What is the frequency of a sine wave of voltage which has a 45V peak and which continuously increases from 0V at t = 0s to 24V at t = 46.2ms?
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- The peak to peak distance of a sinusoidal waveform displayed on a C.R.O. screen is 9 cm and the 'volts/cm' switch is on 25 V/cm. The peak voltage is given by O 2.78 V O 159.07 V O 112.50 V O 225.00 VA sinusoidal alternating current of frequency 25 Hz has a maximum value of 100 A. How long will it take for the current to attain values of 20, and 100 A?A pure sinusoidal current is being rectified. For the given maximum value of half wave rectified current is 50 A, then the rms value of full wave rectification will be 50 (a) A (b) 100 - A TC (c) 100 A (d)70.7 A
- The maximum value of sinusoidal voltage is given by Vm= 1 V, What is the value of Peak to Peak voltage?An alternating current has the frequency of 62 Hz, then the value of angular velocity is The peak to peak distance of a sinusoidal waveform displayed on a C.R.O. screen is 9 cm and the 'volts/cm' switch is on 21 V/cm. The peak voltage is given by 133.62 V O 94.50 V O 2.33 V 189.00 VA pure sinusoidal current is being rectified. For the given maximum value of half wave rectified current is 50 A, then the rms value of full wave rectification will be 50 (a) 100 (b) - - (c) 100 A (d) 70.7 A
- What is the ripple factor of a DC signal having ripple voltage of 2 V and a DC voltage of 50 V? Answer in percentage % O 8% O 1.2% O 2.83% O 4.0%The sinusoidal current waveform having the R.M.S value of 20 A, then the maximum value is Answer and unitA double beam oscilloscope displays 2 sine waveforms A and B. The time/cm switch is on 150µs/cm and the volt/cm switch on 5 V/cm. The width of each complete cycle is 8 cm for both the waveforms. The height of waveforms A and B are 5 cm and 4 cm respectively. Determine their a) frequency, b) phase difference, and c) peak and d) r.m.s value of voltages. The difference between the two waveforms is 0.7 cm.
- A double beam oscilloscope displays 2sine waveforms A and B. The time/cm switch is on 100us/cm and the volt/cm switch on 2V/cm. The width of each complete cycle is 5cm for both the waveforms. The height of waveforms A and B are 2cm and 2.5cm respectively. Determine their a) frequency, b) phase difference, and c) peak and r.m.s value of voltages. The difference between the two waveforms is 0.5 cm.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-A 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?