An electric bulb is connected to a power supply of $220~\mathrm{V}$. If the current drawn by the bulb from the supply is $500~\mathrm{mA}$, the power of the bulb is:

Given $V = 220~\mathrm{V}$ and $I = 500~\mathrm{mA} = 0.5~\mathrm{A}$. From the chapter, electric power is $P = VI$ (Eq. 11.22), the same relation used in Example 11.12 of the chapter (220 V, 0.50 A bulb). So $P = 220 \times 0.5 = 110~\mathrm{W}$.

(A) 11 W: arises from an arithmetic slip (dividing instead of multiplying, or using $I=50~\mathrm{mA}$) — incorrect.

(C) 220 W: obtained by ignoring the current altogether and just taking $V$, or by doubling — incorrect, since $P=VI$ requires multiplying by $I=0.5$, not $1$.

(D) 1100 W: results from mistakenly using $I = 5~\mathrm{A}$ instead of $0.5~\mathrm{A}$ (misplacing the decimal in converting mA to A) — incorrect.