Power in Electric Circuits — Question 2

Question 2

A zero-mean current still heats a resistor

A constant resistor R carries $I(t)=I_d+I_a\cos\omega t$. Average over a whole number of periods.

1. Derive average power and root-mean-square current.

2. Compare equal-peak sinusoidal and symmetric square-wave currents of peak magnitude $I_a$ and zero mean.

Question 2 - Solution

A zero-mean current still heats a resistor

(a) Method and calculation

$$\langle I^2\rangle=I_d^2+2I_dI_a\langle\cos\omega t\rangle+I_a^2\langle\cos^2\omega t\rangle=I_d^2+I_a^2/2.$$$$\overline P=R(I_d^2+I_a^2/2),\quad I_{\mathrm{rms}}=\sqrt{I_d^2+I_a^2/2}.$$

(b) Interpretation and further calculation

$$P_{\sin}=RI_a^2/2,\quad P_{\square}=RI_a^2.$$$$P_{\square}/P_{\sin}=2,\quad\langle I\rangle=0\ \text{for both}.$$

Physical check

Resistive heating depends on the mean square, not the signed mean current.