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The current generator provides a constant current \$I\$ and the circuit is at règime conditions. At \$t=0\$, the switch T is being closed.

I have to calculate the voltage \$v_c(t)\$ by using the Laplace (unilateral) transform.

schematic

simulate this circuit – Schematic created using CircuitLab

After having calculated the initial condition (at \$t=0^-\$) \$I_0\$ and \$V_0\$, I can draw the circuit in the frequency domain, calculate \$V_c(s)\$ and antitrasform to get \$v_c(t)\$.

schematic

simulate this circuit

In the above circuit \$R_1\$ doesn't appear because it's shortcircuited, but why \$I\$ doesn't appear too? Could anybody explain-me just this, please?

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The current source doesn't influence the circuit after the switch has closed because all the current it provides will flow through the switch itself, therefore it is not useful to the computation of \$V_c\$ in any way.

To see why, think of two impedances in parallel with the current source: one is the switch, the other is the equivalent of the circuit of all the other elements to the right of the switch. Then apply the current divider formula and you'll see that the zero-impedance (ideal) switch will hog all the current from the source.

schematic

simulate this circuit – Schematic created using CircuitLab

\[ I_x = I \cdot \dfrac {Z_{sw}} {Z_{sw} + Z_x} \]

\$I_x = 0\$ if the switch is closed, i.e. if \$Z_{sw} = 0\$.

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