Question

A series RLC circuit withR = 6 ohm, C = 0.02 Farad and L = 0.1 has no applied voltage. Find the subsequent current in the circuit if the initial charge, on the capacitor is q0 and the initial current is zero.

Answer 1

Task:

$R = 6 \Omega; \\ L = 0.1 \text{H}; \\ C = 0.02 \text{F}; \\ i(0) = 0 \text{A}; \\ q_C(0) = q_0$

Find: $i(t)$

Solution:

Current through the capacitor and voltage across the inductance can be described as:

$i_{C}=C \cfrac{d v_{C}}{d t}\ (1)\\ v_{L}=L \cfrac{d i_{L}}{d t}\ (2)$

In this case all the elements are connected in series, so $i_C = i_L = i$ . According to the second Kirchhoff’s circuit law:

$v_{R}+v_{L}+v_{C}=0\ (3)$

After inserting (1) and (2) into (3) the last equation can be rewritten as:

$v_R + v_L + v_C = Ri + L \cfrac{di}{dt} + \cfrac{1}{C} \int\limits_0^{t_0} idt = 0\ (4)$

After differentiation of the left and right parts of equation (4) it is possible to get:

$\cfrac{d}{dt} (Ri + L \cfrac{di}{dt} + \cfrac{1}{C} \int\limits_0^{t_0} idt) = \cfrac{d0}{dt} \Leftrightarrow R \cfrac{di}{dt} + L \cfrac{d^2 i}{dt} + \cfrac{1}{C} i = 0.1\cfrac{d^2 i}{dt} + 6\cfrac{di}{dt} + 50i=0$

It is a second order ordianary differential equation with constant coefficients. To solve this equation it is necessary to find the roots of the characteristic polynomial:

$0.1 \lambda^2 + 6 \lambda + 50 = 0$

According to the Vieta's formuals:

$\begin{cases} \lambda_1+\lambda_2 = -\cfrac{6}{0.1} = -60 \\ \lambda_1 \cdot \lambda_2 = \cfrac{50}{0.1} = 500 \end{cases} \Bigg| \Rightarrow \begin{cases} \lambda_1 = -50 \\ \lambda_2 = -10 \end{cases}$

So, the general solution for $i(t)$ is equal to:

$i(t) = C_1 e^{\lambda_1 t} + C_2 e^{\lambda_2 t} = C_1 e^{-50t} + C_2 e^{-10t}\ (5)$

To get the partial solution it is necessary to solve the Cauchy's problem. First of all its is necessary to get initial condition for $\cfrac{di}{dt}$ . According to (4):

$L \cfrac{di}{dt} = -Ri-u_C \Rightarrow \cfrac{di}{dt} (0) = \cfrac{1}{L} (-Ri(0)-u_C(0)) \ (6)$

To get $u_C(0)$ it is convenient to use the defenition of capacity:

$C = \cfrac{q_C}{u_C} \Rightarrow u_C(0) = \cfrac{q_C(0)}{C} = \cfrac{q_0}{C}\ (7)$

After inserting (7) into (6) the latter can be rewritten as:

$\cfrac{di}{dt}(0) = \cfrac{1}{L} (-Ri(0) - \cfrac{q_0}{C})\ (8)$

According to the task, $i(0) = 0$, so, according to (8):

$\cfrac{di}{dt}(0) = -\cfrac{1}{LC} q_0 = -500q_0$

Now it is possible to solve the Cauchy's problem:

$\begin{cases} i(0) = C_1 e^{-50\cdot 0} + C_2 e^{-10 \cdot 0} = C_1 + C_2 = 0 \\ \cfrac{di}{dt} (0) = \cfrac{d}{dt} (C_1 e^{-50t} + C_2 e^{-10t})(0) =-50C_1 -10C_2 =-500q_0 \end{cases}$

So, the system for $C_1$ and $C_2$ is:

$\begin{cases} C_1+C_2=0\\ -50C_1-10C_2=-500 q_0 \end{cases} \Leftrightarrow \begin{cases} C_2 = -C_1\\ -40C_1=-500 q_0 \end{cases} \Leftrightarrow \begin{cases} C_1 = 12.5 q_0\\ C_2 = -12.5q_0 \end{cases} \ (9)$

After inserting (9) in (5) it is possible to get partial solution:

$i(t) = 12.5q_0 e^{-50t} -12.5q_0 e^{-10t}$

Answer: $i(t) = 12.5q_0 e^{-50t} -12.5q_0 e^{-10t}$ .