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i(0+) and di/dt(0+) with sinusoidal source

Short answer

i1(0⁺) = V0 sin φ / R, i2(0⁺) = 0, di1/dt(0⁺) = (V0/R)[ω cos φ − sin φ/(RC)], di2/dt(0⁺) = V0 sin φ / L.

In the given circuit, switch K is closed at time t = 0. Find i1(0⁺), i2(0⁺), di1(0⁺)/dt and di2(0⁺)/dt. [Figure: source Vo sin(ωt + φ) with switch K (closes at t = 0), feeding two parallel branches: a resistor R in series with a capacitor C (current i1), and a resistor R in series with an inductor L (current i2)]

Assumptions: the source is v(t) = V₀ sin(ωt + φ), the capacitor is initially uncharged, and the inductor current is initially zero. Branch 1 is R + C (current i₁), branch 2 is R + L (current i₂). Both branches are connected directly across the source after K closes.

Values at t = 0⁺

  • Capacitor voltage v_C(0⁺) = 0 (continuous).
  • Inductor current i₂(0⁺) = 0 (continuous).
  • Source voltage v(0⁺) = V₀ sin φ.

Branch 1 (R–C): v = R i₁ + v_C

i₁(0⁺) = [v(0⁺) − v_C(0⁺)]/R = V₀ sin φ / R

Branch 2 (R–L): v = R i₂ + L di₂/dt

i₂(0⁺) = 0
di₂/dt (0⁺) = [v(0⁺) − R i₂(0⁺)]/L = V₀ sin φ / L

Derivative of i₁

Differentiate v = R i₁ + v_C, noting dv_C/dt = i₁/C:

dv/dt = R di₁/dt + i₁/C
di₁/dt = (1/R)[dv/dt − i₁/C]

with dv/dt (0⁺) = ωV₀ cos φ:

di₁/dt (0⁺) = ωV₀ cos φ / R − V₀ sin φ /(R²C)

Answer:

  • i₁(0⁺) = V₀ sin φ / R
  • i₂(0⁺) = 0
  • di₁/dt (0⁺) = (V₀/R)[ω cos φ − sin φ/(RC)]
  • di₂/dt (0⁺) = V₀ sin φ / L

✓ Answer checked by Claude Opus 5.5 (high effort)

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