Question

(1) Develop an equation that relates the rms voltage of a sine wave to its peak-to-peak voltage. a. If a sine wave has a peak-to-peak value of 1.5V, a frequency of 3kHz, and a phase of 0 radians, what is the rms voltage

Answer 1

(A) Equation will be
`v=v_msin\omega t=0.75sin(18840t)`

(B) RMS value of voltage will be 0.530 volt

Step-by-step explanation:

We have given peak to peak voltage of ac wave = 1.5 volt

Peak to peak voltage of ac wave is equal to 2 times of peak voltage

So
`2v_(peak)=1.5volt`

`v_(peak)=(1.5)/(2)=0.75volt`

Frequency of ac wave is given f = 3 kHz

So angular frequency
`\omega =2\pi f` = 2×3.14×3000 = 18840 rad/sec

So expression of equation will be
`v=v_msin\omega t=0.75sin(18840t)` ( As phase difference is 0 )

Now we have to find the rms value of voltage

So rms voltage will be equal to
`v_(rms)=(v_(peak))/(√(2))=(0.75)/(1.414)=0.530volt`