Question

The time/div and voltage/div axes of an oscilloscope have been erased. A student connects a 1 kHz, 5 V p-p square wave calibration pulse to channel 1 of the scope and observes the screen to be as shown in the upper trace of the figure. An unknown signal is connected to channel 2 (lower trace) of the scope. If the time/div and V/div on both channels are the same, the amplitude (p-p)and period of the unknown signal are respectively

• A. 5 V, 1 ms
• B. 5V, 2 ms
• C. 7.5 V, 2 ms
• D. 10 V, 1 ms

Answer 1

The correct option is C 7.5 V, 2 ms

Peak-Peak (p-p) division of upper trace voltage=2 and value of (p-p) voltage=5 V.

$\therefore \frac{Voltage}{Divisions}=2.5V$

Now it will be same for unknown voltage (p-p) division of unknown voltage =3

$\therefore p-pvoltage=3\times 2.5=7.5V$

Frequency of upper trace=1 kHz

$\therefore Timeperiod=\frac{1}{{10}^3}=1ms$

$Divisionofx-axis\left(upper\right)=4$
$Divisionofx-axis\left(lower\right)=8$

$\therefore Periodofunknownsignal=2ms$