A) \[80\,\,W\]
B) \[60\,\,W\]
C) \[40\,\,W\]
D) \[20\,\,W\]
Correct Answer: A
Solution :
Key Idea: Power dissipated is inversely proportional to equivalent resistance. Let \[R\] be the resistance of individual component. When such resistances are connected in series, the equivalent resistance is \[{{R}_{eq}}=R+R+R+R=4R\] When, they are connected in parallel, equivalent resistance is \[\frac{1}{R_{eq}^{'}}=\frac{1}{R}+\frac{1}{R}+\frac{1}{R}+\frac{1}{R}=\frac{4}{R}\] \[\Rightarrow \] \[R_{eq}^{'}=\frac{R}{4}\] From Joule's law, the power dissipated is \[\Rightarrow \] \[P=\frac{{{V}^{2}}}{{{R}_{eq}}}\] Hence, \[\frac{{{P}_{s}}}{{{P}_{p}}}=\frac{R_{eq}^{'}}{{{R}_{eq}}}=\frac{R}{4}\times \frac{1}{4R}=\frac{1}{16}\] \[\Rightarrow \] \[{{P}_{p}}=16\times {{P}_{s}}\] Given, \[{{P}_{s}}=5\,\,W\] \[{{P}_{p}}=16\times 5=80\,\,W\]
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