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Laplace: loop currents with 10 V charged capacitor

Short answer

Mesh analysis in s-domain with initial capacitor voltage 10 V and zero inductor current. Poles at s=−1,−6 (overdamped). Inverse transform yields i1(t)=72e^−6t−2e^−t A, i2(t)=84e^−6t−14e^−t A.

Using Laplace transform method, find the loop currents i1 and i2 for t > 0 in the circuit shown in the following figure. [Figure: a capacitor C = 1 F charged to 10 V (+ on top plate) in the left mesh (current i1) with a switch (t = 0) in the top wire; an inductor L = 1/6 H in the middle shared branch; a resistor R = 1/7 Ω in the right mesh (current i2). The scan prints R = 1/7 F, taken here as 1/7 Ω.]

Assumptions: the resistor value is taken as R = 1/7 Ω (the scan shows 1/7 "F"); both mesh currents are clockwise; the capacitor (C = 1 F, v_C(0) = 10 V, + on top) discharges; the inductor L = 1/6 H is the shared branch with zero initial current.

s-domain mesh equations

Mesh 1: 10/s = (1/s) I1 + (s/6)(I1 − I2)
Mesh 2: 0 = (s/6)(I2 − I1) + (1/7) I2

Mesh 2: I2 (s/6 + 1/7) = (s/6) I1 → I2 = 7s I1 / (7s + 6).

Then (s/6)(I1 − I2) = s I1/(7s + 6), and mesh 1 gives

10/s = I1 (7s + 6 + s²) / [ s (7s + 6) ]
I1(s) = 10 (7s + 6) / (s² + 7s + 6) = 10 (7s + 6) / [(s + 1)(s + 6)]
I2(s) = 70 s / [(s + 1)(s + 6)]

Partial fractions

I1 = −2/(s + 1) + 72/(s + 6)
I2 = −14/(s + 1) + 84/(s + 6)

Time domain

i₁(t) = 72 e^(−6t) − 2 e^(−t) A, t > 0

i₂(t) = 84 e^(−6t) − 14 e^(−t) A, t > 0

Check: i₁(0) = i₂(0) = 70 A = 10/(1/7), which is correct because the inductor blocks current changes at t = 0⁺ and the capacitor voltage is across R. Roots −1 and −6 (overdamped). At t = 0.5 s: i₁ = 2.372 A, i₂ = −4.309 A, equal to the numerical inverse Laplace transform.

✓ Answer checked by Claude Opus 5.5 (high effort)

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