STEP-BY-STEP APPROACH FOR TRANSIENT
FIRST-ORDER RL, RC CIRCUITS WITH DC SOURCES
1. For a transient that occurs at t = 0, the solution is of the form x(t ) = x(∞) + (x(0 + ) − x(∞) )e− t / τ , t > 0 .
The variable “x” represents the voltage or current that is the target of your solution. “ x (∞ ) ” is the steady
state solution after the transient when the transient part of the response has died out. “ x(0 + ) ” is the initial
state of the solution just after the transient has occurred. “τ” is the time constant that governs the decay of
the transient part of x.
2. Draw a schematic for t < 0 and find the initial current for an inductor iL(0-) or the initial voltage for a
capacitor vC(0-). Inductors for this analysis should be modeled as short circuits. Capacitors should be
modeled as open circuits.
3. Find the initial value of the target voltage or current at t = 0+. Take advantage of the fact that energy
storage dictates that neither the voltage of a capacitor nor the current through an inductor can change
instantaneously. This means that vC (0 + ) = vC (0 − ), iL (0 + ) = i L (0 − ) . Draw another schematic for t = 0+ and
replace capacitors with independent voltage sources of value VS = vC(0+). Replace inductors with
independent current sources of value IS = iL(0+). Solve for x(0+).
4. Find the final steady state value of the target voltage or current at t = ∞ . Once again, replace
inductors with their steady state models of short circuits and capacitors with their steady state models of
open circuits. Solve for x (∞ ) .
5. Compute the equiva lent resistance RTh at the terminals of the L or C. Shut off independent sources. If
only resistors are involved, reduction may be used. If there are dependent sources in the network, then
you will have to use an external excitation at the terminals and solve for RTh = V/I. For a RC circuit,
τ = RTh C seconds. For a RL circuit , τ = L / RTh seconds.
6. Formulate the solution of the target voltage or current by inserting your computed values
⇒ x(t ) = x(∞) + (x(0 + ) − x(∞))e − t / τ , t > 0 . Note that after about time t = 5τ seconds, the solution is pretty
close to its steady state final value.