Question

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 6.25 V but must record the rms voltage on a report. What value should be reported?

Answer 1

The rms voltage on a report is 4.42 V.

Step-by-step explanation:

Given that,

Peak voltage = 6.25 V

We need to calculate the rms voltage

The rms voltage is equal to the peak voltage divided by square root of 2.

Using formula of rms voltage

`V_(rms)=(v_(0))/(√(2))`

Where,
`V_(0)` = Peak voltage

Put the value into the formula

`V_(rms)=(6.25)/(√(2))`

`V_(rms)=4.42\ V`

Hence, The rms voltage on a report is 4.42 V.

Answer 2

`V_(rms)= 4.42 V`

Step-by-step explanation:

A technician uses an aosilloscope to measure the AC voltage across a resistor in a circuit

Oscillating voltage to be a sine wave with peak voltage

`V_(max)= 6.25 V`

we need calculate rms value of voltage

by formula we can write

`V_(rms)=(V_(max))/(√(2) )`

now putting value

`V_(rms)=(6.25)/(√(2) )`

=4.42 V

therefore,
`V_(rms)= 4.42 V`