Alternating Current — Question 2

Question 2

Zero mean voltage can deliver a nonzero mean heating rate

A resistor R carries $i(t)=I_p\sin\omega t$. Average over one complete period.

1. Derive mean current, rms current, and mean power directly from the integrals.

2. Find energy dissipated over n complete cycles.

Question 2 - Solution

Zero mean voltage can deliver a nonzero mean heating rate

(a) Method and calculation

$$\langle i\rangle=\frac{I_p}{2\pi}\int_0^{2\pi}\sin\theta\,d\theta=0.$$$$\boxed{I_{\mathrm{rms}}^2=\frac{I_p^2}{2\pi}\int_0^{2\pi}\sin^2\theta\,d\theta=I_p^2/2}.$$

(b) Interpretation and further calculation

$$\langle P\rangle=R\langle i^2\rangle=RI_p^2/2.$$$$\boxed{H_n=\langle P\rangle nT=\pi n RI_p^2/\omega}.$$

Physical check

Heating depends on squared current, so opposite half-cycles both contribute positively.