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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 given below. 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
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- 5 V, 1 ms
- 5V, 2 ms
- 7.5 V, 2 ms
- 10V, 1 ms
Correct Option: C
p– p divison of upper trace voltage = 2
and value of p-p voltage = 5V.
| ∴ | = 2.5 | |
| divisions |
Now it will same for unknown voltage, p-p division of unknown voltage = 3
So p-p voltage = 3 × 2.5 = 7.5 voltage Frequency of upper trace = 1 kHz.
| ∴ Time Period = | = 10-3 = 1 ms. | |
| 10³ |
and divison on x-axis = 4
| Thus, | = | ||
| Divison | 4 |
and divison of Lower trace voltage on x-axis = 8
| Thus, | = | ||
| Divison | 4 |
| Period of unknown signal, |  |  | × 8 = 2 ms | |
| Division |
