Alternating Current — Question 5

Question 5

Equal peaks do not imply equal heating for triangular and sinusoidal sources

A symmetric triangular voltage rises linearly from -Vp to +Vp over half a period and falls linearly back during the other half.

1. Compute its rms value from a voltage integral over a uniform linear sweep.

2. Compare its resistor heating with that of a sine wave of the same peak and with a square wave of the same peak.

Question 5 - Solution

Equal peaks do not imply equal heating for triangular and sinusoidal sources

(a) Method and calculation

$$\langle v^2\rangle=\frac1{2V_p}\int_{-V_p}^{V_p}v^2dv=V_p^2/3.$$$$\boxed{V_{{\mathrm{tri}},\mathrm{rms}}=V_p/\sqrt3}.$$

(b) Interpretation and further calculation

$$P_{\mathrm{tri}}=V_p^2/(3R),\quad P_{\mathrm{sin}}=V_p^2/(2R),\quad P_{\mathrm{sq}}=V_p^2/R.$$$$\boxed{P_{\mathrm{tri}}/P_{\mathrm{sin}}=2/3,\quad P_{\mathrm{tri}}/P_{\mathrm{sq}}=1/3}.$$

Physical check

The time spent near the peak determines mean-square voltage.