Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN: 9780133923605
Author: Robert L. Boylestad
Publisher: PEARSON
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- -write legibly-show complete solutionarrow_forwardA voltage wave of 250 kHz has a Vmax-5 V and Vmin=-35 V. The duty cycle of this wave is 0.875. This voltage waveform has been applied across a 2.5 µH inductor for a very long time so that steady state has been achieved. Diagrams of the wave parameters and circuit are shown in this figure: L What is the average voltage of this wave? Average voltage = 35 V Voltage (arb, unit) 1.25 Minimum current: -3.5€ Amps 1 0.75 0.5 0.25 0 4.25 05 Have you integrated the wave over full cycle? 4.75 4 -1.25 Vmax 0.5 Vmin What is the average current flowing through the inductor? Average current = 3.56 A 15 Period/s On average, there is no power consumed by the inductor. Have you considered this? What are the maximum and minimum currents that flow through the inductor? Maximum current: 0.50 Amps D= 25 TH TH Have you used the relationship for current and voltage in an inductor? Have you considered the slope of the current wave?arrow_forward:Converting the following sine function sin(3t - 76°) 184- :to a cosine function will result in the following A cos(3t + BO) ?What is the value of Aarrow_forward
- Show step by step solution and write legibly.arrow_forwardR = 250 and the sinusoidal alternating emf revice operates at Em = 40 v and fo = 55 Hz a) What is VRct) and amplitude Ve of UR(E) ? b) and the currentarrow_forward1- The wavelength of a 50Hz sinusoidal waveform is A-40ms B-100ms C-50ms D-25ms 2- The intensity of the signal waveform is A-regular waveform B-Period C-Wavelength D- Amplitude 3- The average value of an AC waveform is equal to A- 0.637* Vmax B-0.707 * Vmax C-1/√2 *Vmax - 4 Radian is Equal to A-Degree x Π 180 Π π 180 D-1.11 B-Radian x C- Degree x D- π 180arrow_forward
- HW1//: find the amplitude, period, and frequency for the following waveform. a. 20 sin 377t b. 5 sin 754t c. 10° sin 10,000r d. 0.001 sin 942t f. ) sin 6.283t e. -7.6 sin 43.6t HW/2 Find the phase relationship between the waveforms of each set: a. v= 4 sin(wt + 50°) i = 6 sin(wt + 40°) 80°) i = 5 x 10- sin(wt 60°) i = 0.1 sin(wt + 20°) b. v= 25 sin(wt 10°) c. v= 0.2 sin(wt d. v= 200 sin(wt 210°) i- 25 sin(wwt – 60°) HW/3 Repeat Problem 29 for the following sets: a. v= 2 cos(wr - 30°) i = 5 sin(wt + 60°) c. v= -4 cos(wt + 90°) i = -2 sin(wt + 10°) b. v -1 sin(wr + 20°) i - 10 sin(wwt - 70°)arrow_forwarddetermine which waveform is lagging in each of the following pairs. (d) -sin 5t, cos (5t+2)(e) sin 5t + cos (5t-45) here's the continuation. thank you again.arrow_forwardConsider the sinusoidal wave below. What is the frequency of the wave? 20 V - 15 V 10 V 5 V O V .t (μs) 10 20 30 -5 V - -10 V - -15 V – -20 V - tearrow_forward
- The pictuere shows double-beam c.r.o. waveform traces. Find the ff:• Amplitude of P• Peak-to-peak value of Q• Periodic time of both waveforms• Frequency of both waveforms• R.m.s. value of P• R.m.s. value of Q• Phase displacement of waveform Q relativeto waveform Parrow_forwardThe RC circuit of the figure is fed from the constant magnitude variable frequency sinusoidal voltage sourceVin. At 100HZ, the R and L element each as voltage drop Urms · If the frequency of the source is changed to 50HZ, then voltage drop across R is. www Vin 000arrow_forwardA series RL circuit connected to an AC voltage sinusoidal source of 100 V peak voltage and 1 kHz frequency. If R = 1 ohm and L = 1 mH, then... a) The rms current through the circuit leads the source voltage b) The rms current through the circuit is 11.114 A 279.043° c) The rms current through the circuit is 15.718 A z-80.96° O d) The rms current through the circuit is 15.718 A 270.56°arrow_forward
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