passive-networks
reaches steady-state at \$t = 0^{-} \$, so I am trying to calculate the initial conditions of \$i\$ (current through inductor) and \$v\$ (voltage across capacitor). It is easy to find that \$i(0) = 10A\$, but I am not able to find \$v(0)\$. What am I missing here?
The current through an inductor is continuous so
$$i_L(0+) = 2A$$
The voltage across a capacitor is continuous so
$$v_C(0+) = 0V$$
Now, the inductor and the resistor are in series so they have identical currents. So, armed with just KVL and Ohm's Law,
$$v_L(0+)=?$$
What you have is a series RLC circuit. One of the most common introductory circuit in EE. For example, look under Series RLC Circuit in this wikipedia page:
en.wikipedia.org/wiki/RLC_circuit
These are equations copied from the same page for the under-damped case. Given R, C along with measured \$\omega_d\$, L can be solved from these equations. (You don't even need to measure the attenuation factor \$\alpha\$.)
If the RLC do want to oscillate, the node where D1 cathode is connected will go negative. Therefore, depending on how D1 is driven, it may conduct during the negative half cycle and kill the oscillation.
Best Answer
Just think about how you derived the current of 10 amps. There is a 100 volt supply and there is a 10 ohm resistor - to calculate 10 amps it must mean that all the voltage is across the resistor so, what voltage is across the inductor?
Or, put another way, V = L di/dt and, because the circuit is in steady state (your words not mine), this implies the current is also "steady" so, using that formula, what is the voltage across the inductor?