Alternating Current — Question 8

Question 8

Different harmonics add their mean-square contributions independently

A current is $i(t)=3\cos\omega t+4\sin2\omega t$ amperes through a 2.00-ohm resistor. Average over the fundamental period.

1. Compute rms current using harmonic orthogonality.

2. Find mean heating and compare with a 5.00-A peak sinusoid.

Question 8 - Solution

Different harmonics add their mean-square contributions independently

(a) Method and calculation

$$\langle i^2\rangle=9\langle\cos^2\omega t\rangle+16\langle\sin^22\omega t\rangle+24\langle\cos\omega t\sin2\omega t\rangle.$$$$\boxed{I_{\mathrm{rms}}^2=(9+16)/2=12.5\,A^2}.$$

(b) Interpretation and further calculation

$$\boxed{\langle P\rangle=2(12.5)=25.0\,W}.$$$$I_{\mathrm{rms,sine}}=5/\sqrt2=I_{\mathrm{rms}}\Rightarrow\boxed{P_{\mathrm{sine}}=P}.$$

Physical check

Equal rms heating does not mean equal waveforms or equal instantaneous peaks.