SATHEE: Electromagnetic Induction and Alternating Current 4 Question 21

Electromagnetic Induction and Alternating Current 4 Question 21

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

$(1996, 3 M)$

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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

$i = i_{0} (1 - e^{- t / τ_{L}})$

Here, $i_{0} =$ Maximum value of current

$τ_{L} =$ Time constant $= L / R$

$τ_{L} = \frac{10 H}{2 Ω} = 5 s$

Therefore, $i = i_{0} / 2 = i_{0} (1 - e^{- t / 5})$

$\begin{aligned} or & \frac{1}{2} & = 1 - e^{- t / 5} or e^{- t / 5} = \frac{1}{2} \\ or & - t / 5 & = \ln \frac{1}{2} or t / 5 = \ln (2) = 0.693 \\ ∴ & t & = (5) (0.693) or t = 3.465 s \end{aligned}$

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Electromagnetic Induction and Alternating Current 4 Question 21

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