Question

A solenoid has an inductance of $10$ henry and a resistance of $2$ ohm. It is connected to a $10$ volt battery. How long will it take for the magnetic energy to reach $1/4$ of its maximum value?

• A. $2.142\,s$
• B. $3.465\,s$
• C. $0.693\,s$
• D. $4.345\,s$

Answer 1

The Correct Option is B

The current grows in $L-R$ circuit till steady state is reached At steady state, energy stored in inductor is maximum At steady state, current $=I_{0}$ At some instant of time, current $= I$ $\therefore$ Maximum energy $=\frac{1}{2}LI_{0}^{2}$ Energy stored when current is $I=\frac{1}{2}LI^{2}$ $\therefore \frac{U_{max}}{U}=\frac{2\,LI_{0}^{2}}{2\,LI^{2}}$ or $ \frac{U_{max}}{\left(\frac{U_{max}}{4}\right)}=\left(\frac{I_{0}}{I}\right)^{2}$ or $\left(\frac{I_{0}}{I}\right)^{2}=4$ or $I=\frac{I_{0}}{2} \ldots\left(i\right)$ For growth of current in $L-R$ circuit, $I=I_{0} \left[1-e^{-\frac{R}{L}t}\right]$ or $\frac{I_{0}}{2}=I_{0}\left[1-e^{-\frac{2}{10}t}\right]$ or $\frac{1}{2}=1-e^{-t /5}$ or $e^{-t /5}=\frac{1}{2}$ or $t=5$ ln $2$ or $t=5\times0.693$ or $t=3.456\,s$