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 i_L(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:

i_L(t) = i_C(t) = 0

• For 0 ≤ t < τ switch S₁ is On. There is a closed net representing a RL circuit. Applying Kirchoff equations we have:

$$\left\{ \begin{matrix} E - R_{3}i_{3} - L\frac{di_{1}}{\text{dt}} - R_{1}i_{1} = 0 \\ \text{\ \ \ i}_{2} = 0\ \ = = = > \ \ i_{1} = i_{2} = i_{L}\text{\ \ \ } \\ \end{matrix} \right.\ $$

We obtain a First Order Linear Differential equation with initial conditions:

$$\left\{ \begin{matrix} L\frac{di_{L}}{\text{dt}} + 2Ri_{L} = E \\ i_{L}\left( 0 \right) = 0 \\ \end{matrix} \right.\ $$

To find the solution, first we consider the associated homogenous equation and the characteristic equation:

$L\frac{di_{L}}{\text{dt}} + 2Ri_{L} = 0$ Lλ + 2R = 0 $\lambda = - \frac{2R}{L}$

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 i_L is zero:

2Ri_L = 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:

i_L(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:

$$\left\{ \begin{matrix} E - R_{3}i_{3} - L\frac{di_{L}}{\text{dt}} - R_{1}i_{L} = 0 \\ R_{3}i_{3} - v_{C} - R_{2}C\frac{dv_{C}}{\text{dt}} = 0 \\ i_{L} = C\frac{dv_{C}}{dt} + i_{3} \\ \end{matrix} \right.\ $$

Considering the current i_L(t), we obtain a Second Order Linear Differential equation

$$2RLC\frac{d^{2}i_{L}}{dt^{2}} + \left( 5R^{2}C + L \right)\frac{di_{L}}{\text{dt}} + 2Ri_{L} = E$$

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:

$$i_{L}^{'}\left( \tau \right) = \left. \ \frac{di_{L}(t)}{\text{dt}} \right|_{t = \tau}\ = \left. \ \frac{E}{L}\ \left( e^{- \frac{2R}{L}t} \right) \right|_{t = \tau} = \frac{E}{L}\ \frac{1}{e} = 295\ A/sec$$

Finally we obtain a Second Order Linear Differential equation with initial conditions

$$\left\{ \begin{matrix} 2RLC\frac{d^{2}i_{L}}{dt^{2}} + \left( 3R^{2}C + L \right)\frac{di_{L}}{\text{dt}} + 2Ri_{L} = E \\ i_{L}\left( \tau \right) = \frac{E}{2R}\ \left( 1 - \frac{1}{e} \right)\text{\ \ \ \ } \\ i_{L}^{'}\left( \tau \right) = \frac{E}{L}\ \frac{1}{e} \\ \end{matrix} \right.\ $$

To find the solution, first we consider the associated homogenous equation and the characteristic equation:

$2RLC\frac{d^{2}i_{L}}{dt^{2}} + \left( 5R^{2}C + L \right)\frac{di_{L}}{\text{dt}} + 2Ri_{L} = 0$ 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:

λ_1, 2 = α ± jβ = −202, 6 ± j32, 3

so:

i_L(t) = e^αt(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 i_L is zero:

2Ri_L = 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 i_L(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 i_L^′(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(t) = e^−202, 6t(−1.19*cos(32,3t)+22,429*sin(32,3t)) + 2 for t > 2,5 msec

Laplace method

• For t < 0 there is no closed net. As a consequence there is no current flowing in the circuit:

i_L(t) = i_C(t) = 0

• For 0 ≤ t < τ switch S₁ is On. There is a closed net representing a RL circuit.

The impedance of the circuit is:

Z = R₃ + sL + R₁= 2R + sL

So the current flowing in the circuit is:

$$I\left( s \right) = \frac{V(s)}{Z} = \ \frac{1}{2R + sL}V\left( s \right) = \frac{1}{2R + sL}*\frac{E}{s} = \frac{E}{L}\ \frac{1}{s\left( \frac{2R}{L} + s \right)} = \frac{E}{L}\ \left( \frac{A}{s} + \frac{B}{\left( \frac{2R}{L} + s \right)} \right)$$

where:

$$A = \operatorname{}{\frac{1}{\left( \frac{2R}{L} + s \right)} = \frac{2R}{L}}$$

$$B = \operatorname{}{\frac{1}{s} = - \frac{2R}{L}}$$

Definitively:

$$I\left( s \right) = \frac{E}{2R}\left( \frac{1}{s} - \frac{1}{s + \frac{2R}{L}} \right)$$

Applying the inverse Laplace transformer we obtain:

$$i_{L}\left( t \right) = i\left( t \right) = \frac{E}{2R}\ \left( 1 - e^{- \frac{2R}{L}t} \right)$$

that is the same result obtained with the classic method.

• For t > τ switch S₁ and S₂ are On. Let’s calculate the total impedance step by step:

$$Z_{1} = R_{2} + \frac{1}{\text{sC}}$$

$$Z_{2} = \frac{R_{3}\left( R_{2} + \frac{1}{\text{sC}} \right)}{R_{3} + \left( R_{2} + \frac{1}{\text{sC}} \right)} = R\frac{RCs + 1}{2RCs + 1}$$

$$Z_{3} = R\frac{RCs + 1}{2RCs + 1} + R + SL = \frac{2RLCs^{2} + \left( 3R^{2}C + L \right)s + 2R}{2RCs + 1}$$

The current is:

$$I\left( s \right) = \frac{V(s)}{Z_{3}} = \ \frac{2RCs + 1}{2RLCs^{2} + \left( 3R^{2}C + L \right)s + 2R}*\frac{E}{s}$$

Replacing all the known values, we have:

$$I\left( s \right) = \ 100\frac{9,5*10^{- 3}s + 1}{s\left( 1,1875*10^{- 3}s^{2} + 481,25*10^{- 3}s + 50 \right)} = \frac{100}{1,1875*10^{- 3}}\left( \frac{9,5*10^{- 3}s + 1}{s\left( \left( s + 202,6 \right)^{2} + {32,3}^{2} \right)} \right) = \ \frac{100}{1,1875*10^{- 3}}\left( \frac{A}{s} + \frac{Bs + C}{\left( s + 202,6 \right)^{2} + {32,3}^{2}} \right)$$

Decomposing the polynomial and using inverse Laplace transformer we have

i_L(t) = e^−202, 6t(−1.17*cos(32,3t)+22,4*sin(32,3t)) + 2