(h) Reverse the positions of R and L and measure VL (p.p) (to ensure that the oscillator (or generator) and the oscilloscope have a common ground). 10 kHz E=8V (p-p) L 000 10 mH R measured R 1 ΚΩ + Channel 2 Channel 1 Vert.: 1 V/div. Vert.: 1 V/div. Hor.: 20 μs/div. Hor.: 20 us/div.
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I need help, can you illustrate what the question wants what me to do with the circuit I am getting confused about the question
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- The waveform displayed on an oscilloscope is as shown in Figure The 'time/cm' switch is set to 0.2 ms/cm, and the 'volts/cm' switch is set to 30 V/cm. Determine the (1) amplitude of waveform Q, (i) peak to peak value of waveform P, (ii) frequency of waveform P and (iv) phase angle difference between P and Q in Degrees. Upload your Answers steps in the "Final Answer Submission" Link provided in the Moodle. (i) Amplitude of waveform Q = (ii) Peak to peak value of waveform P = (iii) Frequency of waveform P = (iv) Phase angle difference between P and Q in Degrees =What is the maximum voltage across the capacitor? (Hint: Observe from the oscilloscope graph) Function Generator Set up:Frequency = 100 Hz,Amplitude = 5 Vp [Peak voltage = 5 V i.e. Peak- peak = 10 V]Waveform = square wave Oscilloscope Set up:Time base scale: 2 ms/div(Input signal) Channel A scale: 2 V/ div(Output Signal) Channel B scale: 2V/ divThe waveform displayed on an oscilloscope is as shown in Figure The 'time/cm' switch is set to 2 ms/cm, and the 'volts/cm' switch is set to 1 V/cm. Determine the (i) amplitude of waveform Q (i) peak to peak value of waveform P. (ii)i frquency of waveform P and (iv) phase angle difference between P and Q in Degrees. Upload your Answers steps in the "Final Answer Submission" Link provided in the Moodle. () Amplitude of waveform Q = (ii) Peak to peak value of waveform P = (ii) Frequency of waveform P = (iv) Phase angle difference between P and Q in Degrees =
- The waveform displayed on an oscilloscope is as shown in Figure The 'time/cm' switch is set to 0.5 ms/cm, and the 'volts/cm' switch is set to 0.2 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. Upload your Answers steps in the "Final Answer Submission" Link provided in the Moodle. (i) Amplitude of waveform Q = (ii) Peak to peak value of waveform P = (iii) Frequency of waveform P =It is the voltage supply that limits the AC voltage level. a. Square wave input b. Bias c. Output voltage d. Capacitor voltageThe R.M.S. value of a half-wave rectified current is 10A, its value for full-wave rectification would be amperes (a) 20 (b) 14.14 (c) 20/z (d) 40/z
- 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.From the following oscilloscope trace, provide the following parameters: (a) Period (b) Frequency (c) Peak-to-peak voltage (d) Peak voltage (e) RMS voltageTopic: Half Wave Rectification
- Identify which measurement this oscilloscope time difference reading is useful for. Oscilloscope-XSC1 Time 250.000 us 1.261 ms Channel_A 169.064 V 168.513 V -550. 584 mV Channel_B 89. 282 V 95.005 V 5.722 V Reverse T2-T1 1.011 ms Save Ext. trigger Timebase Scale: 200 us/Div : Scale: 100 V/Div X pos. (Div): 0 Channel A Channel B Scale: 100 V/Div Y pos. (Div): 0 Trigger Edge: F Level: 0 A B Ext Y pos. (Div): 0 V Y/T Add B/A A/B AC DC AC DC Single Normal O period O none of the above O phase shiftGiven an oscilloscope input resistance (R2 in Figure 3 b) is 1MegΩ. Find the probe resistance (R1) so that the ratio of the source voltage (Vs) to the oscilloscope input voltage (Vin) is Vs/Vin = 10. R1=?The waveform displayed on an oscilloscope is as shown in Figure The 'time/cm' switch is set to 1 ms/cm, and the 'volts/cm' switch is set to 1 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. P Upload your Answers steps in the "Final Answer Submission" Link provided in the Moodle. (i) Amplitude of waveform Q = (ii) Peak to peak value of waveform P = (iii) Frequency of waveform P = (iv) Phase angle difference between P and Q in Degrees =