A bulb of ${\rm{220 V}}$ and ${\rm{300 W}}$ is connected across ${\rm{110 V}}$ circuit, the percentage reduction in power is

$\begin{array}{*{20}{l}} {{\rm{75}}\,\,{\rm{\% }}} \end{array}$

$\begin{array}{*{20}{l}} {{\rm{100}}\,\,{\rm{\% }}} \end{array}$

$\begin{array}{*{20}{l}} {{\rm{25}}\,\,{\rm{\% }}} \end{array}$

$\begin{array}{*{20}{l}} {{\rm{70}}\,\,{\rm{\% }}} \end{array}$

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JEE PHYSICS MCQ: CURRENT ELECTRICITY

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Application +4 | -1

A bulb of

\mathrm{220 V}

and

\mathrm{300 W}

is connected across

\mathrm{110 V}

circuit, the percentage reduction in power is

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Correct option: D

\begin{array}{llllllllllllllllllll} {\mathrm{75}\,\,\mathrm{\% }} \end{array}

Option D (75 %) is correct because the power is proportional to the square of the voltage, so reducing the voltage by half reduces the power to one-fourth.

JEE PHYSICS MCQ: CURRENT ELECTRICITY

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