Past exam of the MPhil in micro- and nanotechnology enterprise course of the University of Cambridge / 2019 / Paper 1 / Section B / 12 / v / a

The duty cycle $D=t_{\mathrm{active}}/(t_{\mathrm{active}}+t_{\mathrm{sleep}})$ is the fraction of time spent active. At a constant supply voltage and with switching energy neglected,For $D=0.01$, the average current is $0.01(25\mathrm{mA})+0.99(6.1\mu \mathrm{A})=256.039\mu \mathrm{A}$. Thus $\bar{P}=512.078\mu \mathrm{W}\simeq 0.512\mathrm{mW}$. Multiplying only the active power by the duty cycle would omit the sleep contribution.