capacitive Let XL & Xc be the inductive reactances XL = 5 WL = 2TfL . Xc = the atte {"W = 277]

Page of NR ZIR V=IXL Vca IXc = volt) substituting in Pythagoras equ ® u(t)= VER)²+(IX L-IX c)? VCE) > I VR?+cX2- Xcs? ult) = I Z) there z= JR ² + (x2-xg Inapedance of RLC ckt: Impedance triangle of RLC series (kt ; e XL Z 17X95XL-YC . Z= JR? + x² z=FR²+ (XL-XJ?

Page 6 Ref С€ол® The phone angles : cond = 1 sin q = * -t Hypotenue sino = opposite Hypotamme opposite tano = tang = XL-Xc 1 R sino - caso Adjacenti seme . Ne= volt) from ear ② Volt>= Vc = N2-vu (tuh_up? Nolta Ixi-Vutja GR? Volt) = I x L - FE z ²-ERR Volt)= IXL - I √ 2² - R2 Volt)= I (XL- V 22 R²)