Question

交变电流有效值怎么计算？

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Answer 1

由正弦式交变电流的对称性，即丛谈烂只需计算半周期的热效应。将正弦式交变电流的半渗漏周期分成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/；侍陆2)*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