Electronic – Second-Order RLC Circuit and Short Circuits

circuit analysishomework

I was given this circuit to analyze.

Second-Order RLC Circuit

The first thing is I know that \$5u(t) = 5\$ for \$t>0\$ and \$5u(t) = 0\$ for \$t < 0\$. My book suggests to solve for initial conditions first (solve for \$v(0)\$ in the capacitor and solve for \$i(0)\$ in the inductor).

Here is where I got totally confused. For \$t < 0\$, the voltage source is short-circuited, giving me this circuit:
Drawing for t < 0

Which I argue that \$v(0^-)\$ = 0 because it is parallel to the short circuit. But, for \$t > 0\$, I drew this circuit:

Drawing for t > 0

Which I argue that \$v(0^+)\$ = 5 because it is parallel to the short circuit.

I know that for capacitors, the voltage cannot suddenly drop or increase, proving my analysis to be totally wrong. I checked using LTSPice,
LTSpice of v(t)

which shows that the correct value for \$v(0^-) = v(0^+) = v(0) = 5V\$. However, I cannot derive what seemingly to be a linear function for \$v(t)\$ when \$t < 0\$.

How do I obtain the general equation for \$v(t)\$ for the capacitor or \$i(t)\$ for the inductor?

EDIT: I think I got my LTSpice wrong. The graph should look like this:
enter image description here

which shows a discontinuity at \$t = 0\$. I don't understand why this happens.

Best Answer

A voltage source has zero internal resistance, which means that a capacitor across it will follow any variation of voltage, resulting in an infinite valued current through it. In LTspice (note the spelling) and SPICE, in general, the voltage source has a limited resistance due to the machine precision, so plotting the current through the capacitor will show a limited value, but still very large. It will also be a function of the derivative of the finite slope that will make for the rising time of the source (it has to have a finite value, it can't be zero). E.g. for a 1 V / 1 µs rise the current will be a pulse of 1 MA for 1 µs. This means that your first picture is simplified to a series RL.