The time taken for the magnetic energy to rea | vedclass

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(C) The magnetic energy stored in an inductor is given by $U = \frac{1}{2} L i^2$.Let $U_{max}$ be the maximum energy,which occurs when the current re

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$\frac{L}{R} \ln 2$

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(C) The magnetic energy stored in an inductor is given by $U = \frac{1}{2} L i^2$.Let $U_{max}$ be the maximum energy,which occurs when the current reaches its maximum value $i_0$. Thus,$U_{max} = \frac{1}{2} L i_0^2$.We are given that the energy reaches $25\%$ of its maximum value:$U = 0.25 U_{max} = \frac{1}{4} (\frac{1}{2} L i_0^2) = \frac{1}{2} L (\frac{i_0}{2})^2$.Comparing this with $U = \frac{1}{2} L i^2$,we find that the current $i$ must be $\frac{i_0}{2}$.The expression for current in an $RL$ circuit during charging is $i = i_0 (1 - e^{-Rt/L})$.Setting $i = \frac{i_0}{2}$,we get: $\frac{i_0}{2} = i_0 (1 - e^{-Rt/L})$.$\frac{1}{2} = 1 - e^{-Rt/L} \Rightarrow e^{-Rt/L} = \frac{1}{2}$.Taking the natural logarithm on both sides: $-\frac{Rt}{L} = \ln(\frac{1}{2}) = -\ln 2$.Therefore,$t = \frac{L}{R} \ln 2$.

The time taken for the magnetic energy to reach $25\%$ of its maximum value,when a solenoid of resistance $R$ and inductance $L$ is connected to a battery,is:

Similar Questions

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