Initial currents i1, i2, i3 at t=0+ with C1, C2, L

Assumptions (figure values not given, so the answer is in symbols): the series resistor in the source branch is R (unlabelled); all energy-storing elements are initially relaxed (i_L = 0, v_C1 = v_C2 = 0) and the switch closes at t = 0 with V applied. Mesh currents are clockwise: i₁ in the source loop (V, R, R₁, L), i₂ in the upper loop (R₁, C₁) and i₃ in the lower loop (L, C₂). The current in R₁ is i₁ − i₂ and in L is i₁ − i₃.

Equivalent circuit at t = 0⁺

• L with zero current: open circuit, so i_L = i₁ − i₃ = 0.
• C₁ and C₂ uncharged: short circuits, so the top wire, the capacitor junction and the bottom wire are at the same potential.

Solution

1. The switch connects the R₁-L junction to the C₁-C₂ junction. With both capacitors shorted, the top wire, the bottom wire and both junctions are at the same potential: V_top = V_bottom.
2. So the whole source voltage V appears across the series resistor R in the source branch:

V = R·i₁   →   i₁(0⁺) = V/R

3. The voltage across R₁ is zero (its two ends are joined through the shorted C₁ and the closed switch), so its current is zero: i₁ − i₂ = 0, hence i₂ = i₁.
4. The inductor is open: i₁ − i₃ = 0, hence i₃ = i₁.

Answer: i₁(0⁺) = i₂(0⁺) = i₃(0⁺) = V/R. There is no current in R₁ or L; the whole current V/R flows through the series resistor and the short-circuited capacitors.