VI 1 120 <20° VRMS ZTL 3 – j10 Ω ZTL ZTL 3 + j40 Ω Z a Z Z
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- A non-sinusoidal current source (iT=2+1.5sin1000t+ 1cos30o0t- 0.5sin4000t) A is applied to the * ?circuit shown. Find the total power İL iT ic -J9N J4Q 39 W O 20 W O 55 W 28 W )!( هذا السؤال مطلوبFor the circuit of the following figure, if R=26 0, Zc=-J26 0, and ZL=J30 0, and if the value of the nodal voltage V is 1.743-27 47° V, find the value of the reactive power supplied by the 3L-41 V voltage source. Ze R 15/24 V E ZL 341 V a 0.0377 VAR b.00736 VAR C0.0471 VAR d. 0.0589 VAR leR. Ro b. load the circuit in the a lo ad F ormed by figure, there i's in seric s Ro and the in ductance Lo c onnec tin g the resistor bet ween terminals a ond b The uoldage amplitude slt) the OS= 0'degree period T= 200 micro- 's JE 150 volt phose ongle Since R- 300 S and c = 1 pf seconds load p ase on gle OL= 30 c ons ton t, find the R. a.)if the lo a d o draw the max imum a llow the an dLo Jalues that au er a ge Power from the circuit- the auerage 4he load- Power c alculate
- DC-AC Circuits Express R1 using Rx, Ry, and Rz when performing wye to delta conversion R3 Rx Rz R2 R1 Ry O A. (Rx+Ry)(Ry+Rz)(Rx+Rz) Rx В. RyRz Rx+Ry+Rz O C. Ry+Rz RXRYRZ O D. RXRYRZ Ry+Rz OE. RxRy+RyRz+RXRZ RxIf the voltage Vs = 800 mV, then the current flowing in the primary winding Ip, is: 1:8 -j202 Select one: a. Ip =40<0° mA O b. Ip =320<-90° mA O c. Ip = 0.320<90° A %3D O d. None of these ll4. Solve for I, I2, Iz and I4 in the ckt below using Cramer's Rule. ISMAT169NUMI 252 METHO 169NUMERICALIM SANDANALYSISM THOANDANALY RICLMET 169NUICA MAT169NUMERICA 幸 DDSANDA MAT169NUMERI THODSANDANALYSSMAT ANALYSISMAT 9NUM METHODSA 12V (1) I2 1 YSISMAT169N CALMETHODSANDA 351 ETHO SISMATI6 OSANDANAL RICALMETHODSAND NDAN SAND ANALYSISMAT 丰 RICAM MERICALME NDANALYSISM NUMERICAL METHOSAN SAND ALYSI METH JALYS 69NUMERI ISN I) I3 ERICALM ooDANAL TRICALMET NDANASA SMAT169NUAALYSISMAT169NUMERICALMETHODSA AT 169NU SANDA MERICLMETHODSANDA
- Determine the nodal voltage V1 if l=4L25° A, and R=4 2, for the following figure: -j2 Q 20/-90° V j12 Ref. Oa. 26.56 L 72.507 V Ob. 22.56 L 68.507 V Cc. 28.56 L 74.507 V Od. 24.56 L 70.507 V When the voltage source is compared with the current source which of the fol owing is true Ca. The current source angle leads the voitage source ang e by 135 Ob. The current source angle ags the vo tage source ang e by115 Oc. The current source angle ags the voltage sourde angle by 135 Co. The current source ang e leads the vo tage source angle by115 wwConsider the circuit below: Simplify the circuit and write the super position circuits for I1, 12, and 13 2 AZ45° 1220 v 4 AZ0° O 20 3 VZ-90°VI Find Is, In, Ia, Is, Is, V. ,Vz R4 Rs V. Is 43 RS Ia Y Ra R2 5A VR. R. Ve YL Is Ic IL 1,07 01 f: 50 H2 2mfC Find Is, Ic, I, Vai, VRz, Ye, VL (ANSWER IN RMS)
- Vs= 40 V Z1 = 1 0hm Z2 = j2 Ohm Z3 = 2 Ohm Z4 = -j4 Ohm Z5 = 1 Ohm Z6 =j2 Ohm Z7 = 2 Ohm Z8 = -j4 Ohm Vs + 1 Find Vo. * Z1 O (2.41 +j 0.49) V O (1.41-j1.51) V O (36.12-j18.83) V O (1.31 -j41.66) V O (35.78-j17.68) V Z2 Z3 Z4 Z5 www Z6 m Z7 Z8 o+ Vo 10Application: Construct the circuit in figure. R= 2,2KN, L = 1 mH, and C= 10 nF. L Necesssary Formulae: 1 Xc 2.π.f.C X.-2. π.f. L By using the formulea perform the calculations of Xc and XL then fill the table. f (Hz) Xc XL 50000 60000 70000 80000 90000 100000 110000 120000 130000 140000"Three impedeances, A = 6+j3 ohms, B = 5+j6 ohms and C=2-j8 ohms are connected in parallel. Determine the equivalent impedance Zab" %3D O 3.94 j0.776 ohms O 3.94tj0.776 ohms O 4.93+j7.76 ohms none