Physics - EMI/AC Question with Solution | TestHub

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

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

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

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

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

$\begin{array}{rcl}\frac{d}{dt}\left(E_{\text{induction }}\right)& =& 2\left[1-e^{\frac{-Rt}{L}}\right]·e^{\frac{-Rt}{L}}·\frac{R}{L}·\frac{E^{2}L}{2R^2}\\ & =& \frac{E^2}{R}\left[e^{\frac{-Rt}{L}}-e^{\frac{-2Rt}{L}}\right] ...\left(3\right)\end{array}$

For maximum rate,

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

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

$\begin{array}{rcl}\frac{d}{dt}\left[\frac{{\displaystyle E^2}}{{\displaystyle R}}\left[e^{\frac{{\displaystyle -Rt}}{{\displaystyle L}}}-e^{\frac{{\displaystyle -2Rt}}{{\displaystyle L}}}\right] \right]& =& 0\\ & \Rightarrow & e^{\frac{-Rt}{L}}=2e^{\frac{-2Rt}{L}} \\ & \Rightarrow & e^{\frac{Rt}{L}}=2 ...\left(5\right)\end{array}$

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

$\begin{array}{rcl}\frac{d}{dt}\left(E_{\text{induction }}\right)& =& \frac{E^2}{R}\left[2^{-1}-2^{-2}\right]\\ & =& \frac{E^2}{4R}\end{array}$

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

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

Hence, the required ratio is given by

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