EENG 2610: Circuit Analysis
Class 12: First-Order Circuits
Oluwayomi Adamo
Department of Electrical Engineering
College of Engineering, University of North Texas
Transient Analysis
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Circuits in transition
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Transient analysis: study of circuit behavior in transition phase.
Transition phase is caused by a sudden change in circuit
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Suddenly apply or remove voltage or current source,
Open or close a switch in the circuit.
Transition is affected by capacitor or inductor in circuit
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No transient analysis is needed for resistive circuit network.
Because capacitor and inductor can store energy, the circuit
response to a sudden change will go through a transition period
before settling down to a steady-state value.
First-Order and Second-Order Circuits
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First-order circuits contain only a single capacitor or inductor
Second-order circuits contain both a capacitor and an inductor
Two techniques for transient analysis that we will learn:
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Differential equation approach
Step-by-step approach
Laplace transform method is a much simpler method for
transient analysis – you will learn it in another course.
First-order differential equation
Solution of transient analysis requires to
solve a first-order differential equation:
dx (t )
 ax (t )  f (t )
dt
General solution to the first-order differential equation:
x(t )  x p (t )  xc (t )
x p (t ) : particular integral solution (or forced response)
xc (t ) : complementary solution (or natural response)
xc(t) is the solution to the homogeneous equation: dx (t )  ax (t )  0
dt
Now we only consider f (t) is constant, that is, f (t) = A :
dx (t )
 ax (t )  A
dt
Solution:
x(t )  K1  K2et /
 : time constant of the circuit
 x ( 0)  K 1  K 2 ,
 x ( )  K ,

1

   1 / a,
 K1  A / a,
A drop of 63.2%
xc (5 )  K2  0.67%  K2 1%
The Differential Equation Approach
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State-variable approach
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Step 1: Find state-variables
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Voltage across the capacitor and current through inductor are
called state-variables.
State-variables cannot change instantaneously.
Step 2: Find initial value of state
variables at t  t0 (usually t0  0)
Step 3: Write KCL equation for the voltage across the
capacitor and/or KVL equation for the current through
the inductor for t  t0
Step 4: Solve first-order differential equation
Example 7.1: Calculate the current i(t ) for t  0.
Assume the switch has been in position 1 for a long time.
t  0
t0
Example 7.2: Find output voltage vo (t ) for t  0.
iL (t )
t  0
iL (t )
t0
Step-by-Step Approach
for First-Order Circuits
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Step1: We assume a solution for the variable x(t) (either voltage or
current) of the form
x(t )  K  K et /
1
2
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Step 2: Assuming that the original circuit has reached steady state
before switch actions, draw this previous circuit with the capacitor
replaced by an open circuit or the inductor replaced by a short
circuit. Solve for the voltage across the capacitor, vC(0-), or the
current through the inductor, iL(0-), prior to switch action.
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Step 3: Voltage across a capacitor and the current through an
inductor cannot change in zero time: vC(0+)=vC(0-), iL(0+)=iL(0-).
Therefore, draw the circuit valid for t=0+ with the switches in their
new positions. Replace a capacitor with a voltage source of value
vC(0+) or an inductor with a current source of value iL(0+). Solve for
the initial value of the variable x(0+).
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Step 4: Assuming that steady state has been reached after the
switches are thrown, draw the equivalent circuit, valid for t  5 ,
by replacing the capacitor by an open circuit or the inductor by a
short circuit. Solve for the steady-state value of the variable x()
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Step 5: Find Thevenin equivalent resistance RTh at the terminals
of the storage element by looking into the circuit from the
terminals of the storage element. The time constant for a circuit
containing a capacitor is   RTh C , and for a circuit containing an
inductor the time constant is   L / RTh
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Step 6: Evaluate the constants K1, K2
in step 1 using the relations:
 x(0)  K1  K 2

 x()  K1
Then, we can find that the solution is
x(t )  x()  [ x(0)  x()]et /
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Important: This solution form applies only to a first-order circuit
having constant, DC sources.
If switch action occurs at any time t0, the step-by-step analysis yields the
following equations:
 x(t0 )  K1  K 2

 x (  )  K1
x(t )  x()  [ x(t0 )  x()]e(t t0 ) / ,
for t  t0
The function is essentially time-shifted by t0 seconds.
Example 7.3: The circuit is in steady state prior to time t=0,
when the switch is closed. Calculate the current i(t) for t>0.
t  0-
t  0
t
Find RTh
RTH