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A technician at a semiconductor facility is using an oscilloscope to measure the AC voltage across a resistor in a circuit. The technician measures the oscillating voltage to be a sine wave with a peak voltage of 3.25 V . However, the technician must record the RMS voltage on a report. What value should be reported?

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Answer:

RMS voltage is 2.29 V

Explanation:

Given that,

The peak voltge of the voltage = 3.25 V

We need to find the RMS voltage reported by the technician.

We know that the relation between the peak and the RMS voltage is given by the formula as follows :

[tex]V_{rms}=\dfrac{V_o}{\sqrt2}\\\\V_{rms}=\dfrac{3.25}{\sqrt2}\\\\V_{rms}=2.29\ V[/tex]

So, the RMS voltage is 2.29 V.

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