Search any question & find its solution
Question:
Answered & Verified by Expert
A voltage of peak value $283 \mathrm{~V}$ and varying frequency is applied to a series $\mathrm{L}, \mathrm{C}, \mathrm{R}$ combination in which $\mathrm{R}=3 \mathrm{ohm}, \mathrm{L}=25 \mathrm{mH}$ and $\mathrm{C}=400 \mu \mathrm{F}$. The frequency (in Hz) of the source at which maximum power is dissipated in the above is
Options:
Solution:
1298 Upvotes
Verified Answer
The correct answer is:
$50.3$
Average power of an LCR circuit is $\mathrm{P}=\mathrm{E}_{\mathrm{V}} \mathrm{I}_{\mathrm{v}} \cos \phi .$ Maximum power is dissipated at resonance when $\mathrm{X}_{\mathrm{L}}=\mathrm{X}_{\mathrm{C}}$. The resonance frequency is given by
$v=\frac{1}{2 \pi \sqrt{\mathrm{LC}}}$
$=\frac{1}{2 \times 3.14 \times \sqrt{25 \times 10^{-3} \times 400 \times 10^{-6}}}$
$=\frac{1}{2 \times 3.14 \times 5 \times 20 \times 10^{-5} \sqrt{10}}$
$=\frac{10^{5}}{3.14 \times 200 \sqrt{10}}$
$=\frac{10^{3}}{6.28 \times \sqrt{10}}=\frac{159.2}{3.16}=50.31 \mathrm{~Hz}$
$v=\frac{1}{2 \pi \sqrt{\mathrm{LC}}}$
$=\frac{1}{2 \times 3.14 \times \sqrt{25 \times 10^{-3} \times 400 \times 10^{-6}}}$
$=\frac{1}{2 \times 3.14 \times 5 \times 20 \times 10^{-5} \sqrt{10}}$
$=\frac{10^{5}}{3.14 \times 200 \sqrt{10}}$
$=\frac{10^{3}}{6.28 \times \sqrt{10}}=\frac{159.2}{3.16}=50.31 \mathrm{~Hz}$
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.