In the given figure, an inductor and a resistor are connected in series with a battery of emf E volt. Eᵃ/(2 b) J / s represents the maximum rate at which the energy is stored in the magnetic field (inductor). The numerical value of b/a will be __________.

Solution

Physics sir · 4 steps

Energy flows into the coil at i(E - iR); this peaks when half the battery voltage sits across L.

1. Rate of storage. The energy in the inductor is U = 1/2Li², so dU/dt = Lidi/dt. Loop rule: Ldi/dt = E - iR, which gives dU/dt = Ei - i²R.
2. Find the maximum. Set d/di(Ei - i²R) = E - 2iR = 0. The rate peaks at i = E/2R, half of the final current.

   (dU/dt)ₘₐₓ = E·E/2R - E²/4R² R = E²/4R

3. Put in numbers. With R = 25 Ω, the maximum rate is E²/100 J s⁻¹. Matching this with Eᵃ/2b gives a = 2 and 2b = 100, so b = 50. The value of L does not affect the maximum.
4. So. b/a = 50/2 = 25.

The common slip: 50. Matching E²/100 to Eᵃ/b and forgetting the 2 in 2b gives b = 100, so b/a = 50, not 25.

The idea to keep. Power into an inductor is Ei - i²R, a downward parabola in i. It peaks at i = E/2R with value E²/4R, whatever the inductance is.

Answer from the NTA final answer key. Solution by PrepWiser.