In the given figure, an inductor and resistor are connected in...

The formula to calculate the current flowing through an L-R circuit is given by

$i = \dfrac{E}{R}\left[1 - e\dfrac{- Rt}{L}\right] . . . \left(1\right)$

The energy stored within the inductor can be calculated as follows:

$\begin{cases}{E}_{\text{induction}} & = & \dfrac{1}{2}L{i}^{2} \\ & = & \dfrac{1}{2}L\dfrac{{E}^{2}}{{R}^{2}}{\left[1 - e\dfrac{- Rt}{L}\right]}^{2} . . . \left(2\right)\end{cases}$

The rate of change of energy stored within the inductor is given by

$\begin{cases}\dfrac{d}{dt}\left({E}_{\text{induction}}\right) & = & 2\left[1 - e\dfrac{- Rt}{L}\right] \cdot e\dfrac{- Rt}{L} \cdot \dfrac{R}{L} \cdot \dfrac{{E}^{2}L}{2{R}^{2}} \\ & = & \dfrac{{E}^{2}}{R}\left[e\dfrac{- Rt}{L} - e\dfrac{- 2Rt}{L}\right] . . . \left(3\right)\end{cases}$

For maximum rate,

$\dfrac{{d}^{2}E}{d{t}^{2}} = 0 . . . \left(4\right)$

From equation (3) and (4), it can be written that

$\begin{cases}\dfrac{d}{dt}\left[\dfrac{{E}^{2}}{R}\left[e\dfrac{- Rt}{L} - e\dfrac{- 2Rt}{L}\right]\right] & = & 0 \\ & ⇒ & e\dfrac{- Rt}{L} = 2e\dfrac{- 2Rt}{L} \\ & ⇒ & e\dfrac{Rt}{L} = 2 . . . \left(5\right)\end{cases}$

From equation (3) and (5), it can be written, under maximum energy stored condition for the circuit, that

$\begin{cases}\dfrac{d}{dt}\left({E}_{\text{induction}}\right) & = & \dfrac{{E}^{2}}{R}\left[{2}^{- 1} - {2}^{- 2}\right] \\ & = & \dfrac{{E}^{2}}{4R}\end{cases}$

Substitute the values of the resistance and compare with the given expression to obtain the values of the parameters.

$\begin{cases}a & = & 2 \\ b & = & 2R \\ & = & 50\end{cases}$

Hence, the required ratio is given by

$\dfrac{b}{a} = 25$