(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
- From the following oscilloscope trace, provide the following parameters: (a) Period (b) Frequency (c) Peak-to-peak voltage (d) Peak voltage (e) RMS voltageA 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.Topic: 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 shiftThe 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 =Given 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=?