In the given figure, an inductor and a resistor are connected in series with a battery of emf \(E\) volt. \(\frac{E^a}{2 b }\,J / s\) represents the maximum rate at which the energy is stored in the magnetic field (inductor). The numerical value of \(\frac{ b }{ a }\) will be ............

Rate of energy storing \(=\frac{d E}{d t}=L I \frac{d I}{d t}\)

Now we Know for \(R-L\) circuit

\(I=\frac{E}{R}\left(1-e^{-t \frac{R}{L}}\right)\)

So \(\frac{d I}{d t}=\frac{E}{L} e^{-\frac{t R}{L}}\)

\(\frac{d E}{d t}=\frac{E^2}{R}\left(1-e^{-\frac{t R}{L}}\right)\left(e^{-t \frac{R}{L}}\right)\)

Time at which rate of power storing will be \(\max\)

\(t =\frac{L}{R \ln 2}\)

So \(\frac{d E}{d t}=\frac{E^2}{R}\left(1-\frac{1}{2}\right) \times \frac{1}{2}\)

\(\Rightarrow \frac{E^2}{4 R}=\frac{E^2}{100}=\frac{E^2}{2 \times 50}\)

\(a=2, b=50\)

So \(\frac{b}{a}=25\)