Steady state with s closed. C is open I_s = 4. This source current is divided between two resistors Ω and 1 Ω. Across 1 Ω, C is open, I in 1 Ω is = 1 A. Hence voltage across C is 1 V. The initial charge on C is V₀ = 1 V. At t = 0⁺, s is opened bringing R = 1 Ω into the circuit. C is replaced by 1 V source with positive polarity left. The circuit has two sources I_s = 4 and V₀ = 1. The currents in the resistor R = and R = 1 can be determined by superposition. For R = Due to I_s only (short V₀) I = I_s x= 3 Due to V₀ only (open I_s) I == Downwards Adding, current through R = is 3= .