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Laplace: i1, i2 with 2 V charged capacitor

Short answer

Set up mesh equations in s-domain with initial capacitor voltage 2 V and zero inductor current. Solve for I1(s) and I2(s), partial fractions give overdamped poles −1 and −8. Inverse transform yields i1(t)=18.29e^−8t−0.286e^−t A, i2(t)=20.57e^−8t−2.571e^−t A.

Using Laplace Transform method in the circuit shown in figure below. Find the current i1 and i2 for t > 0. [Figure: a capacitor C = 1 F charged to 2 V (+ on top plate) in the left mesh (current i1) with a switch (t = 0) in the top wire; an inductor L = 1/8 H in the middle shared branch; a resistor R = 1/9 Ω in the right mesh (current i2)]

Assumptions: both mesh currents i₁ (left) and i₂ (right) are clockwise, the capacitor (C = 1 F, +ve on top plate, v_C(0) = 2 V) discharges through the circuit, and there is no initial inductor current. The source-free circuit has C in the left mesh, L = 1/8 H shared, and R = 1/9 Ω in the right mesh.

s-domain mesh equations

The charged capacitor becomes a source V₀/s = 2/s in series with 1/(sC) = 1/s.

Mesh 1: 2/s = (1/s) I1 + (s/8)(I1 − I2)
Mesh 2: 0 = (s/8)(I2 − I1) + (1/9) I2

From mesh 2: I2 (s/8 + 1/9) = (s/8) I1, so I2 = 9s I1 / (9s + 8).

Substitute in mesh 1 ( (s/8)(I1 − I2) = (s/8) I1 × 8/(9s + 8) = s I1/(9s + 8) ):

2/s = I1 [ 1/s + s/(9s + 8) ] = I1 (9s + 8 + s²) / [ s (9s + 8) ]
I1(s) = 2 (9s + 8) / (s² + 9s + 8) = 2 (9s + 8) / [(s + 1)(s + 8)]
I2(s) = 18 s / [(s + 1)(s + 8)]

Partial fractions

I1 = −(2/7)/(s + 1) + (128/7)/(s + 8)
I2 = −(18/7)/(s + 1) + (144/7)/(s + 8)

Time domain

i₁(t) = 18.29 e^(−8t) − 0.286 e^(−t) A, t > 0

i₂(t) = 20.57 e^(−8t) − 2.571 e^(−t) A, t > 0

Check: i₁(0) = 18.29 − 0.286 = 18 A and i₂(0) = 20.57 − 2.571 = 18 A. At t = 0⁺ the inductor carries no current, so the 2 V capacitor drives 2/(1/9) = 18 A through R ✓. The system is overdamped (roots −1 and −8). Values at t = 0.5 s: i₁ = 0.1616 A, i₂ = −1.183 A, matching the numerical inverse transform.

✓ Answer checked by Claude Opus 5.5 (high effort)

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