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In given $L - R$ circuit
Magnitude of rate of change of current is $\frac{\varepsilon }{{2L}}$ when energy stored in inductor is $\frac{{L{\varepsilon ^2}}}{{8{R^2}}}$
Magnitude of rate of change of current is $\frac{\varepsilon }{{L}}$ when energy stored in inductor is $\frac{{L{\varepsilon ^2}}}{{4{R^2}}}$
Magnitude of rate of change of current is $\frac{\varepsilon }{{2L}}$ when energy stored in inductor is $\frac{{L{\varepsilon ^2}}}{{4{R^2}}}$
All the options may be correct.
Solution
$\frac{1}{2} \mathrm{LI}^{2}=\frac{\mathrm{L} \varepsilon^{2}}{8 \mathrm{R}^{2}}$
$I=\frac{\varepsilon}{2 R}$
$\mathrm{IR}=\frac{\varepsilon}{2}$
So $\frac{\mathrm{LdI}}{\mathrm{dt}}=\frac{\varepsilon}{2}$
$\frac{\mathrm{d} \mathrm{I}}{\mathrm{dt}}=\frac{\varepsilon}{2 \mathrm{L}}$
