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Suppose a series RLC circuit that is critically damped.

Circuit is RLC in series with a DC voltage source \$ V_S \$ and a switch initially open. Capacitor and inductor are both discharged.

at t=0 the switch is closed.

I am trying to analyze what happens during the transient phase.

The circuit equation is

\$ i(t) = (At + B)e^{-\alpha t} \$

I want to find A and B.

So I apply the initial conditions.

The first condition is current when t=0. We know that the inductor will resist the initial current, so i(0) = 0.

If this is true and I apply that to the equation, to find A/B, I get

\$ i(t) = (At + B)e^{-\alpha t} \$

\$ 0 = (At + B)e^{-\alpha t} \$

at t=0

\$ 0 = (0 + B)e^{0} \$

\$ B = 0 \$

The other condition, di/dt at t=0.

We know that current will be zero, because the inductor will guarantee that, so

\$ \frac{di}{dt} = \frac{V}{L} = 0 \$

so,

\$ i(t) = (At + B)e^{-\alpha t} \$

\$ \frac{di}{dt} = 0 = -\alpha A t e^{-\alpha t} -\alpha B e^{-\alpha t} \$

when t= 0

\$ \frac{di}{dt} = 0 = -\alpha B \$

\$ B = 0 \$

again... I don't get it.

What am I doing wrong?

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    \$\begingroup\$ $$\begin{align*}\text{D}\left[\left(A \,t+B\right)e^{-\alpha\, t}\right]&=\text{D}\left[A \,t+B\right]e^{-\alpha\, t}+\left(A \,t+B\right)\text{D}\left[e^{-\alpha\, t}\right]\\\\&=A\,e^{-\alpha\, t}\,\text{d} t-\alpha\,t\,A\,e^{-\alpha\, t}\,\text{d} t-\alpha\,B\,e^{-\alpha\, t}\,\text{d} t \end{align*}$$ \$\endgroup\$
    – jonk
    Commented Aug 29, 2018 at 0:18
  • \$\begingroup\$ thanks. ahh, I see... DUUUH. You are right. But what about the first condition, i(0) = 0. This will always give B=0. If this is true, I don't see how the general form for the equation can be \$ i(t) = (At + B)e^{-\alpha t} \$ and not \$ i(t) = Ate^{-\alpha t} \$ and if B is 0, A will be zero, even by your equation \$\endgroup\$
    – Duck
    Commented Aug 29, 2018 at 0:27

1 Answer 1

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$$ i_{(0+)}= i_{(0-)}= 0 $$ $$ v_{L_{(0-)}} = 0 $$ $$ v_{L_{(0+)}} = L \left[ \frac{di}{dt} \right]_{t=t_{0+}} $$ Note that \$\left[ \frac{di}{dt} \right]_{t=t_{0+}} \neq 0 \$

From \$ i_{(0+)} \$ , \$ B = 0 \$

From the derivative of response \$i(t)\$: $$\left[ \frac{di}{dt} \right]_{t=t_{o+}}= \frac{v_{L_{(0+)}}}{L} = A$$

From the KVL (\$V\$ is the source voltage and \$V_{C_{(0+)}}\$ is the voltage on capacitor in \$t=t_{o+}\$):

$$ v_{L_{(0+)}} = -V_{C_{(0+)}} + V -Ri_{(0+)}$$

or

$$ v_{L_{(0+)}} = V$$

Therefore:

$$ A=\frac{V}{L}$$

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  • \$\begingroup\$ thanks but every book say that the current equation for the critically damped series RLC circuit is always \$ i(t) = (At + B)e^{-\alpha t} \$. If B is always zero for this kind of circuit, the books should say that the current equation for this kind of circuit was \$ i(t) = Ate^{-\alpha t} \$ \$\endgroup\$
    – Duck
    Commented Aug 29, 2018 at 1:10
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    \$\begingroup\$ But the initial current is not always zero (B=0). For example, this circuit may have been brought into this initial state by an appropriate combination of comutating switches such that this initial current is non zero. The same thing apllies to initial capacitor voltage. \$\endgroup\$ Commented Aug 29, 2018 at 1:18

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