Search any question & find its solution
Question:
Answered & Verified by Expert
A standard filament lamp consumes $100 \mathrm{~W}$ when connected to $200 \mathrm{~V}$ a.c. mains supply. The peak current through the bulb will be :
Options:
Solution:
2605 Upvotes
Verified Answer
The correct answer is:
$0.707 \mathrm{~A}$
Power,
$\begin{aligned}
\mathrm{P} & =\mathrm{I}_{\mathrm{rms}} \mathrm{V}_{\mathrm{rms}} \\
\mathrm{I}_{\mathrm{rms}} & =\frac{\mathrm{P}}{\mathrm{V}_{\mathrm{rms}}}=\frac{100}{200}=\frac{1}{2} \\
\mathrm{I}_0 & =\sqrt{2} \mathrm{I}_{\mathrm{rms}} \\
\mathrm{I}_0 & =\sqrt{2} \times \frac{1}{2} \\
\mathrm{I}_0 & =\frac{1}{\sqrt{2}}=0.707 \mathrm{~A}
\end{aligned}$
$\begin{aligned}
\mathrm{P} & =\mathrm{I}_{\mathrm{rms}} \mathrm{V}_{\mathrm{rms}} \\
\mathrm{I}_{\mathrm{rms}} & =\frac{\mathrm{P}}{\mathrm{V}_{\mathrm{rms}}}=\frac{100}{200}=\frac{1}{2} \\
\mathrm{I}_0 & =\sqrt{2} \mathrm{I}_{\mathrm{rms}} \\
\mathrm{I}_0 & =\sqrt{2} \times \frac{1}{2} \\
\mathrm{I}_0 & =\frac{1}{\sqrt{2}}=0.707 \mathrm{~A}
\end{aligned}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.