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

Answer:

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

(B) RMS value of voltage will be 0.530 volt

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 $2v_{peak}=1.5volt$

$v_{peak}=\frac{1.5}{2}=0.75volt$

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 $v_{rms}=\frac{v_{peak}}{\sqrt{2}}=\frac{0.75}{1.414}=0.530volt$