JEE MAIN - Physics (2023 - 13th April Morning Shift - No. 24)
In the given figure, an inductor and a resistor are connected in series with a battery of emf E volt. $$\frac{E^{a}}{2 b} \mathrm{~J} / s$$ represents the maximum rate at which the energy is stored in the magnetic field (inductor). The numerical value of $$\frac{b}{a}$$ will be __________.
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Answer
25
Explanation
$$
E=\frac{1}{2} L I^2
$$
Rate of energy storing $=\frac{d E}{d t}=L I \frac{d I}{d t}$
Now we Know for $R-L$ 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 $\max$
$$ \mathrm{t}=\frac{L}{R \ln 2} $$
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 $\frac{b}{a}=25$
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 $\max$
$$ \mathrm{t}=\frac{L}{R \ln 2} $$
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 $\frac{b}{a}=25$
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