An inductor L = 4 H and resistor R = 25 Ω are in series with a DC source of emf E. If (aE²)/(2b) J s⁻¹ represents the maximum rate at which magnetic energy is stored, find b/a.

Question

G1 Physics · Accepted Answer

For an LR growth circuit: I = I₀(1 - e⁻ᵗᐟτ) where I₀ = E/R, τ = L/R Magnetic energy: U = (1/2)LI² Rate of storage: P = dU/dt = LI(dI/dt) Let x = e⁻ᵗᐟτ. Then: I = I₀(1-x) dI/dt = (I₀/τ)x So, P = (LI₀²/τ)x(1-x) This is maximum at x = 1/2. Therefore, Pₘₐₓ = LI₀²/(4τ) Since τ=L/R: Pₘₐₓ = I₀²R/4 = E²/(4R) The question writes this as: aE²/(2b) Thus, a/(2b) = 1/(4R) which gives b/a = 2R = 50/2 = 25 Final value: 25