Question
In given $L - R$ circuit

Answer

$\frac{1}{2} \mathrm{LI}^{2}=\frac{\mathrm{L} \varepsilon^{2}}{8 \mathrm{R}^{2}}$

$I=\frac{\varepsilon}{2 R}$

$\mathrm{IR}=\frac{\varepsilon}{2}$

So $\frac{\mathrm{LdI}}{\mathrm{dt}}=\frac{\varepsilon}{2}$

$\frac{\mathrm{d} \mathrm{I}}{\mathrm{dt}}=\frac{\varepsilon}{2 \mathrm{L}}$

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