Given:
V(0⁺) = 90 V
I_L(0⁺) = - 30 mA

Find:
a) The initial current in each branch of the circuit.
b) V(t) for t > 0
c) i_L(t) for t > 0

a)

I_R(0⁺) = V(0⁺) / 2000 = 90 / 2000 = 45 mA
I_R(0⁺) = 45 mA

I_C(0⁺) + I_L(0⁺) + I_R(0⁺) = 0
I_C(0⁺) = - I_L(0⁺) - I_R(0⁺) = 30 mA - 45 mA
I_C(0⁺) = - 15 mA

a = 1 / 2RC = 1 / [2 (2000) (10 n)] = 25 K

Since a > w₀, the circuit is overdamped, thus the voltage across the parallel circuit has the form

V(t) = A₁ e^{S1 t} + A₂ e^{S2 t} V

To find the values of s₁ and s₁, we know that

and inserting the values for a and w0 from above yields

s₁ = 40 K
s₂ = 10 K

We also know that for an overdamped parallel circuit that

V(0⁺) = A₁ + A₂

and

I_C(0⁺) / C = s₁ A₁ + s₂ A₂

This yields the two equations

90 = A₁ + A₂
- 15 m / 10 n = - 1.5 M = 40 K A₁ + 10 K A₂

Solving these two equations yields

A₁ = 20
A₂ = 70

Thus

V(t) = 20 e- 40,000 t + 70 e- 10,000 t V

c)

There are two methods to get i_L(t) once we have V(t). I will use both methods, which will also serve as a check of the work.

Method 1: Kirchoff's Current Law (I think this is the easiest in most cases.)

First obtain i_R(t) and i_C(t), then I_L(t) = - I_R(t) - I_C(t)

i_R(t) = V(t) / R = [20 e^{- 40,000 t} + 70 e^{- 10,000 t}] / 2000
i_R(t) = 10 e^{- 40,000 t} + 35 e^{- 10,000 t} mA

Next, find i_C(t) = C dv/dt

i_C(t) = 10 n [20 (- 40,000) e^{- 40,000 t} + 70 ( - 10,000) e^{- 10,000 t}]
i_C(t) = - 8 e^{- 40,000 t} - 7 e^{- 10,000 t} mA

Thus

I_L(t) = - I_R(t) - I_C(t)
I_L(t) = - 2 e- 40,000 t - 28 e- 10,000 t mA

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Method 2: Integration

For this method we will use the relationship between the inductor voltage and the current through it

I_L(t) = [- 2 m e^{- 40,000 t} - 28 m e^{- 10,000 t} + 2 m + 28 m] - 30 mA
I_L(t) = - 2 e- 40,000 t - 28 e- 10,000 t mA

which agrees with our previous result.