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QJEEM23Phb6bc: JEE Main 2023 Physics - Class 12 JEEM23Phb6bc Solution: In the given figure, an inductor and a resistor are connected…

In the given figure, an inductor and a resistor are connected in series with a battery of emf E volt. Ea2b J/s\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 ba\frac{b}{a} will be __________.

JEE Main 2023 (Online) 13th April Morning Shift Physics - Alternating Current Question 38 English

Step-by-Step Verified Solution

E=12LI2E=\frac{1}{2} L I^2

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

Now we Know for R−LR-L circuit

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

So dIdt=ELe−tRL\frac{d I}{d t}=\frac{E}{L} e^{-\frac{t R}{L}}

dEdt=E2R(1−e−tRL)(e−tRL)\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⁡\max

t=LRln⁡2\mathrm{t}=\frac{L}{R \ln 2}

So dEdt=E2R(1−12)×12\frac{d E}{d t}=\frac{E^2}{R}\left(1-\frac{1}{2}\right) \times \frac{1}{2}

⇒E24R=E2100=E22×50\Rightarrow \frac{E^2}{4 R}=\frac{E^2}{100}=\frac{E^2}{2 \times 50}

a=2,b=50a=2, b=50

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

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