Average Value of a Function with Respect to Time

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mill4864

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Household electricity is supplied in the form of alternating current that varies from 155V to -155V with a frequency of 60 cycles per second (Hz). The voltage is thus given by the equation E(t) = 155sin(120?t) where t is the time in seconds. Voltmeters read the RMS (root-mean-square) voltage, which is the square root of the average value of [E(t)]^2 over one cycle.

(a) Calculate the RMS voltage of household current.
(b) Many electric stoves require an RMS voltage of 220V. Find the corresponding amplitude A needed for the voltage E(t) = Asin(120?t).

For (a), I squared the E(t) equation, integrated it from 0 to 1/60 (because the frequency is 60 cycles per second, so the time of one cycle would be 1/60s, I believe), and took the square root. The value I got (about 14.15V) just didn't seem correct. (The value before taking the square root, 4805/24, or about 200.21V, seems much closer to the value they talk about in part (b), not that that means anything, but it's just an observation.)

For (b), I changed the equation E(t) = 155sin(120?t) into E(t) = Asin(120?t) and squared it ( [E(t)]^2 = (A^2)sin^2(120?t) ). Then I integrated it from 0 to 1/60 for the same reason I did in part (a), getting a result of (A^2)/120. Then I put a square root sign over (A^2)/120, set it equal to 220V, and solved for A, getting A = about 2409.979. An amplitude A = about 2409.979 is a lot larger than the amplitude A = 155 in the given equation E(t) = 155sin(120?t), which leads me to believe my answer is incorrect.
 
mill4864 said:
Household electricity is supplied in the form of alternating current that varies from 155V to -155V with a frequency of 60 cycles per second (Hz). The voltage is thus given by the equation E(t) = 155sin(120?t) where t is the time in seconds. Voltmeters read the RMS (root-mean-square) voltage, which is the square root of the average value of [E(t)]^2 over one cycle.

(a) Calculate the RMS voltage of household current.
(b) Many electric stoves require an RMS voltage of 220V. Find the corresponding amplitude A needed for the voltage E(t) = Asin(120?t).

For (a), I squared the E(t) equation, integrated it from 0 to 1/60 (because the frequency is 60 cycles per second, so the time of one cycle would be 1/60s, I believe), and took the square root. The value I got (about 14.15V) just didn't seem correct. (The value before taking the square root, 4805/24, or about 200.21V, seems much closer to the value they talk about in part (b), not that that means anything, but it's just an observation.)

For (b), I changed the equation E(t) = 155sin(120?t) into E(t) = Asin(120?t) and squared it ( [E(t)]^2 = (A^2)sin^2(120?t) ). Then I integrated it from 0 to 1/60 for the same reason I did in part (a), getting a result of (A^2)/120. Then I put a square root sign over (A^2)/120, set it equal to 220V, and solved for A, getting A = about 2409.979. An amplitude A = about 2409.979 is a lot larger than the amplitude A = 155 in the given equation E(t) = 155sin(120?t), which leads me to believe my answer is incorrect.
Click to expand...
the average value of the function over [a,b] is 1/(b-a) times the integral from a to b. therefore before you took the square root you should have multiplied by 1/(1/60-0) or just 60.
 
\(\displaystyle a) \ RMS \ = \ \bigg[\frac{1}{1/60-0}\int_{0}^{1/60}(155)^{2}sin^{2}(120\pi t)dt\bigg]^{1/2}\)

\(\displaystyle = \ \bigg[ 60(155)^{2}\int_{0}^{1/60}sin^{2}(120\pi t)dt\bigg]^{1/2}\)

\(\displaystyle Let \ u \ = \ 120\pi t, \ then \ \frac{du}{120\pi} \ = \ dt\)

\(\displaystyle Ergo, \ we \ have \ \bigg[\frac{155^{2}}{2\pi}\int_{0}^{2\pi}sin^{2}(u)du\bigg]^{1/2}\)

\(\displaystyle = \ \bigg[\frac{155^{2}}{4\pi}\int_{0}^{2\pi}(1-cos(2u))du\bigg]^{1/2}\)

\(\displaystyle = \ \bigg[\frac{155^{2}}{4\pi}[u-sin(u)cos(u)]_{0}^{2\pi}\bigg]^{1/2}\)

\(\displaystyle Hence, \ RMS \ = \ \bigg[\bigg(\frac{155^{2}}{4\pi}\bigg)(2\pi)\bigg]^{1/2} \ = \ \frac{155}{\sqrt2} \ = \ 109.60V\)

\(\displaystyle b) \ 220 \ = \ \frac{A}{\sqrt2}, \ therefore \ A \ = \ 220\sqrt2 \ = \ 311.127\)
 
Thank you very much! And what program did you use to type up those equations? Something like that might be very helpful for when I have to type up my Physics lab reports.
 
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