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peak to peak is 2x peak voltage in a symmetrical waveform.

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A class A amplifier features an amplifying element (Transistor, MOSFET, Tube, etc) that must remain in conduction (not saturated or cut off) during the entire wave form.

• RMS voltage does not represent the entire input amplitude range.
• Neither does average power.
• Resting voltage (or quiescent operation) also does not tell us anything except the DC state of the amplifier.

ONLY when looking at peak voltage will one be able to evaluate the capability of a class A amplifier.

Last edited by kevinruggles. Register to edit

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1. PV = 30V

   RMS V = (0.5)((2)^(1/2))*PV

   RMS V = 15*sqrt (2)

2. I = V/R = (.3)*sqrt (2)

3. P = I^2*R = ((.3)*sqrt (2))^2 * 50 = 9

OR

PV = 30V RMS E = PV/sqrt(2), = 21V

Power = E*I

You know E and R. E=IR, R=E/I. So, Power also = E^2/R

((21)^2)/50 = 9 watts

or, the easiest way to remember this is to take the PV and multiply it by 0.707. This gives you the approximate RMS voltage. Square the product, then divide it by the resistance to determine the power.

PV = 30V Vrms = 0.707PV given as 300.707 = 21.21 Vrms 21.21*21.21 = 449.86 449.86/50 = 8.997W (round up to 9W for the closest answer)

Last edited by boothe. Register to edit

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We are given the RMS value of the sinusoidal wave form. We must first compute the peak value:

Peak = RMS / (1/sqrt(2)), which is to say: ** Peak Voltage = RMS Voltage/(0.707)**

Which for our problem is: Peak = 65V/0.707 = 91.94V

But this is not the answer! The question asks for Peak-to-Peak, which is 2*Peak.

For our problem: Peak-To-Peak = 2* 91.94V = 183.88 Volts

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In electrodynamics, circular polarization of an electromagnetic wave is a polarization in which the electric field of the passing wave does not change strength but only changes direction in a rotary manner.

https://en.wikipedia.org/wiki/Circular_polarization

Memory aid: circularly: like a wheel rotating

Last edited by robertov416. Register to edit

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the key here is you need to find allowable power (watts). Knowing this you can eliminate the two answers which do not refer to a wattmeter. Next thing to look at is what kind of allowable power and that is maximum, and of the two remaining answers the peak-reading wattmeter will give you your maximum watts reading, the average will only give you an average of the maximum and minimum watts out.

Last edited by beringhill. Register to edit

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Average power dissipation is computed based on the RMS value, not peak or peak-to-peak values.

To answer this question, first determine the RMS voltage of the RF cycle:

RMS = Peak Voltage / sqrt(2) = 35V/1.414 = 24.745V

Then, use that value in Ohm's law to compute the average power: P = E^2 / R

P = 24.745^2 / 50 = 612.5V / 50 Ohms = 12.25 Watts

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RMS voltage is simply Peak Voltage (Vp) divided by the square root of 2. The square root of two is 1.414. If the RMS voltage is 34 then Vp = Sqrt of 2 * RMS or 48 = 1.414 * 34

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Peak = RMS x 1.414

Last edited by screecy. Register to edit

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Vpeak = Vrms * sqrt(2)

sqrt(2) ~= 1.414

US Household Vrms = 120V

so,

Vpeak = 120 x 1.414 = 169.7V

Peak to peak voltage = 2 x Vpeak = 339V

So, correct answer, rounding up, is 340V

Last edited by bennettj1087. Register to edit

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Don't get tripped-up by thinking this question is looking for something "fancy" like "PEAK voltage of a standard U.S. household electrical power outlet" (170V AC) or "PEAK TO PEAK voltage of a standard U.S. household electrical power outlet" (340V AC)

The value of "120V AC" commonly associated with standard U.S. household electrical power outlets is, in fact, the RMS value, which typically goes unsaid.

Last edited by kc5qnj. Register to edit

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(A). The RMS value of a 340 volt peak-to-peak pure sine wave is 120V AC.

In a sine wave, the RMS value is given as:

a / sqrt(2) given: a as amplitude

For this question: (340 / 2) * 0.707 ~= 120

Divide 340 in half to account for half wave and 0.707 is an approximation of sqrt(2)

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