A) \[P={{\left( \frac{{{V}_{0}}}{V} \right)}^{2}}{{P}_{0}}\]
B) \[P={{\left( \frac{V}{{{V}_{0}}} \right)}^{2}}{{P}_{0}}\]
C) \[P=\left( \frac{V}{{{V}_{0}}} \right){{P}_{0}}\]
D) \[P=\left( \frac{{{V}_{0}}}{V} \right){{P}_{0}}\]
Correct Answer: B
Solution :
As we are quite well know that the power of an electric bulbs is \[P=Vi=V\frac{V}{R}=\frac{{{V}^{2}}}{R}\] Hence \[\frac{P}{{{P}_{0}}}=\frac{{{V}^{2}}}{R}\times \frac{R}{V_{0}^{2}}={{\left( \frac{V}{{{V}_{0}}} \right)}^{2}}\] Thus \[P={{P}_{0}}{{\left( \frac{V}{{{V}_{0}}} \right)}^{2}}\]
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