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Electronics Engineering QUESTION #11728
Question 1
An oscilloscope probe is set to 10× attenuation but the scope is set to 1× in software. The measured voltage on screen is $5\,\text{V}$. The actual circuit voltage is:
  • $0.5\,\text{V}$
  • $5\,\text{V}$
  • $50\,\text{V}$✔️
  • $500\,\text{mV}$
Correct Answer Explanation
A 10× probe attenuates the signal by a factor of 10 before it reaches the scope input. If the scope software is not compensated (set to 1×), it does not multiply by 10. Actual voltage $= \text{displayed} \times \text{probe factor} = 5\,\text{V} \times 10 = \mathbf{50\,\text{V}}$. Always set the scope channel to match the probe attenuation ratio.