Discharging RLC. Kirchhoff’s second law

capacitorinductorkirchhoffs-lawspassive-networksresistors

I have been trying to solve this homework problem for who-knows how many hours. I could break it down enough as to specify the problem I'm facing into a single question:

For the following circuit with a charged capacitor, is \$ v_C + v_R + v_L = 0 \$ the right term? Or should it be \$ -v_C + v_R + v_L = 0 \$? Why?

RLC circuit

I am getting really confused because everywhere, the RLC circuit is solved with a tension source, hence the formula is \$ E = v_C + v_R + v_L \$, but I have doubts if it's as simple as E = 0 since in this particular case the source itself of the power is the capacitor, entering the current through the negative side.

Keep reading only if you said the \$ v_C \$ should be negative. I explain how I tried this and didn't work at all about what I've tried in the problem and how I got to that particular question:

Here is the general circuit. The DC source charges the capacitor and in t=0 it's opened. Therefore, for all t>0, we get the simplified circuit of the previous image.
Full circuit

The problem considering the \$ v_C \$ negative is in the general solution of the differential equation:

process to get the general solution
general solution

From this solution, we can see that the current would increase non-stop, which doesn't make any sense (from the initial conditions \$ C_1 \neq 0 \$ and \$ C_2 \neq 0 \$). Therefore, using the right signs yelds to a wrong solution, using all v positive yelds to a good solution (both exponents negative).

Best Answer

Label Vc and Vl in your diagram... current always flows from higher potential to lower potential (voltage). So depending on the direction your draw your current flowing, the voltage always drops in the direction of that current flow.

Simply put, in your first picture there is only one current loop. The current flowing through all the elements must be equal. I'll call the upper left corner node Vc and the upper right corner node Vl. Then namely:

(Vc - 0) / Zc = (Vl - Vc) / R = (0 - Vl) / Zl

Here Zc and Zl are the complex impedance of the capacitor and inductor respectively.

Alternatively, if we assume a current i0 is flowing through the circuit clockwise, and remember that voltage always drops across an element when current flows through it, then:

Vc = 0 - i0 * Zc
Vl = Vc - i0 * R
0 = Vl - i0 * Zl

Therefore:

0 = Vc - i0 * R - i0 * Zl = - i0 * Zc - i0 * R - i0 * Zl

Changing the sign:

i0 * Zc + i0 * R + i0 * Zl = 0

So yes, all elements should have the + sign.

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