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"110伏特，60瓦特之燈炮"表示如果該燈炮接的電源是110v，它確實可以表現60w的功率。
(但兩個燈炮串聯後，每個燈泡的電位差僅55v)

先計算一個燈炮電阻R
P=V^2/R =>R=V^2/P=110*110/60 =1210/6(Ω)
電流V=IR
110=I*(1210/6 +1210/6)=(1210/3) I
=> I = 3/11 (A)

一個燈泡損耗之電功率=I^2*R=(3/11)^2*(1210/6)=15 W

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或先計算一個燈炮電阻R
R=V^2/P=110*110/60 =1210/6(Ω)

P=V^2/R =55^2/(1210/6)=15W

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