Three Simple Circuits — Question 6

Question 6

Pure L and C elements weight voltage harmonics in opposite ways

A voltage $v=V_1\cos\omega t+V_3\cos3\omega t$ is applied separately to an ideal capacitor and ideal inductor. Use zero DC inductor current.

1. Find the ratio of third-harmonic to fundamental current amplitude in each element.

2. Find rms current in each case.

Question 6 - Solution

Pure L and C elements weight voltage harmonics in opposite ways

(a) Method and calculation

$$I_{C,1}=\omega CV_1,\quad I_{C,3}=3\omega CV_3\Rightarrow\boxed{I_{C,3}/I_{C,1}=3V_3/V_1}.$$$$I_{L,1}=V_1/(\omega L),\quad I_{L,3}=V_3/(3\omega L)\Rightarrow\boxed{I_{L,3}/I_{L,1}=V_3/(3V_1)}.$$

(b) Interpretation and further calculation

$$\boxed{I_{C,\mathrm{rms}}^2=\tfrac12(\omega C)^2(V_1^2+9V_3^2)}.$$$$\boxed{I_{L,\mathrm{rms}}^2=\frac1{2(\omega L)^2}(V_1^2+V_3^2/9)}.$$

Physical check

Differentiation emphasizes high-frequency voltage components; integration suppresses them.