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Consider the following circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

\$E\$ is a constant voltage source. The switch is closed at time \$t = 0\$. The request is to calculate the generic equation for \$i_L(t)\$.

Using the PSC convention i get the following Kirchhoff equations:

1) \$L{ {d i_L} \over {dt}} -v_c = 0\$ (KVL)

2) \$v_c -E -i_R R= 0\$ (KVL)

3) \$-i_R -i_C -i_L = 0\$ (KCL)

After few substitutions and algebraic manipulations I get the differential equation which describe the circuit:

\$-{d^2i_L \over dt^2} -{1 \over CR}{di_L \over dt}-{1 \over CL}i_L = -{E \over CLR}\$

(Replace \$i_C\$ with the capacitor law on the 3. Replace \$i_R\$ form the 2 in the 3 then replace \$v_c\$ from the 1 in the 3.)

The answer is correct but I want to find the same equation using a different method. By assuming the state variable as known I can replace \$L\$ with a current source and \$C\$ with a voltage source (I just applied the Substitution theorem). Then using the Superposition theorem I obtain these three circuits:

schematic

simulate this circuit

The current directions are the same of the original circuit. I get these two equations:

\$v_L = 0 + v_C + 0\$

\$i_C = -i_L - {v_C \over R} -{E \over R}\$

After replacing \$v_L\$ and \$i_C\$:

\$-L{di_L \over dt} = 0 + v_C + 0\$

\$-C{dv_C \over dt} = -i_L - {v_C \over R} -{E \over R}\$

However by replacing \$v_C\$ in the second equation with \$-L{di_L \over dt}\$ I can't get the same equation obtained with the Kirchhoff laws. Not only the signs are different but also the coefficients of each term are different.

What am I missing ? Can this method applied for every AC circuit ?

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    \$\begingroup\$ You appear to be saying that IL(t) is -E/CLR and that makes no sense at all. \$\endgroup\$ – Andy aka Jan 23 at 21:29
  • \$\begingroup\$ @Andyaka The answer given for this exercise is correct. The only difference is the sign of each term which is positive on the solution given on the textbook while on my equation all terms are negative. I suppose this is because I used PSC (passive sign convention) while the professor used the ASC (active sign convention). \$\endgroup\$ – Bemipefe Jan 24 at 7:35
  • \$\begingroup\$ I think Andy aka means that, without going into details, you can easily see that the term \$E/CLR\$ is wrong because that term is not a current! So, it cannot be a term that describes \$i_L(t)\$. The term does not have the units A, but A/sec. \$\endgroup\$ – Huisman Jan 24 at 11:32
  • \$\begingroup\$ @Huisman Does it really matter ? The current is described by a differential equation (a second order one). This doesn't mean that the current at time \$t\$ is equal to such constant. \$\endgroup\$ – Bemipefe Jan 24 at 13:09
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    \$\begingroup\$ @Bemipefe You're right the only error was in the first KVL. And no, it doesn't matter whether the units are amperes or /$A/s^s/$. It does matter that equation is consistent: if one term is in /$A/s^s/$, all terms have to be in /$A/s^s/$ \$\endgroup\$ – Huisman Jan 25 at 20:11
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The answer is correct but I want to find the same equation using a different method. By assuming the state variable as known and assuming the circuit in a stable stage I can replace \$L\$ with a current source and \$C\$ with a voltage source (I don't know the name of this method in English).

The method you're (probably) referring to is called the "step-by-step approach"

See:

The method uses the fact that the solution to a first order circuit can be described as $$ x(t) = x(\infty) + \Big( x(0^+)-x(\infty) \Big) e^{{t} \over {\tau}} $$

and that \$x(0^+)\$ and \$x(\infty)\$ can be found by replacing the capacitor or inductor by a voltage source /open circuit or current source / short circuit.

What am I missing ?
Can this method applied for every AC circuit?

This method can only by used for first order circuits (having only DC sources), and the circuit you're showing is second order.

UPDATE
As commented OP says the Substitution Method has been used.

Substitution Theorem states that the voltage across any branch or the current through that branch of a network being known, the branch can be replaced by the combination of various elements that will make the same voltage and current through that branch. In other words, the Substitution Theorem says that for branch equivalence, the terminal voltage and current must be same.

Source: CircuitGlobe.com

The link from the comment states:

Substitution theorem states that if an element in a network is replaced by a voltage source whose voltage at any instant of time is equals to the voltage across the element in the previous network then the initial condition in the rest of the network will be unaltered or alternately if an element in a network is replaced by a current source whose current at any instant of time is equal to the current through the element in the previous network then the initial condition in the rest of the network will be unaltered.

Using the first definition, you cannot replace a capacitor with a voltage source, because the terminal voltage and current of a capacitor are not equal to those of a voltage source at all time. You trying to obtain a differential equaton that is valid for all time.

Using the second definition, it is already highlighted. You can use this method to find the initial conditions in the test of the network (just like in the step-by-step approach). Although initial conditions are needed to gain the Particular Solution, it will in no way provide information for the second order equation itself.

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  • \$\begingroup\$ No the method that I'm referring to is not the "step-by-step approach". That method is used to solve the circuit when the current in the circuit is constant ( as you said it can't be applied with AC sources). Anyway I believe you can also applies it to second order circuits. Having said that, I discovered that the method I used is called Substitution Theorem. electrical4u.com/substitution-theorem \$\endgroup\$ – Bemipefe Jan 26 at 19:36

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