Question

Help fill in table

The CH2 waveform vr(t) is shown in red. 3. Now, double the function generator frequency (to f=1250Hz about 1250Hz) to display exactly two complete cycles in 8 horizontal divisions. Do not change the amplitude of e(t). Again, both waveforms should be centered vertically on the scope grid. Adjust the vertical POSITION knobs if necessary. Carefully sketch and label both waveforms on the grid. Use the same settings (VOLTS/DIV AND SEC/DIV from part A4.) as with the 625 Hz waveforms. Copy the settings here: CH1 Vertical scale: 1 VOLTS/DIV CH2 Vertical scale: 50m VOLTS/DIV Horizontal Scale: 0.2m SEC/DIV vc(t) is shown in blue Vr(t) is shown in red. 1. Copy the measured resistance of R from page 1: We can now calculate the current at each frequency for the rising portion of vc using Ohm's law, ic = ir = VR/R The resistor voltage amplitude VR (on CH2) can be more accurately obtained by tempora aligning the bottom level of the waveform with one of the grid lines and counting divisions to the to level. Using one-half of this peak-to-peak value gives us the amplitude or positive value of VR. VR = VR,peak peak/2. Use the above measured value of R. Watch units! Frequency VR, (peak-peak) VR (zero to peak) ic = IR = VR/R 100 625 Hz mV mV MA R = 100 1250 Hz MA mV mV How does the amplitude of ic (the CH2 waveform) change if the amplitude of Vc (CH1 waveform) is reduced from 8V peak to peak to 4V peak to peak? (Do not change the frequency.) Peak-to peak here means from the lower (negative) level to the upper (positive) level of the VR waveform.

Answer 1

VR ( zo ropeak) 625 HZ VR (peak-peak) loomu loomi 2 VR (0-1) = Somu ů cep = VR/R ( - Somy risca 1oon = 0.5mA