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402 ⏐⏐⏐ CAPACITORS
The initial and final values of the voltage were drawn first, and
then the transient response was included between these levels. For
the current, the waveform begins and ends at zero, with the peak
value having a sign sensitive to the defined direction of iC in Fig.
10.49.
Let us now test the validity of the equation for vC by substituting
t � 0 s to reflect the instant the switch is closed.
e �t/t � e �0 � 1
and vC � 24 V � 20 Ve �t/t � 24 V � 20 V � 4 V
When t � 5t,
e �t/t � 0
and vC � 24 V � 20 Ve �t/t � 24 V � 0 V � 24 V
10.10 INSTANTANEOUS VALUES
On occasion it will be necessary to determine the voltage or current at
a particular instant of time that is not an integral multiple of t, as in the
previous sections. For example, if
vC � 20(1 � e �t/(2�10�3 )
)
the voltage vC may be required at t � 5 ms, which does not correspond
to a particular value of t. Figure 10.30 reveals that (1 � e �t/t ) is
approximately 0.93 at t � 5 ms � 2.5t, resulting in vC � 20(0.93) �
18.6 V. Additional accuracy can be obtained simply by substituting t �
5 ms into the equation and solving for vC using a calculator or table to
determine e �2.5 . Thus,
vC � 20(1 � e �5ms/2ms )
� 20(1 � e �2.5 )
� 20(1 � 0.082)
� 20(0.918)
� 18.36 V
The results are close, but accuracy beyond the tenths place is suspect
using Fig. 10.30. The above procedure can also be applied to any other
equation introduced in this chapter for currents or other voltages.
There are also occasions when the time to reach a particular voltage
or current is required. The procedure is complicated somewhat by the
use of natural logs (loge, or ln), but today’s calculators are equipped to
handle the operation with ease. There are two forms that require some
development. First, consider the following sequence:
vC � E(1 � e �t/t )
� vC
� � 1 � e
E
�t/t
1 � � vC
� � e
E
�t/t
C
loge�1 � �v �
E � � logee �t/t
C
loge�1 � �v �
E � �� t �
t