Two bulbs of ${\rm{100}}\,\,{\rm{watt}}$ and ${\rm{200}}\,\,{\rm{watt}}$ working at ${\rm{220}}\,\,{\rm{V}}$ are joined in series with ${\rm{220}}\,\,{\rm{V}}$ supply. Total power consumed will be

$\begin{array}{*{20}{l}} {{\rm{65}}\,\,{\rm{W}}} \end{array}$

$\begin{array}{*{20}{l}} {{\rm{33}}\,\,{\rm{W}}} \end{array}$

$\begin{array}{*{20}{l}} {{\rm{300}}\,\,{\rm{W}}} \end{array}$

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

JEE PHYSICS MCQ: CURRENT ELECTRICITY

1 Medium

1 of 1

Application +4 | -1

Two bulbs of

\mathrm{100}\,\,\mathrm{watt}

and

\mathrm{200}\,\,\mathrm{watt}

working at

\mathrm{220}\,\,\mathrm{V}

are joined in series with

\mathrm{220}\,\,\mathrm{V}

supply. Total power consumed will be

View answer and explanation

Correct option: A

\begin{array}{llllllllllllllllllll} {\mathrm{65}\,\,\mathrm{W}} \end{array}

Option A (

\mathrm{65}\,\,\mathrm{W}

) is correct because the total power consumed is calculated by finding the equivalent resistance and then using the power formula.