(16) Vc $ !! Vs (t) = V₂ Sin wt V₁ (t) = vc sin (wt + $c) RC option. Vs (t) . ->> V₁ = Vs Vc = 0.573 · V₁ = 0·4672 V₁ V₁ = VG B²=3K R 0=-22.5° 10= =639 = -26.56° 0.5%š, 0=45° 1² ] Ve(t) +
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- VR. VR2 R2 R. Ve YL AIs Ic 6ト0°Y f: 50 H2 2mfC 15mH3 L Find= Is, Ic, I-, Vai, Vaz, Ye, VL (ANSWER IN RMS)For the circuit of the following figure, if R=14Q, Zc=-J26 Q, and ZL=J10 (Q, find the value of the current through ZL. Zc V1 1.5 /24° V ZL -) 3-41° V Oa. 173.759 L -81.69 ° mA Ob. 175.759L-71.69 ° mA Oc. 175.759L-81.69 ° mA Od. 173.759 L -71.69 ° mAQUESTION 3 An inductor, a capacitor, and a resistor are connected in parallel with a function generator. The inductor draws 100mA RMS, the capacitor draws 150mA RMS, and the resistor draws 200mA RMS. Find the magnitude of the total current. O 450mA RMS O 544mA RMS O 352mA RMS O 206mA RMS
- Evaluate 8/10° +62-20° 9/80° -4/50° 0.50 +j2.24 -0.50-j2.24 O 50-j224 O -50-j224 and express the result in rectangular form.4. A capacitor now is connected in parallel with R1 to reduce the ripple. Find the value of the capacitance needed to make the following ripple index less than 5% through PSpice simulation study. V -×100% -V o,min 0,max rpp RFP = -x100% = V dc dc D1 Dbreak R1 V1 1k VAMPL = 169.71 FREQ = 50 VoEXAMPLE 2.0 FIND ALL THE ROOTS OF x. ((3275°- j3794)2- 4e [(12- 42° + el30")• – (2 – j5)*]) (7/15° + 249°)* – (j73479 – j723)3 + [(3 - j2)(64°)]2
- 3) Given the circuit with an AC sinusoidal current source and an AC sinusoidal voltage source: ZR1 $60k0 <0 deg V₁" I₂" IR1 V1- Zc 85 kohms <-90 deg Ic VR2™ VL™ VIC ZR1 = 60 k < 0 degrees Zc = 85 k0 <-90 degrees ZR2 = 20 kn 0 degrees Z₁ = 50 kn < 90 degrees www ZR2 20k0 <0 deg Isa 1.2 mA <0 deg 12- + Esb All sinusoidal complex phasor currents and voltages are peak phasors. Isa 1.2 mA <0 degrees represents a current source peak phasor Esb = 80 V < -150 degrees represents a voltage source peak phasor The current source and voltage source have angular velocity w 2000 rad/s At the previously stated angular velocity, the schematic shows the element complex impedances as follows: ZL 50 kohms < 90 deg 80 V<-150 deg Use the Superposition theorem and circuit analysis to find all of the circuit peak complex phasor voltages and currents requested below:Three impedances, Z1 =10020.Q, Z2 =63.2518.43.2 and Z3 =1002-90•Q are connected in star. Convert the star to an equivalent delta connection. And draw the star network and also the equivalent of delta connectionX 23) WhatsApp Meet hpd-sdbt-qho yl la- Chrome a web.whatsapp.com YouTube Gmal M Course Jalllbi th Select one: O True O False the machine is generating 355 V at 60 H The e.m.f generated by a synchronous generator at a frequency of 40H2 and field current of 2.7A is field current of 3A. ENG 4 ProBook 4340 SISP POWAN
- Prob. I and Vc For the following RMS phasor circuit, a coworker has determined values for 50 Ohms 50 Ohms i,- 225 2253.13° = 190.92 + 254.56 VRMS R1 + R2 Vc = 237.17/26.565° = 212.13+ 106.07 VRMS 100 2 -3Z98.13° =-0.4243 + J2.970 ARMS C -j60 Ohms For this problem provide at least 3 digits of precision for the answers. The results for the following calculations are to be answered in - a) Find the current I b) Determine the real (P) and reactive(Q) power DELIVERED by the source c) Determine the real (P) and reactive (Q) power ABSORBED by each impedance ( resistors R1 and R2, inductor and capacitor). The values for P and Q can be positive or negative. Remember S =P+jQ, so do not include the j with your answer for Q. Each of these answers will require two parts. For the phasor a magnitude and angle will be needed. For the powers, The real power P and the reactive power Q are required.I PERFORm THE FOLLOWING OPERATIONS. (EXPRESS THE ANSWERS IN RECTANGULAR PORM, & Stow THE COMPLET SOLUTIONS). a) (4+j3) C6-je). (34j3) -(2+j3) c) (6 /20°.) (201-46")sjs) 2 30° 2. 860 140 Q LO) +J00 tj400) 3tj9Two impedances are connected in parallel across a 220-volt source. If Z1= 4 cis (30) and Z2 = 5 cis (60) ohms. Find the equivalent admittance ( Siemen ) of the circuit. O 0.435 cis(- 43.3) O 2.3 cis (43.3) O 0.835 cis (-53.3) O 0.835cis (53.3)