A standard electrical outlet provides a 120-Hz sinewave. How do I determine the time required for one cycle, Vp-p, and Vav?

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valentin68 | College Teacher | (Level 3) Associate Educator

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The figure attached below shows one period for a sinusoidal AC Voltage supply.

By DEFINITION the period (or time required for one cycle) of a wave is

Period = 1/Frequency   or    `T =1/F`

Knowing that the frequency for US standard household AC voltage is F =60 Hz we obtain

`T=1/60 =0.0167 sec =167 ms`

When you specify that the value of the AC voltage is 120 V you tell its Root Mean Square value `V_(rms)` . This is the equivalent voltage of a DC source that will produce the same heating effect as the AC source can produces on the same load resistance.

The peak to peak value of the AC voltage Vpp is the voltage difference between the positive peak and the negative peak of the sine wave. If we define the peak value Vp as

`V_(pp) = V_p -(-V_p) =2*V_p`

the relation between `V_(rms)` and `V_p` is

`V_(rms) = sqrt((1/T)*int_0^(2*pi)(Vp^2*sin^2(theta)*d(theta))) =V_p/sqrt(2)`

Therefore the peak voltage is

`V_(p) = sqrt(2)*V_(rms) =sqrt(2)*120 =169.7 "Volts"`

and the peak to peak voltage is

`V_(pp) =2*V_p =339.4 "Volts"`

Now the average value of the voltage `V_(avg)` is the average of the sine wave over one half of the period (the average value over an entire period is zero).

`V_(avg) = (1/pi)*int_0^pi(V_p*sin(theta)*d(theta)) =(2/pi)*Vp=0.637*Vp` `V_(avg) = 0.637*169.7 =108.03 "Volts"`

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