For a source which has load has the following characteristics: • Vrms = 125 20° V • P= 400 W • pf = 0.925 lagging What is the rms current and reactive power?
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A: The detailed solution is provided below.
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A: Circuit diagram is given as, Comparing with standard equation.
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Q: The apparent power entering a certain load Z is 250 VA at a power factor of 0.8 leading. Q1) If the…
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A: The given circuit is:
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Q: Puestion Thap The apparent power entering certain load z is 250VA at 퍼 the power factor of RMS…
A: Power entering the load is 250 VA RMS input voltage is 125 V. So,by formula P=VrmsIrms Putting the…
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Q: Electric Circuit Analysis 2020/2021 Dr. Yaseen H. Tahir Example: Sketch the phasor diagrams of the…
A: The solution can be achieved as follows.
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A: Circuit diagram is given as,
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A: Given data The rms voltage is Vrms = 140 V Frequency is f = 50 Hz Resistance is R = 100 ohm…
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- An industrial load consisting of a bank of induction motors consumes 50 kW at a power factor of 0.8 lagging from a 220-V,60-Hz, single-phase source. By placing a bank of capacitors in parallel with the load, the resultant power factor is to be raised to 0.95 lagging. Find the net capacitance of the capacitor bank in F that is required.The rms value of v(t)=Vmaxcos(t+) is given by a. Vmax b. Vmax/2 c. 2Vmax d. 2VmaxShown in the figure below is an "RL" circuit drive by an AC power source. The AC power source has an RMS voltage of Vps (RMS) = 9.84 Volts and is running at a frequency of f = 8.585e+04 Hz. The resistor has a resistance of R = 2170 and the inductor has an inductance of L = 3.54e-03 Henries. Vps R ww Write the FORMULA for the total impedance of the circuit Ztot = Determine the numerical value of Ztot = 2890.5 Determine the numerical value of $z= = 41 Determine the current through the circuit: • I(PEAK) = 4.81E-3 • I(RMS) = 3.404E-3 Determine the voltage across the resistor: • VR(PEAK) = 7.387 • VR(RMS) = 5.22 ✔✔ Amps ✔Amps Write the FORMULA for the phase of the total impedance of the circuit z... = | tan-1 2701 R x Volts X Volts Determine the voltage across the inductor: • VL(PEAK) = 9.184 • VL(RMS) = 6.49 ✔ Volts Volts L R²+ WL- ✔ degrees 2 If a second circuit were connected in parallel with the inductor, this circuit would be considered as: O a low-pass filter O a capacitive switcher…
- Shown in the figure below is an "RC" circuit drive by an AC power source. The AC power source has an RMS voltage of Vps (RMS) = 11.58 Volts and is running at a frequency of f = 1.326e+04 Hz. resistor has a resistance of R = 3750 2 and the capacitor has an capacitance of C = 3.17e-09 Farads. Vps R ww Write the FORMULA for the total impedance of the circuit Ztot = Write the FORMULA for the phase of the total impedance of the circuit tot = Determine the numerical value of Ztot = Determine the numerical value of Pztot= Determine the current through the circuit: • I(PEAK) = • I(RMS) = Determine the voltage across the resistor: Amps Amps S2 C degreeswhat is the reactive power absorbed by the load ZL in VAR? Assume source is in rmsDetermine the rms current in the figure for Vrms = 16 V and C = 74µF V f=10 kHz
- 1) the RMS value of the current is? /2) the peak value of the voltage is?/3) the value of the reactive power is?In an experiment, the function generator is adjusted to generate a square voltage at a certain frequency. This voltage is displayed on an oscilloscope and the reading is 16 V peak-to-peak. The rms value of this voltage is .What will be the rms value of this voltage if the frequency is doubled? Select one: O 16 Vrms 32 Vrms O 8 Vrms . 8 Vrms ..... O 5.657 Vrms 5.657 Vrms ....... O 8 Vrms . 16 Vrms O 16 Vrms 8 Vrms ......The rms value of the resultant current in a wire carries a dc current of 10 A and a sinusoidal alternating current of peak of value 20 A is-
- ESSAY: Describe what happens to the power in this RLC series circuit as the capacitance decreases. Explain in general terms why the observed change should occur. Understand the effects of basic electrical components on the average power of the system and the concept of power factor. Differentiate between average power and real power.7 Find the average and effective values of voltage for sinusoidal waveform shown in Figure. Vm = 100 V 2x x14 9714 [Ans: 27.17 V, 47. 67 V]In an experiment, the function generator is adjusted to generate a square voltage at a certain frequency. This voltage is displayed on an oscilloscope and the reading is 10 V peak-to-peak. The rms value of this voltage is .What will be the rms value of this voltage if the frequency is reduced to the half value? 3.536 Vrms . 3.536 Vrms O 5 Vrms . 1.7677 Vrms O 3.536 Vrms. 1.7677 Vrms O 5 Vrms . 5 Vrms ...... O 10 Vrms . 10 Vrms ......