正弦交变电流的有效值是如何推导的?

Question

如题所述

Answer 1

由正弦式交变电流的对称性,即只需计算半周期的热效应.将正弦式交变电流的半周期分成n份(n→∞),每份Δt=T/(2n)
I=f(t)=Im*sin(2π/T)* Δt
则,
I1=f(t1)=f(Δt)=Im*sin(π/n)
I2=f(t2)=f(2Δt)=Im*sin(2π/n)
……
In=f(tn)=f(nΔt)=Im*sin(nπ/n)
所以,
W(T/2)=I1²*R*Δt+ I2²*R*Δt+……+In²*R*Δt
=Im²*R*(T/2n)*(sin² (π/n)+ sin² (2π/n)+……+sin² (nπ/n))
= Im²*R*(T/4)= I²*R*(T/2)
I=(√2/2)*Im
补充证明：(sin² (π/n)+ sin² (2π/n)+……+sin² (nπ/n))=n/2
sin² (π/n)=cos²(π/2-π/n)
sin² (2π/n)= cos²(π/2-2π/n)
……
sin² (π/2-π/n)= cos²(π/n)
sin² (π/2)= cos²(π)
sin² (π/2+π/n)= cos²(π/2+π/n)
……
sin² (π-π/n)= cos²(π/2+π/n)
sin² (π)= cos²(π/2)
(sin² (π/n)+ sin² (2π/n)+……+sin² (nπ/n))+ (cos² (π/n)+ cos² (2π/n)+……+cos² (nπ/n))=n
(sin² (π/n)+ sin² (2π/n)+……+sin² (nπ/n))=n/2