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Define RMS value of AC. Derive a relation between RMS value of AC voltage & maximum voltatge. |
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Answer» Solution :i. RMS value is ALSO defined as that value of the steady current which when flowing through a given circuit for a given time produces the same amount of heat as produced by the alternating current when flowing through the same circuit for the same time. The effective value of an alternating voltage is represented by `V_(eff).` ii. The alternating current `i=I_(m)sinomegat""ori=I_(m)sintheta,` is represented GRAPHICALLY in Figure. The corresponding squared current wave is also shown by the dotted lines. iii. 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"))` IV. An elementary area of thickness `d""theta` is considered in the first half-cycle of the squared current wave as shown in Figure. Let `i^(2)` be the element `=i^(2)d""theta` Area of one cycle of squared `" wave "=int_(0)^(2pi)i^(2)d""theta`  `=int_(0)^(2pi)I_(m)^(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))/(sqrt(2))[" Base length of one cycle is "2pi]` `I_(RMS)=0.707I_(m).` v. Thus we find that for a symmetrical sinusoidal current rms value of current is `70.7%` of its peak value. Similarly for alternating voltage, it can be shown that `V_(rms)=0.707I_(m).` |
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