Get Answers to all your Questions

header-bg qa

A solenoid of inductance 100 mH and resistance 20\Omega is connected to a cell of emf 10 V. What is the energy stored in the inductor when the time t = 5 ln 2 milli sec.

Option: 1

3.125 \times 10^{-3} \mathrm{~J}


Option: 2

4.25 \times 10^{-3} \mathrm{~J}


Option: 3

4.725 \times 10^{-3} \mathrm{~J}


Option: 4

5.15 \times 10^{-3} \mathrm{~J}


Answers (1)

best_answer

The current at any time can be given as \mathrm{\mathrm{i}=\mathrm{i}_0\left(1-\mathrm{e}^{-\mathrm{t} / \tau}\right)}

\mathrm{\text { Where } \tau=\frac{L}{R}=\frac{100 \mathrm{mH}}{20 \Omega}=5 \times 10^{-3} \mathrm{~s}}

\mathrm{\begin{aligned} \mathrm{i} & =\mathrm{i}_{\circ}\left(\frac{5 \times 10^{-3}}{1-\mathrm{e}(5 \ln 2) \times 10^{-3}}\right) \\ & =\mathrm{i}_{\circ}\left(1-\frac{1}{\mathrm{e}^{\ln 2}}\right)=\mathrm{i}_{\circ}\left(1-\frac{1}{2}\right)=\frac{\mathrm{i}_{\circ}}{2} \end{aligned}}

\mathrm{\text { Where } \mathrm{i}_0=\frac{10 \mathrm{~V}}{20 \mathrm{ohm}}=0.5 \mathrm{Amp} \text {. }}

\mathrm{\therefore \text { The required energy stored }=\quad \frac{1}{2} \mathrm{Li}^2}

\mathrm{\begin{aligned} & =\frac{1}{2} \times\left(100 \times 10^{-3}\right)\left(\frac{1}{4}\right)^2 \\ & =\frac{1}{320} \mathrm{~J}=3.125 \times 10^{-3} \mathrm{~J} . \end{aligned}}

Posted by

Rishi

View full answer

NEET 2024 Most scoring concepts

    Just Study 32% of the NEET syllabus and Score up to 100% marks