Electromagnetic Induction and Alternating Current 4 Question 21
####21. A solenoid has an inductance of $10 \mathrm{H}$ and a resistance of $2 \Omega$. It is connected to a $10 \mathrm{~V}$ battery. How long will it take for the magnetic energy to reach $1 / 4$ of its maximum value?
$(1996,3 \mathrm{M})$
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Answer:
Correct Answer: 21. 3.465 s
Solution:
- $U=\frac{1}{2} L i^{2}$ i.e. $U \propto i^{2}$
$U$ will reach $\frac{1}{4}$ th of its maximum value when current is reached half of its maximum value. In $L-R$ circuit, equation of current growth is written as
$$ i=i_{0}\left(1-e^{-t / \tau_{L}}\right) $$
Here, $i_{0}=$ Maximum value of current
$\tau_{L}=$ Time constant $=L / R$
$$ \tau_{L}=\frac{10 \mathrm{H}}{2 \Omega}=5 \mathrm{~s} $$
Therefore, $\quad i=i_{0} / 2=i_{0}\left(1-e^{-t / 5}\right)$
$$ \begin{aligned} & \text { or } & \frac{1}{2} & =1-e^{-t / 5} \text { or } e^{-t / 5}=\frac{1}{2} \\ & \text { or } & -t / 5 & =\ln \frac{1}{2} \text { or } \quad t / 5=\ln (2)=0.693 \\ \therefore & & t & =(5)(0.693) \text { or } \quad t=3.465 \mathrm{~s} \end{aligned} $$
