Consider the oscilloscope view below. Channel 1 is the voltage waveform of a sin

Question

Consider the oscilloscope view below. Channel 1 is the voltage waveform of a sinusoidal voltage source and Channel 2 is the voltage waveform across the resistor in a series RLC circuit. On this oscilloscope, the scale values per division are given in the top left for each channel and the scale value per division of time is given at the top, to the left of "Auto"(it is set to 20 mu s). The cursors have been set to measure the difference in time between t he two vertical cursors, and that difference is Delta X = 8.8 mu s. Determine the following: a) The approximate amplitude of the source voltage (Channel 1) and output voltage across the resistor (Channel 2) b) The period of the waveforms. c) The frequency of the waveforms in hertz and rad/s. d) Which Channel's waveform is leading and by how many degrees? What type of filter is this (series RLC circuit with output as the voltage across the resistor)? e) The resistance of the resistor assuming the current through it has amplitude 2mA. Assume the source voltage is the reference point (i.e., cosine function with phase offset of 0 degrees).

Explanation / Answer

Answer:- Given that amplitude scale value for channel 1 = 5 mV/division, channel 2 = 2 mV/division, time scale is 20 microsecond/division.

a) To calculate the voltage for channel 1, we count the number of division on vertical axis, form horizontal axix to peak of the wave of channel 1, we get this value as 13 divisions so channel 1 has amplitude of 13 x 5 mV = 65 mV.

Similarly for channel 2, number of divisions on vertical axis in between horizontal axis and peak of channel 2 is 10, so amplitude of channel 2 is 10 x 2 mV = 20 mV.

b) To get the time period of the waveform, take one complete period of wave and count the number of division in between vertical axis and end of one period, then multiply the count with time scale.

so for channel 1 we get number of time division count = 4, hence time period T1 = 4 x 20 microsecond = 80 microsecond.

for channel 2, number of divisions = 4, hence T2 = 80 microsecond.

c) since T1 = T2 = 80 microsecond hence frequency f1 = f2 = 1/80 MHz = 12.5 kHz

in terms of radian per second, just multiply frequency with 2 x pie, i.e w1 = w2 = 2 x 3.14159 x 12.5 x 1000 =78539.75 rad/s .

d) when channel 2 is crossing at vertical axis, channel 1 is at some higher value, hence channel 1 is leading. To calculate the leading amount in degree we use the given difference time of 8.8 microsecond.

since time difference is 8.8 microsecond, the diffrence in degree is-

= (2 x 180 x 8.8 ) / ( T1 x 1000000) degree

= 39.6 degree.

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