Electronics and Communication Engineering - Exam Questions Papers - Discussion

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= .