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Laplace: current i(t) with two capacitor branches

Short answer

Combine parallel RC branches to get equivalent series R = 12.5 Ω and C = 1 F. Source current in s-domain is 4/(s+0.08). Inverse transform gives i(t) = 4 e^(−0.08t) A, t > 0.

Using laplace transformation technique, find the expression for current i(t) in the network shown below for t > 0 when the switch is closed at t = 0. Assume zero initial charge across the capacitors. [Figure: 50 V source with a switch (t = 0) and a 10 Ω resistor carrying current i, feeding two parallel branches: a capacitor (value not labelled in the scan) in series with a 5 Ω resistor, and a 5 Ω resistor in series with a 0.5 F capacitor]

Assumption: the unlabelled capacitor is 0.5 F, the same as the other branch (the scan does not show it). Both capacitors are initially uncharged. Source = 50/s.

Branch impedances

  • Branch 1: Z₁ = 5 + 1/(0.5 s) = 5 + 2/s
  • Branch 2: Z₂ = 5 + 2/s (identical)

The two identical branches in parallel:

Z_p = (5 + 2/s)/2 = 2.5 + 1/s

Source current

I(s) = (50/s) / (10 + 2.5 + 1/s)
     = 50 / (12.5 s + 1)
     = 4 / (s + 0.08)

i(t) = 4 e^(−0.08t) A, t > 0

This is a single exponential because the circuit reduces to R_eq = 12.5 Ω in series with C_eq = 1 F: τ = 12.5 s.

Check: i(0⁺) = 50/12.5 = 4 A (both capacitors act as shorts, so 10 + 5∥5 = 12.5 Ω) ✓, i(∞) = 0 ✓. At t = 0.1 s, i = 3.968 A, which matches the numerical inverse Laplace transform.

If the unlabelled capacitor were 1 F instead: I(s) = 50(10s + 3)/(125s² + 45s + 2), giving i(t) = 3.874 e^(−0.0519t) + 0.126 e^(−0.308t) A.

✓ Answer checked by Claude Opus 5.5 (high effort)

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