Transient Analysis
DC Steady-State
ELEC 308
Elements of Electrical Engineering
Dr. Ron Hayne
Images Courtesy of Allan Hambley and Prentice-Hall
Transient Analysis
 Scope of study:

Circuits that contain sources, switches, resistances,
inductances, and capacitances
 Transients:

Time-varying currents and voltages resulting from
sudden application of sources, usually due to switching
 Transient Analysis:



Involves using circuit concepts from Chapters 1 & 2
Current-voltage relationships for inductances and
capacitances involve derivatives and integrals
Create circuit equations that are differential equations
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First-Order RC or RL Circuits
 First-Order RC Circuits

Contains DC sources, resistances, and a
SINGLE capacitance
 First-Order RL Circuits

Contains DC sources, resistances, and a
SINGLE inductance
 Frequently used in timing applications

Due to time constant
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First-Order Circuit Algorithm
1. Apply KCL, KVL, and/or Ohm’s Law to write
the circuit equation.
2. If equation contains integrals, differentiate
each term to produce a PURE differential
equation.
3. Assume a solution of the form K1+K2est.
4. Substitute the solution into the diff. eqn. to
determine the values of K1 and s.
5. Use the initial conditions to determine the
value of K2.
6. Write the final solution.
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Discharge of a Capacitance
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Discharge of a Capacitance
Writinga KCL equat ion at topnode aft erswitch closed, we have:
dvC t  vC t 
C

0
dt
R
Multiplying by theresist anceR :
dvC t 
RC
 vC t   0
dt
Generalformof solutionis given by
vC t   Ke st
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Discharge of a Capacitance
Substitute our generalsolution:
RCKse st  Ke st  0
Solving for s :
1
RCs  1  s 
RC
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Discharge of a Capacitance
Use our initialconditionsto solve for K :
 
vC 0   Vi  Ke s 0  K  Vi
Substituing thisin our solutiongives us our completesolution:
vC t   Vi e t RC
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Time Constant
 Time interval τ = RC is called the time constant of
the circuit
 After t=5τ, vC(t)≈0
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Charging a Capacitance
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Charging a Capacitance
WritingKCL equation at node between resistance& capacit ance,
dvC t  vC t   Vs
C

0
dt
R
Rearranging,
RC
dvC t 
 vC t   Vs
dt
Generalformof solutionis given by
vC t   K1  K 2 e st
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Charging a Capacitance
Substitute our generalsolutionand solve for s and K1 :
1
s
RC
and
K1  Vs
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Charging a Capacitance
Use our initialconditionsto solvefor K 2 :
 
vC 0   0  Vs  K 2e s 0  K 2  Vs
T hus thecompletesolution:
vC t   Vs  Vs e t RC  Vs  Vs e t 
 First term is STEADY-STATE RESPONSE

Or FORCED RESPONSE
 Second term is TRANSIENT RESPONSE
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Charging a Capacitance
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DC Steady State
 Transient terms in the expressions for current
and voltages in RLC circuits decay to zero
with time
 For DC sources, steady-state currents and
voltages are CONSTANT
 For steady-state conditions with DC sources:


CAPACITANCES behave like OPEN circuits
INDUCTANCES behave like SHORT circuits
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Capacitance in DC Steady-State
 Remember current through a capacitance:
dv C t 
iC t   C
dt
 If voltage is constant, current is _________.

 CAPACITANCE behaves just like an ____
circuit
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Inductance in DC Steady-State
 Remember voltage across an inductance:
di L t 
v L t   L
dt
 If current is constant, voltage is _________.

 INDUCTANCE behaves just like a ______
circuit
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Steady-State DC Analysis
 Find vx and ix for t >> 0
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Exercise 4.3
 Find va and ia for t >> 0
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RL Transient Analysis
 Find i(t) and v(t)
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RL Transient Analysis
 Time interval τ = L/R is called the time constant of
the circuit
 After t=5τ, i (t)≈2
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Example 4.3
 Find i(t) and v(t)
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Example 4.3
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Exercise 4.6
 Find i(t) and v(t)
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Summary
 Transient Analysis


First Order RC Circuits
First Order RL Circuits
 DC Steady State
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