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JEE Main 2023 (13 Apr, Shift 1) · Inductors & DC Transients (Mixed)

Inductors & DC Transients (Mixed) question from JEE Main 2023 (13 Apr, Shift 1)

Q6JEE Main 2023 · 13 APR · SHIFT 1Difficult

Inductors & DC Transients (Mixed)

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}{2b}\text{ 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 __________.

Figure

Numerical answer type. Work it out, then open the answer.

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Answer: 25

  1. Rate of energy storage P=ddt(12Li2)=Lididt=Ei−i2RP = \frac{d}{dt}\left(\frac{1}{2}Li^2\right) = Li\frac{di}{dt} = Ei - i^2R
  2. Maximum when dPdi=E−2iR=0⇒i=E2R\frac{dP}{di} = E - 2iR = 0 \Rightarrow i = \frac{E}{2R}, giving Pmax=E24RP_{max} = \frac{E^2}{4R}
  3. With R=25 ΩR = 25\ \Omega: Pmax=E2100=E22×50P_{max} = \frac{E^2}{100} = \frac{E^2}{2 \times 50}, so a=2a = 2, b=50b = 50
  4. ba=25\frac{b}{a} = 25
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