JEE PHYSICS MCQ: CURRENT ELECTRICITY

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Application +4 | -1

Two bulbs marked

\mathrm{200}\,\,\mathrm{V - }\,\,\mathrm{100}\,\,\mathrm{W}

and

\mathrm{200}\,\,\mathrm{V}\,\mathrm{ - }\,\mathrm{200}\,\,\mathrm{W}

are joined in series and connected to a power supply of

\mathrm{200}\,\,\mathrm{V}

. The total power consumed by the two will be near to

View answer and explanation

Correct option: B

\begin{array}{llllllllllllllllllll} {\mathrm{66}\,\,\mathrm{W}} \end{array}

Option B (

\begin{array}{llllllllllllllllllll} \mathrm{66}\,\,\mathrm{W} \end{array}

) is correct because it's calculated by finding the equivalent resistance of the bulbs in series and then using the power formula.

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\mathrm{\rho }

3\,\,{\mathrm{m}^3}

3\,\,\mathrm{\Omega }

4.7\,\, \times \,\,{10^6}\,\,\mathrm{\Omega }\,\, \pm \,\,5\,\,\%

22.5\,cm