# First-Order Transient Circuits

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```							Transient Analysis - First Order
Circuits

Switches, Transient Response,
Differential Equations

Kevin D. Donohue, University of Kentucky   1
Transient Response
   DC analysis of a circuit only provides a description of
voltages and currents in steady-state behavior.

   When the applied voltage or current changes at some
time, say t0, a transient response is produced that dies
out over a period of time leaving a new steady-state
behavior.

   The circuit’s differential equation must be used to
determine complete voltage and current responses.

Kevin D. Donohue, University of Kentucky     2
Examples
Describe v0 for all t. Identify transient and steady-state
responses.
VDC                t=0           R             +
C   v0
-
t=0
Show:
             t 
VDC 1  exp
              volts     for t  0
v0 (t )              RC  
 0 volts                        for t  0


For steady-state response, let t  , for transient response subtract

Kevin D. Donohue, University of Kentucky            3
Instantaneous Voltage and Current
Changes in Capacitors and Inductors:
    What would be the required                                          ic

current, ic , in this circuit for the         VDC             t=0            +
voltage on the capacitor to                                              C   vC
-
change instantaneously?

iL
     What would be the required
t=0
voltage, vL , in this circuit for the         IDC                            +
L   vL
current in the inductor to change                                             -
instantaneously?

Conclusion: If the source cannot produce infinite instantaneous
power, then neither the capacitor voltage, nor the inductor current can
change instantaneously.

Kevin D. Donohue, University of Kentucky                       4
Switch Notation and Initial Conditions:

In order to denote the time right before t=0 (limit from the left as
t0), and the time right after t=0 (limit from the right as t0),
the following notation will be used:

Let t=0+ be the moment after the switch
t=0                    is closed and t=0- be the moment before
the switch is closed.

For circuits with practical sources,
the voltage across a capacitor cannot                v c ( 0  )  vc ( 0  )
change instantaneously,

and the current in an inductor
cannot change instantaneously                   i L (0  )  i L (0  )

Kevin D. Donohue, University of Kentucky                     5
Complete Solution by the Differential
Equation Approach
5 major steps in finding the complete solution:
 Determine initial conditions on capacitor voltages
and/or inductor currents.
 Find the differential equation for either capacitor
voltage or inductor current (mesh/loop/nodal ….
analysis).
 Determine the natural solution (complementary
solution).
 Determine the forced solution (particular solution).
 Apply initial conditions to the complete solution to
determine the unknown coefficients in the natural
solution.
Kevin D. Donohue, University of Kentucky   6
Example
Find the complete solution for iL for vs  10V

t=0            25 W                +
vs                                       0.25 H    vL
-

Show for t  0:     iL  0.4(1  exp(100t ))

Kevin D. Donohue, University of Kentucky        7
Example
Find the complete solution for vc when is  1 mA

t=0

is          t=0                                   +
100 W           1 mF   vc
-
R

Show for t  0:       vc  0.1 exp( 10 t )

Kevin D. Donohue, University of Kentucky          8
Step-by-Step Method
The solution of circuits containing energy storage elements can be divided into a
element is present, the Thévenin resistance can be obtained with respect to the
terminal of the energy storage element and used to compute the time constant for the
transient component.

 t
   Assume solution is of the form x(t )  K1  K 2 exp   
 
   Assume steady-state before the switch is thrown and let either vc (0 )  vc (0 ) or
iL (0 )  iL (0 ) , and find initial condition for quantity of interest x(0 )

   Let K1 = steady-state solution after switch is thrown, K2  x(0 )  K1
and   CRth , or   L Rth

Kevin D. Donohue, University of Kentucky                         9

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