In the given figure, an inductor and a resistor are connected in series with a battery of emf E volt. Ea2bJ/s represents the maximum rate at which the energy is stored in the magnetic field (inductor). The numerical value of ba will be _____.

E=12LI2Rate of storing energy=dE/dt=LIdldtNow for L-R circuitI=ER1−e−tRL So dIdt=ELe−tRLdEdt=E2R1−e−tRLe−tRLTime at which it will be maximumt=LRln2substituting in above equation So dEdt=E2R1−12×12⇒E24R=E2100=E22×50a=2,b=50 So ba=25

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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 _____.

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answer is 25.

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Detailed Solution

$E = \frac{1}{2} {LI}^{2}$

$Rate of storing energy$

$= dE / dt = LI (\frac{dl}{d t})$

Now for L-R circuit

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

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

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

Time at which it will be maximum

$t = \frac{L}{R} \ln (2)$

substituting in above equation

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

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

$\begin{aligned} a = 2, b = 50 \\ So \frac{b}{a} = 25 \end{aligned}$

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