Let’s consider the following circuit:
R = R1 = R2 = R3 = 25 Ω
C = 190 µF
L = 125 mH
E = 100 V
where switches S1 and S2 are closed at t = 0 and t = τ (where τ is the time constant of the LR circuit), according the following behavior:
$$\left\{ \begin{matrix}
S1 = Off & S2 = O\text{ff} & dla\ t < 0 \\
S1 = On & S2 = Off & dla\ 0\ \leq t < \tau \\
S1 = On & S2 = On & dla\ t\ \geq \ \tau \\
\end{matrix} \right.\ $$
The goal is to find the solution of the above circuit, basically the current iL(t) through the inductor and the voltage v(t)C on the capacitor. Knowing v(t)C we can easily find the current i1(t). We will obtain the solution with 2 different methods:
classic method
Laplace method
Classic method
For t < 0 there is no closed net. As a consequence there is no current flowing in the circuit:
iL(t) = iC(t) = 0
For 0 ≤ t < τ switch S1 is On. There is a closed net representing a RL circuit. Applying Kirchoff equations we have:
$$\left\{ \begin{matrix}
E - R_{3}i_{3} - L\frac{di_{1}}{dt} - R_{1}i_{1} = 0 \\
\text{\ \ \ i}_{2} = 0\ \ = = = > \ \ i_{1} = i_{2} = i_{L}\text{\ \ \ } \\
\end{matrix} \right.\ $$
We obtain a First Order Linear Differential equation with initial conditions:
$$\left\{ \begin{matrix}
L\frac{di_{L}}{dt} + 2Ri_{L} = E \\
i_{L}\left( 0 \right) = 0 \\
\end{matrix} \right.\ $$
To find the solution, first we consider the associated homogenous equation and the characteristic equation:
$L\frac{di_{L}}{dt} + 2Ri_{L} = 0$ Lλ + 2R = 0 ∖ t $\lambda = - \frac{2R}{L}$
so
$$i_{L}\left( t \right) = Ce^{\text{λt}} = Ce^{- \frac{2R}{L}t}$$
To find a particular solution we assume that the derivative of the current iL is zero:
2RiL = E $i_{L} = \frac{E}{2R}$
So:
$$i_{L}\left( t \right) = \text{Ce}^{- \frac{2R}{L}t} + \frac{E}{2R}$$
To calculate the value of constant C, we use the initial condition:
$0 = C + \frac{E}{2R}$ $C = - \frac{E}{2R}$
Finally the solution of the differential equation is:
$i_{L}\left( t \right) = \frac{E}{2R}\ \left( 1 - e^{- \frac{2R}{L}t} \right)$ for 0 ≤ t ≤ τ
where τ is:
$$\tau = \left| \frac{1}{\lambda} \right| = \frac{L}{2R}$$
Replacing the values for R, L and E we have:
iL(t) = 2 (1−e−400t) for 0 ≤ t ≤ 2,5 msec
For t > τ switch S1 and S2 are On. Applying Kirchoff equations we have:
$$\left\{ \begin{matrix}
E - R_{3}i_{3} - L\frac{di_{L}}{dt} - R_{1}i_{L} = 0 \\
R_{3}i_{3} - v_{C} - R_{2}C\frac{dv_{C}}{dt} = 0 \\
i_{L} = C\frac{dv_{C}}{dt} + i_{3} \\
\end{matrix} \right.\ $$
Considering the current iL(t), we obtain a Second Order Linear Differential equation
$$2RLC\frac{d^{2}i_{L}}{dt^{2}} + \left( 5R^{2}C + L \right)\frac{di_{L}}{dt} + 2Ri_{L} = E$$
The initial conditions can be obtained by the solution:
$i_{L}\left( t \right) = \frac{E}{2R}\ \left( 1 - e^{- \frac{2R}{L}t} \right)$ dla 0 ≤ t ≤ τ
when t = τ. So we obtain:
$$i_{L}\left( \tau \right) = \frac{E}{2R}\ \left( 1 - e^{- \frac{2R}{L}\tau} \right) = \frac{E}{2R}\ \left( 1 - \frac{1}{e} \right) = 1,26\ A$$
and:
$$i_{L}^{'}\left( \tau \right) = \left. \ \frac{di_{L}(t)}{dt} \right|_{t = \tau}\ = \left. \ \frac{E}{L}\ \left( e^{- \frac{2R}{L}t} \right) \right|_{t = \tau} = \frac{E}{L}\ \frac{1}{e} = 295\ A/sec$$
Finally we obtain a Second Order Linear Differential equation with initial conditions
$$\left\{ \begin{matrix}
2RLC\frac{d^{2}i_{L}}{dt^{2}} + \left( 3R^{2}C + L \right)\frac{di_{L}}{dt} + 2Ri_{L} = E \\
i_{L}\left( \tau \right) = \frac{E}{2R}\ \left( 1 - \frac{1}{e} \right)\text{\ \ \ \ } \\
i_{L}^{'}\left( \tau \right) = \frac{E}{L}\ \frac{1}{e} \\
\end{matrix} \right.\ $$
To find the solution, first we consider the associated homogenous equation and the characteristic equation:
$2RLC\frac{d^{2}i_{L}}{dt^{2}} + \left( 5R^{2}C + L \right)\frac{di_{L}}{dt} + 2Ri_{L} = 0$ 2RLCλ2 + (5R2C+L)λ + 2R = 0
Substituting values for R, L and C we have:
1, 1875 * 10−3λ2 + 481, 25 * 10−3λ + 50 = 0
and solving the second order equation, we obtain two complex solutions:
λ1, 2 = α ± jβ = −202, 6 ± j32, 3
so:
iL(t) = eαt(c1cos(βt)+c2sin(βt)) = e−202, 6t(c1cos(32,3t)+c2sin(32,3t))
To find a particular solution we assume that the derivative of the current iL is zero:
2RiL = E $i_{L} = \frac{E}{2R}$
So:
$$i_{L}\left( t \right) = e^{- 202,6t}\left( c_{1}\cos\left( 32,3t \right) + c_{2}\sin\left( 32,3t \right) \right) + \frac{E}{2R}$$
To calculate the value of constant c1 and c2, we use the initial condition on iL(τ):
1,26 = e−202, 6 * 2, 5 * 10−3(c1cos(32,3*2,5*10−3)+c2sin(32,3*2,5*10−3)) + 2
-1,226 = c1cos(0,08075) + c2sin(0,08075)
0, 81565 * c1 + 0, 00115 * c2 = −1
and the initial condition on iL′(τ):
295 = −202, 6 * e−202, 6 * 2, 5 * 10−3(c1cos(32,3*2,5*10−3)+c2sin(32,3*2,5*10−3)) + e−202, 6 * 2, 5 * 10−3(− 32, 3 * c1sin(32,3*2,5*10−3)+32, 3 * c2cos(32,3*2,5*10−3))
122, 2474 * c1 + 19, 6623 * c2 = 295
We obtain the following system of equations:
$\left\{ \begin{matrix} 0,81565*c_{1} + 0,00115*c_{2} = - 1 \\ 122,2474*c_{1} + 19,6623*c_{2} = 295 \\ \end{matrix} \right.\ $ $\left\{ \begin{matrix} c_{1} = - 1,19 \\ c_{2} = 22,429 \\ \end{matrix} \right.\ $
Finally the solution of the differential equation is:
$i_{L}\left( t \right) = e^{- 202,6t}\left( - 1.19*\cos\left( 32,3t \right) + 22,429*\sin\left( 32,3t \right) \right) + \frac{E}{2R}$ for t > 2,5 msec