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11.2 
ZEro-Input rEsponsE of 
RC
 cIrcuIt
The differential equations derived above can be used to solve for v
C
(t) for all t provided the input 
source function is known for all t. However, we do not know the input source function for all t. The 
input source function is usually known only for t > 0. It may contain a discontinuity at t 
=
0 too. The 
input source function is generally unknown for t < 0.
Therefore, the effect of all the unknown currents which went through the capacitor from infinite past 
to t 
=
0
-
is given in a condensed manner in the form of an initial value specification for v
C
(t) at t 
=
0
-
. 

---

Zero-InputResponseof
RC
Circuit

11.3
Let us denote this value as V
o
. Now we can obtain the total response of the circuit by adding the two 
response components – zero-input response and zero-state response. 
Zero-input response
istheresponsefor
t

≥
0
+
whentheinputisheldatzerofrom
t

=
0
onwards.
Zero-state response
istheresponsefor
t

≥
0
+
whentheinputisheldzerofrom
infinitepastto
t

=
0
-
(orequivalently,initialconditionat
t

=
0
-
is0)
 
andthenaspecified
inputsourcefunctionisappliedfrom
t

=
0onwards.
Zero-input response is the same as the so-called source-free response. The describing differential 
equation in this situation is 
dv t
dt
v t
t
v
V
C
C
C
o
RC
for all 
with
( )
( )
( )
+
=
≥
=
+
−
1
0
0
0
(11.2-1)
Physically this amounts to connecting a charged capacitor to a resistor to form a closed loop at
t 
=
0. The initial voltage across the capacitor appears across the resistor from t 
= 
0 onwards. The 
resistor demands a current of v
C
(t)/R and this current flow is of suitable polarity to discharge the 
capacitor. As the capacitor discharges more and more, its voltage comes down, resulting in the 
discharge current also going down. Thus the capacitor keeps discharging; but at slower and slower 
rates as time increases. The discharge current divided by capacitance value gives rate of decrease 
of voltage across the capacitor. Hence, the initial rate of decrease of voltage is V
o
/
t
V/s, where 
t
is 
the time constant (
=
RC) of the circuit. If the initial rate of decrease were maintained throughout, the 
voltage across capacitor would have gone to zero in one time constant.
There was no impulse current source in this circuit and hence the voltage across capacitor does not 
change instantaneously at t 
=
0. Hence v
C
(t) at t 
=
0
+
is same as v
C
(t) at t 
=
0
-
, i.e., V
o
. Therefore, the 
expression for v
C
(t) is given by the familiar exponential function.
v t
V e
t
t
C
o
V for
( )
=
≥
−
+
t
0
(11.2-2)
The graph of this function is only too well known and is not repeated here.