9.5 Determine the instantaneous power supplied by the source i the circuit in Fig. P9.5. 12/60°v( + 4 Ω j2 Ω -j2 Ω ww ΖΩ
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- 9.31 Determine the impedance Z, for maximum average power transfer and the value of the maximum power transferred to Z for the circuit shown in Fig. P9.31. j1 Ω 6/0° A Figure P9.31 www 1Ω ZLIf the rms phasor of a voltage is given by V=12060 volts, then the corresponding v(t) is given by (a) 1202cos(t+60) (b) 120cos(t+60) (c) 1202sin(t+60)Given a sinusoidal voltage source in series with R,L, and C. The sinusoidal frequency is 600 Hz. R=2 Ohms, L=5 milliHenry, C=200 microFarad, V=70sin(wt) volts. Determine the watts and VARs for the capacitor, resistor, inductor, and source. Also find power factor. Given R1=9 Ohms, R2=1 Ohms, L= 3 milliHenries, and a sinusoidal voltage source of 5 volts amplitude and 60 Hz freqency all in series. Determine the voltage across R1: magnitude (volts), frequency (Hz), and phase (degrees).
- A load absorbs 72044 W at a lagging power factor of pf = 0.67. If the current flowing through the load is Ix = 0.100 A−rms, determine the impedance of the load.Two alternating voltages are given by v1 = 15sinωt volts and v2 = 25 sin (ωt – π/6) volts. Plot both functions on the same axes and hence determine a sinusoidal expression for the resultant vr = v1 + v2. Check your answer using an analytical method. Your manager has asked you to analyse the variation in in results between the two methodsA load with a voltage amplitude of 50 Hz 300 V rms at its terminals consumes 28.98 kW at a back power factor of 0.69. If a 225.87 microfarad capacitor is connected parallel to the terminals of this load, how much will the amplitude of the current drawn from the line decrease?
- Read A 70-Vac source has the following waveform. Determine:a. Vpkb. Vpk-pkc. Vrmsd. Periode. Frequencyf. Angular Velocityg. Phase Angleh. equation of the waveform (in time domain)i. the instantaneous voltage when t = 120 msj. the angle (1st occurrence) after t = 0 when the voltage is +80 Vk. the time (2nd occurrence) after t = 0 when the voltage is –10 VAnswers:a. 98.9949 Vb. 197.9899 Vc. 70 Vd. 100 mse. 10 Hzf. 20π rad/secg. 125.0876°h. v(t) = 98.9949sin(20πt + 125.0876°)i. -29.0880 Vj. 0.99959°k. 63.643 ms Give the solution for J and K ONLY ans (0.99959 degrees and 63.643 ms respectively)A linear circuit supplied by a nonsinusoidal source is shown. Find the average power absorbed by the circuit if the source voltage is expressed as vs(t)=100+50 cos(120лt)+25 cos(240πt)+15 cos(360πt).The impedance ZL in the circuit is adjusted for maximum average power transfer to ZL. The internal impedance of the sinusoidal voltage source is 4+j7 Ω. 1. a) What is the maximum average power delivered to ZL?2. b) What percentage of the average power delivered to the lineartransformer is delivered to ZL?
- The voltage between the ends of a load and the current through it are given by: v (t) = 20 + 60 cos100ti (t) = 1 - 0.5 sin100t Find:The rms values of the voltage and current and the average power dissipated in the load.The 9 Ω resistor in the circuit is replaced with a variable impedance Zo. Assume Zo is adjusted for maximum average power transfer to Zo. 1. a) What is the maximum average power that can be delivered toZo?2. b) What is the average power developed by the ideal voltage sourcewhen maximum average power is delivered to Zo?3. c) Choose single components from Appendix H to form animpedance that dissipates average power closest to the value in part(a). Assume the source frequency is 60 Hz.Determine the rms values of each of the following (a) v(t) = 100 sin(ωt) V Vrms = 70.7V (b) i(t) = 8 sin(377t) A Irms = 5.656V(c) v(t) = 40 sin(ωt + 40o) V Vrms = 28.28V (d) i(t) = 120 cos(ωt) mA Irms = 84.84A Can you double check if my answer is correct?