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Laplace: current through 2 Ω, exponential source

Short answer

Transform exponential source, find parallel impedance. Current in 2 Ω resistor is I2(s)=1/[(s+3)(s+5)(s+6)]. Inverse transform: i2(t) = (1/6)e^(−3t) − (1/2)e^(−5t) + (1/3)e^(−6t) A, t>0.

Using Laplace transform method, find the expression for current through 2 Ω resistor for t > 0 in the circuit shown in figure. [Figure: source V = 1/10 e^(−3t) with a switch (t = 0), a 1 Ω resistor and a 0.1 H inductor in series; then a capacitor (labelled 0.5 Ω in the scan, presumably 0.5 F) and the 2 Ω resistor as parallel shunt branches]

Assumption: the capacitor labelled "0.5 Ω" in the scan is 0.5 F, in parallel with the 2 Ω resistor. All initial conditions are zero. Source V(s) = 0.1/(s + 3).

Parallel branch impedance

Y = sC + 1/R = 0.5 s + 0.5, so Z_p = 1/(0.5 s + 0.5) = 2/(s + 1).

Source current

I(s) = V(s) / (R + sL + Z_p) = [0.1/(s + 3)] / [ 1 + 0.1 s + 2/(s + 1) ]

Voltage across the parallel part and current in 2 Ω

Simplify the total impedance: 1 + 0.1s + 2/(s + 1) = (0.1 s² + 1.1 s + 3)/(s + 1) = 0.1 (s + 5)(s + 6)/(s + 1), so

V_p(s) = 0.1/(s + 3) × 2/[0.1 (s + 5)(s + 6)] = 2 / [ (s + 3)(s + 5)(s + 6) ]
I₂(s) = V_p/2 = 1 / [ (s + 3)(s + 5)(s + 6) ]

Partial fractions

I₂(s) = (1/6)/(s + 3) − (1/2)/(s + 5) + (1/3)/(s + 6)

(K₁ = 1/((2)(3)) = 1/6, K₂ = 1/((−2)(1)) = −1/2, K₃ = 1/((−3)(−1)) = 1/3)

i₂(t) = (1/6) e^(−3t) − (1/2) e^(−5t) + (1/3) e^(−6t) A, t > 0

Check: i₂(0) = 1/6 − 1/2 + 1/3 = 0 ✓ (the capacitor is initially a short across the 2 Ω). At t = 0.5 s, i₂ = 0.01274 A, equal to the numerical inverse Laplace value.

✓ Answer checked by Claude Opus 5.5 (high effort)

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