A 20 Henry inductor coil is connected to a 10 ohm resistance in series as shown in figure. The time at which the rate of dissipation of energy (joule's heat) across the resistance is equal to the rate at which magnetic energy is stored in the inductor is?

Correct option is C. 2ln2
L(dI/dt) = I²R
L(dI/dt) = I²R
For L=20H and R=10Ω
20(dI/dt) = 10I²
Let I = I₀e⁻ᵗ/τ where τ = L/R = 20/10 = 2s
dI/dt = (-I₀/2)e⁻ᵗ/²
Substituting in the equation above, we get:
20(-I₀/2)e⁻ᵗ/² = 10(I₀e⁻ᵗ/²)²
-10e⁻ᵗ/² = 10I₀e⁻ᵗ
-e⁻ᵗ/² = I₀e⁻ᵗ
-1 = I₀e⁻ᵗ/²
Since current can't be negative. Let's take the absolute value
1 = I₀e⁻ᵗ/²
At t=0, I₀ = 1A
1 = e⁻ᵗ/²
Taking natural log on both sides
0 = -t/2
t = 0
Let's consider the energy equations:
Rate of dissipation of energy across resistance = I²R
Rate of change of magnetic energy stored in the inductor = L(dI/dt)(I)
Equating both, we get:
I²R = L(dI/dt)I
IR = L(dI/dt)
In LR circuit, I = I₀(1 - e⁻ᵗ/τ)
Where I₀ = E/R and τ = L/R
Assuming I₀ = 1A
I = (1 - e⁻ᵗ/²)
dI/dt = (1/2)e⁻ᵗ/²
Substituting into IR = L(dI/dt)
(1 - e⁻ᵗ/²)(10) = 20(1/2)e⁻ᵗ/²
10 - 10e⁻ᵗ/² = 10e⁻ᵗ/²
10 = 20e⁻ᵗ/²
1/2 = e⁻ᵗ/²
Taking natural log on both sides:
ln(1/2) = -t/2
-ln2 = -t/2
t = 2ln2