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Derive an experession for the RMS value of AC. |
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Answer» Solution :i. The term RMS REFERS to time-varying sinusoidal CURRENTS and voltage and not used in DC systems. ii. The root mean square value of an alternating current is defined as the square root of the mean of the squares of all currents over one cycle. It is denoted `I_(RMS).` For alternating voltages, the RMS value is given by `V_(RMS).` iii. The alternating current `i=I_(m)sinomegatori=I_(m)sintheta,` is represented graphically in Figure. The corresponding squared lines. iv. The sum of the squares of all currents over one cycle is given by the area of one cycle of squared wave. Therefore, `I_(RMS)=sqrt(("Area of one cycle of squared wave")/("Base length of one cycle"))` v. An ELEMENTARY area of thickness `d""theta` is considered current wave as shown in Figure. Let `i^(2)` be the mid-ordinate of the element. Area of the element `=i^(2)d""theta` Area of one cycle of squared `"wave"=int_(0)^(2pi)i^(2)d""theta`  `int_(0)^(2pi)I^(2)sin^(2)thetad""theta=I_(m)^(2)int_(0)^(2pi)sin^(2)thetad""theta` `=I_(m)^(2)int_(0)^(2pi)[(1-cos2theta)/(2)]d""theta` `" since "sin^(2)theta=(1-cos2theta)/(2)` `=(I_(m)^(2))/(2)[int_(0)^(2pi)d""theta-int_(0)^(2pi)cos2thetad""theta] `(I_(m)^(2))/(2)[theta-(sin2theta)/(2)]_(0)^(2pi)` `=(I_(m)^(2))/(2)[(2pi-(sin2xx2pi)/(2))-(0-(sin0)/(2))]` `=(I_(m)^(2))/(2)xx2pi=I_(m)^(2)pi[becausesin0=sin4pi=0]` Substituting this in equation (1), we get `I_(RMS)=sqrt((I_(m)^(2)pi)/(2pi))=(I_(m)^(2))/(sqrt(2))[" Base length on cone cycle is "2pi]` `I_(RMS)=0.707I_(m)` |
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