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Q.
Three electric bulbs of $200\, W, 200 \,W$ and $400\, W$ are shown in figure. The resultant power of the combination is
Odisha JEEOdisha JEE 2009Current Electricity
Solution:
Bulbs $A$ and $B$ are in parallel, their effective power is
$P'=P_{A}+P_{B}=200 W +200 W =400 W$
$P'$ and bulb $C$ are in series, the resultant power of the combination is
$P_{R}=\frac{P' \times P_{C}}{P'+P_{C}}=\frac{400\, W \times 400\, W }{400\, W +400\, W }=200\, W$
