# Discharging a LC(R) Circuit

Reading through my lecure script I encountered this example which I don't quite understand:

Given is the following Circuit:

At time $t= 0$ the circuit is closed, before that the circuit was open, and we know that the capacitor $C = 10 [nF]$ is charged $U_0 = 100 [V]$.

Here is the direction of the electric tension and of the respective current displayed:

Already a couple of questions come to mind:

Shouldn't the directions of the resistance and the inductance be on the opposite direction of the voltage at the capacitor?

Why is that the direction of the voltage at the capacitor? is it arbitrary?

Now, trying to show the development of the current in function of time, applying kirchhoffs Voltage Law we come up with the following expression:

$$\Sigma_i U_i = 0 \\ RI + \frac{Q}{C} + LI' = 0$$

Again, why everything with a positive sign?

Then the book jumps directly defining radial frequency and damping ratio: $$\frac{1}{LC} = \omega_0^2, \ \frac{R}{L} = 2\beta$$

What are the steps and how do I get to the following solution: $I(t) = e^{-bt}(A\cos\omega t + B\sin\omega t)$

• There are some things missing: what is Q? (presumably charge, but the sign is important) What are the relative sizes of R,L, and C. Those will affect the specific form of the solution (there are 3 forms). Is there an initial non-zero current I? Is there an initial charge Q? Is Q>0 or Q<0? Which side of the capacitor is at the higher potential? Commented Jun 13, 2018 at 21:30

Again, why everything with a positive sign?

By the Passive sign convention (PSC), the (chosen reference direction of) current should enter the labeled positive terminal of a two-terminal circuit element.

To satisfy the PSC, the reference direction of the current should be counter-clockwise rather than clockwise.

But this isn't a fatal error as long as one is aware of the difference in sign for the power (using the PSC, positive power is power delivered to the circuit element while negative power is power supplied by the circuit element).

Keeping the voltage polarity reference as they are, KVL counter-clockwise around the loop yields:

$$U_L + U_R + U_C = 0$$

Note that if you choose to go clockwise around the loop, everything picks up a negative sign but clearly yields the same equation

$$-U_R - U_L - U_C = 0 = U_L + U_R + U_C$$

1. Shouldn't the directions of the resistance and the inductance be on the opposite direction of the voltage at the capacitor?

It doesn't matter. This is a differential equation of 2nd order. After inserting the initial conditions you get the correct answer. E.g. if the direction of the voltage is the other "direction" you get a minus sign.

2. Why is that the direction of the voltage at the capacitor? is it arbitrary?

It depends on your initial condition. However you have load the capacitor.

3. Circuit analysis

• Kirchhoff's voltage law (KVL)

$$u_C + u_R + u_L = 0$$

and using the elementary equation: $$u_R = i*R$$ $$i_C = C *\dfrac{d u_C}{dt} \quad \quad u_c = \frac{1}{C} \int i_C \space dt + u_{C0}$$ $$u_L = L * \dfrac{di_L}{dt}$$

Now we can insert the 3 equations above into KVL with ($i = i_L=i_C=i_R$):

$$R*i + \frac{1}{C} \int i \space dt + u_{C0} + L * \dfrac{di}{dt} = 0$$

Differentiating this equation and reordering results in:

$$\dfrac{d^{2}i}{dt^{2}} + \frac{R}{L} * \dfrac{di}{dt} + \frac{1}{LC} * i = 0$$

Using the ansatz: $i = e^{\lambda t}$ yields to the characteristic equation:

$$\lambda^{2} + \frac{R}{L}*\lambda + \frac{1}{LC} = 0$$

The result of this quadratic equation is:

$$\lambda = -\frac{R}{2L} \pm \sqrt{\Big(\frac{R}{2L}\Big)^2 - \frac{1}{LC}}$$

What you see is the damping factor, and the resonat frequency (root expresion, assuming the value of under the root is negativ --> complex --> oscillating)

The damping factor can therefore be found to be

$$\beta = \cfrac{R}{2L}$$

and for the case $\Big(\frac{R}{2L}\Big)^2 < \frac{1}{LC}$

you get the solution:

$$i = I_0 \cdot e^{-\frac{R}{2L}\cdot t + \varphi_0} \cdot e^{j \cdot \sqrt{\frac{1}{LC}-\Big(\frac{R}{2L}\Big)^2} \cdot t} \quad,$$

where $\varphi$ and $I_0$ are got by initial conditions.

NOTE: $u_c = \frac{Q}{C}$ is only valid if Q keeps constant, which is not the case in your problem --> definitely wrong!!

BTW: Using Laplace makes it very simple and a common way in electrical engineering to make circuit analysis.

• sorry, I forgot some crucial information, the boundary conditions. At $t=0$ the circuit is closed, before that the circuit was open, and we know that the capacitor $C=10[nF]$ is charged $U0=100[V]$. Commented Jun 14, 2018 at 8:26