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Laplace: series RLC with 6e^(-2t) source

Short answer

Write s-domain loop equation with source 6/(s+2) and initial capacitor voltage 2 V opposing. Solve I(s), partial fractions give three exponential modes: i(t)=−8/3 e^−t+6e^−2t−10/3 e^−4t A. Integrate to obtain vC(t)=32/3 e^−t−12e^−2t+10/3 e^−4t V.

In a series RLC, as shown in figure below find the value of current for t > 0, also find the voltage across capacitor for t > 0, using Laplace transform method. [Figure: source v(t) = 6e^(−2t) with a switch (t = 0), a 5 Ω resistor, a 1 H inductor and a ¼ F capacitor with an initial voltage of 2 V (+ on the top plate) in a single loop with current i]

Assumptions: R = 5 Ω, L = 1 H, C = ¼ F, no initial inductor current, and the capacitor is connected so that i charges its + plate, i.e. its initial voltage v_C(0) = 2 V opposes the loop current direction in the usual way (drop convention).

s-domain equation

Source V(s) = 6/(s + 2); impedance Z(s) = R + sL + 1/(sC) = s + 5 + 4/s. The charged capacitor is a series source 2/s opposing the current.

I(s) = [ 6/(s + 2) − 2/s ] / (s + 5 + 4/s)
     = [ 4(s − 1) / (s (s + 2)) ] × s / (s² + 5s + 4)
     = 4 (s − 1) / [ (s + 1)(s + 2)(s + 4) ]

Partial fractions for the current

K1 = 4(−2) / ((1)(3))   = −8/3     (s = −1)
K2 = 4(−3) / ((−1)(2))  = 6        (s = −2)
K3 = 4(−5) / ((−3)(−2)) = −10/3    (s = −4)

i(t) = −(8/3) e^(−t) + 6 e^(−2t) − (10/3) e^(−4t) A, t > 0

(i(0) = −2.667 + 6 − 3.333 = 0 ✓.)

Capacitor voltage

v_C = v_C(0) + (1/C)∫i dt, so in the s-domain Vc(s) = 2/s + 4 I(s)/s:

Vc(s) = 2 (s² + 7s + 22) / [ (s + 1)(s + 2)(s + 4) ]
      = (32/3)/(s + 1) − 12/(s + 2) + (10/3)/(s + 4)

v_C(t) = (32/3) e^(−t) − 12 e^(−2t) + (10/3) e^(−4t) V, t > 0

Check: v_C(0) = 10.667 − 12 + 3.333 = 2 V ✓. Substituting both expressions in the loop equation i′ + 5i + v_C = 6e^(−2t) (L = 1) gives equality at every t; for example at t = 0.5 s both sides equal 2.207. The formulas also agree with a numerical solution of the loop equations (error below 10⁻¹²).

✓ Answer checked by Claude Opus 5.5 (high effort)

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