First-order transient circuits are circuits that contain resistors, switches, DC current and DC voltage sources, and only one type of storage element, either and inductor or a capacitor. It turns out that such circuits can be described by a single first-order differential equation that can be found using nodal or mesh analysis.

In first-order transient circuits, we often assume that the circuit is at steady-state before $t=0$ and the state of the switch (closed or opened) is changed only at $t=0$. In this case, it can be shown that all the potentials and currents in the circuit can be written as

where $x(t)$ is either $I(t)$ or $V(t)$ and

• $X=x(0^+)$ is the value of the potential or current immediately after changing the state of the switch ($t=0^+$)
• $Y=x(0^+)-x(\infty)$ is the value of the voltage or current after a very long time, when the steady-state is reached (this theoretically happens at $t=\infty$ but practically we can assume that it happens for $t>5\tau$)
• $\tau$ is the relaxation time of the circuit

Therefore, to compute the potentials and currents in the circuit, one can compute constants $X$, $Y$ and $\tau$ (or $x(0^+)$, $x(\infty)$, and $\tau$) and replace them in the previous equation. Next, we describe how to compute these constants.

To compute $x(0^+)$ we need to set the state of the switch for $t>0$ and use any DC analysis method (such as nodal or mesh analysis) to compute the values of the potentials and currents in the circuit immediately after changing the state of the switch. One slight inconvenient, is that we do not know the values of the currents and voltages across the inductors and capacitors. However, these values can be computed based on the following observations:

• The current going through any inductor cannot change significantly during the infinitesimally short time that we turn the state on of off, because the voltage across the capacitor would go to infinity (remember that $V(t)=L \dfrac{I(t)}{dt}$). Therefore, the current going through any inductor should not change from $t=0^-$ (this is just before we change the state of the switch) until $t=0^+$ (this is just after we change we change the state of the switch).
• The voltage across any capacitor cannot change significantly during the infinitesimally short time that we turn the switch on or off, because the current going through the inductor would go to infinity (remember that $I(t)=C \dfrac{V(t)}{dt}$). Therefore, the voltage across any capacitor should not change from $t=0^-$ until $t=0^+$.

The above two observations allow us to use the following algorithm to compute $x(0^+)$:

Step 1. Compute the potentials and currents in the circuit for $t<0$. For this purpose we:

Step 2. Compute the potentials and currents in the circuit immediately after changing the state of the switch (at $t=0^+$). For this purpose we:

The potentials or currents in the circuit at $t=\infty$ can be computed by assuming that the circuit reached steady-state. Therefore we:

The relaxation time constant $\tau$ can be found by forming the Thévenin equivalent circuit at the terminals of the storage element (L or C) and computing the Thévenin resistance, $R_{Th}$. Therefore we: