At t = 0, a battery is connected to a series arrangement of a resistor and an inductor. If the inductive time constant is 36.0 ms, at what time is the rate at which energy is dissipated in the resistor equal to the rate at which energy is stored in the inductor's magnetic field?

The rate at which energy is dissipated in the resistor is equal to the rate at which energy is stored in the inductor's magnetic field in 24.95 ms.

Explanation:

The energy stored in the inductor is given as

E₁ = ½LI²

The rate at which energy is stored in the inductor is

(dE₁/dt) = (d/dt) (½LI²)

Since L is a constant

(dE₁/dt) = ½L × 2I (dI/dt) = LI (dI/dt)

(dE₁/dt) = LI (dI/dt)

Rate of Energy dissipated in a resistor = Power = I²R

(dE₂/dt) = I²R

When the rate at which energy is dissipated in the resistor equal to the rate at which energy is stored in the inductor's magnetic field

(dE₁/dt) = (dE₂/dt)

OK (dI/dt) = I²R

L (dI/dt) = IR

Current in a this kind of series setup of inductor and resistor at any time, t, is given as

I = (V/R) (1 - e⁻ᵏᵗ)

k = (1/time constant) = (R/L)

(dI/dt) = (kV/R) e⁻ᵏᵗ = (RV/RL) e⁻ᵏᵗ = (V/L) e⁻ᵏᵗ

L (dI/dt) = IR

L [(V/L) e⁻ᵏᵗ] = R [(V/R) (1 - e⁻ᵏᵗ)

V e⁻ᵏᵗ = V (1 - e⁻ᵏᵗ)

e⁻ᵏᵗ = 1 - e⁻ᵏᵗ

2 e⁻ᵏᵗ = 1

e⁻ᵏᵗ = (1/2) = 0.5

e⁻ᵏᵗ = 0.5

In e⁻ᵏᵗ = In 0.5 = -0.69315

- kt = -0.69315

kt = 0.69315

k = (1/time constant)

Time constant = 36.0 ms = 0.036 s

k = (1/0.036) = 27.78

27.78t = 0.69315

t = (0.69315/27.78) = 0.02495 = 24.95 ms

Hope this Helps!!!

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