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The following waveform was derived from an oscilloscope with the given settings:
| CH1 (Yellow) Settings: | CH2 (Blue) Settings: |
| Volts/Div: 50V | Volts/Div: 100V |
| Time/Div: 5ms | Time/Div: 5ms |
| Probe Setting: X1 | Probe Setting: X1 |
Determine the following:
a. CH1 Peak Voltage
b. CH1 RMS Voltage
c. CH1 Period
d. CH1 Frequency
Step by step
Solved in 2 steps with 1 images
- -) Below are two voltage patterns measured from an oscilloscope. The voltage stage of the CHI probe is 5 V / div, the CH2 probe voltage stage is 10 V / div and the time axis is 2ms / div. According to this information, a) What are the peak to peak and rms voltage values of the voltages in CHI and CH2? b) What is the frequency of the voltages in Hz? c) How many degrees is the phase difference between the two voltages, which voltage is leading?A pure element is connected across an AC signal. CH1 and CH2 measures the voltage and current respectively: CH1 (Yellow) Settings: CH2 (Blue) Settings: Volts/Div: 50V Current Probe Ratio: 100 mV/mA Time/Div: 5ms Volts/Div: 20V Probe Settings: X1 Time/Div: 5ms Probe Settings: X1 If ?? (distance between two ‘purple’ cursors) is 5ms, determine:a. Periodb. Frequencyc. Phase Angle Differenced. CH1 Peak Voltagee. CH1 RMS VoltageIn an oscilloscope, a 20 V sinusoidal signal produces a deflection of 2cm corresponding to a certain setting of vertical gain control. If another voltage produces 7.3 cm deflection for the same setting of the vertical gain control, what is the peak-to-peak and rms value of the voltage?
- The waveform displayed on an oscilloscope is as shown in Figure The 'time/cm' switch is set to 20 ms/cm, and the 'volts/cm' switch is set to 30 V/cm. Determine the (i) amplitude of waveform Q, (ii) peak to peak value of waveform P, (iii) frequency of waveform P and (iv) phase angle difference between P and Q in Degrees and (v) R.m.s value of waveform Q. Note: Please do not handwritingExplain how you would use an oscilloscope to measure an alternating potential difference of the order of 10V peak-to-peak. (You may assume that an alternating PD of 5V peak-to-peak is available).What is the RMS value of a peak-to-peak voltage of 10.0V?What is the RMS voltage? Vrms =_________V note: See Figure 13.3 below: An oscilloscope screen produces the waveform shown. The vertical sensitivity control isset to 20 V/div, and the horizontal sensitivity is set at 10 μs/div
- 24. The waveform displayed on an oscilloscope is as shown in Figure . Determine: (i) amplitude of waveform 'A' (ii) RMS value of waveform 'B' (iii) frequency of waveform 'A' and (iv) phase angle difference between 'A' and 'B' in Degrees.The waveform displayed on an oscilloscope is as shown in Figure The 'time/cm' switch is set to 20 ms/cm, and the 'volts/cm' switch is set to 30 V/cm. Determine the (i) amplitude of waveform Q, (ii) peak to peak value of waveform P, (iii) frequency of waveform P and (iv) phase angle difference between P and Q in Degrees and (v) R.m.s value of waveform Q.NEED ASAP. ANSWERS MUST BE IN 4 DECIMAL PLACES Power brightness of a 104) W, 110 V lamp is to be varied by controlling firing angle of SCR full-wave circuit; the rms value of ac voltage appearing across each scr being 110 V. Find the (a) rms voltage and (b) rms current in the lamp at firingangle of 60o .
- Please create the circuit for the below image to show the 2 cycles of the wave form observed on the Oscilloscope. Thank you.What peak-to-peak value of ac sine wave would do the same work as 50 Vdc? Oscilloscope can display which value? (peak or rms) May I ask if a follow-up question in Bartleby will be forwarded to the expert who previously answered the original question? Or will it be redirected to another expert? Thanks for clarifying.The period of this signal is about ____ ms (milliseconds). Use this picture for 2 additional questions below. The oscilloscope range selector determines the scale on the screen. That means number of Volts per line for amplitude vertical (V/div) and the number of milliseconds per line for time horizontally (ms/div). The frequency of the signal in previous question is about ____ Hz. The peak to peak amplitude of the signal on the oscilloscope on first question is about ___ Volt(s). During sound digitizing the sound frequencies ranged from 200 Hz to 2.2 kHz. Based on Nyquist theorem the sampling frequency should be at least ____ Hz. Select the FM Radio Quality recording format on the picture below and calculate the size of a 30 seconds uncompressed audio file (like a WAV file) in kB (kilo Bytes). This is a ____ mega pixels picture. Based on the same image file properties screen shown in previous question, the size of…