# Analyzing circuit with dirac delta current source without using Laplace transformation

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?

• $v_0(t)$ 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. – 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.