Substitution + Superposition with RLC circuits

Question

Consider the following circuit:

^{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:

^{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 ?

Answer 1

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 book Basic Engineering Circuit Analysis
• the paper Transient Analysis of First-Order Circuits: Approaches and Recommendations).

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.