In the given figure, an inductor and a resistor are connected in series with a battery of emf E volt. represents the maximum rate at which the energy is stored in the magnetic field (inductor). The numerical value of will be _________.

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Solution

Q.
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 _________. [2023]

Ans.
(25)

$E = \frac{1}{2} L I^{2}$

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

$Now we know for an R - L circuit$

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

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

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

$Time at which rate of power storing will be maximum, t = \frac{L}{R \ln 2}$

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

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

$a = 2, b = 50$

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

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