At \$t=0\$ the switch is closed and \$v_o(0)= -5 V\$.

The problem asks for \$v_o(t)\$. I did it using Laplace method and get $$v_o(t)=5+10\,e^{-2t}$$ which does not even satisfy the initial condition, \$v_o(0)= -5 V\$.

Is this how it should be or I did a mistake?

How do I find \$v_o(t)\$ without using Laplace transformation?

  • \$\begingroup\$ \$v_0(t)\$ should be correct if you do your Laplace-domain calculations correctly. Show your work, and how the problem is worded, and we can help you find where you're confused. \$\endgroup\$ – TimWescott Dec 20 '18 at 20:03

At \$\small t=0\$ the impulse of strength \$\small 2\: A\$ deposits \$\small Q=2\:coulomb\$ of charge on the top plate. This is equivalent to adding \$\small 20\:V\$ to the top plate \$ \small \left( V=\frac{Q}{C}=\frac{2}{0.1}=20\:V\right)\$.

The voltage on the capacitor at \$\small t= 0^-\$ is: \$\small v_0(0^-)= -5\:V\$, hence the total initial voltage on the capacitor is: \$\small v_0(0)=20-5 = 15\:V\$

Now, using the general solution for a 1st order system with a step input: $$ v_0(t)= v_0(\infty)+\left[ v_0(0)-v_0(\infty)\right]e^{-t/\tau} $$

Converting the input to a Thevenin source with: \$\small V_{TH}=5\:V\$, and \$\small R_{TH}=5\:\Omega\$, gives: \$\small \tau=0.5\:sec\$,

hence: $$ v_0(t)= 5+\left[ 15-5\right]e^{-2t} $$

or $$ v_0(t)= 5+10e^{-2t} $$

So your LT analysis is correct.


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