In the given figure,an inductor of L = 4 , H | vedclass

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(C) The energy stored in an inductor is given by $U = \frac{1}{2} L I^2$.The rate of energy storage is $P = \frac{dU}{dt} = L I \frac{dI}{dt}$.For an

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$25$

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(C) The energy stored in an inductor is given by $U = \frac{1}{2} L I^2$.The rate of energy storage is $P = \frac{dU}{dt} = L I \frac{dI}{dt}$.For an $R-L$ circuit,the current at time $t$ is $I = \frac{E}{R} (1 - e^{-tR/L})$.The rate of change of current is $\frac{dI}{dt} = \frac{E}{L} e^{-tR/L}$.Substituting these into the power equation:$P = L \left[ \frac{E}{R} (1 - e^{-tR/L}) ight] \left[ \frac{E}{L} e^{-tR/L} ight] = \frac{E^2}{R} (e^{-tR/L} - e^{-2tR/L})$.To find the maximum rate,we differentiate $P$ with respect to $t$ and set it to zero:$\frac{dP}{dt} = \frac{E^2}{R} \left( -\frac{R}{L} e^{-tR/L} + \frac{2R}{L} e^{-2tR/L} ight) = 0$.This implies $e^{-tR/L} = 2 e^{-2tR/L}$,so $e^{tR/L} = 2$,or $t = \frac{L}{R} \ln 2$.At this time,$e^{-tR/L} = \frac{1}{2}$.Substituting this back into the power equation:$P_{max} = \frac{E^2}{R} \left( \frac{1}{2} - (\frac{1}{2})^2 ight) = \frac{E^2}{R} (\frac{1}{4}) = \frac{E^2}{4R}$.Given $R = 25 \, \Omega$,we have $P_{max} = \frac{E^2}{4 	imes 25} = \frac{E^2}{100}$.Comparing this with $\frac{E^a}{2b}$,we get $a = 2$ and $b = 50$.Therefore,$\frac{b}{a} = \frac{50}{2} = 25$.

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