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and
C
t ��t loge�1 � �v �
E �
but �loge� x
y � ��loge� y
x �
Therefore, t � t loge�� E
�
E � �
The second form is as follows:
v C � Ee �t/t
v C
� vC
� � e
E
�t/t
loge� vC
� � logee E
�t/t
loge� vC
� ��
E
and t ��t loge� v
t
�
t
C
�
E
(10.24)
E
or t � t loge�� (10.25)
v
For i C � (E/R)e �t/t :
For example, suppose that
C
E
t � t loge�� (10.26)
iCR
v C � 20(1 � e �t/(2�10�3 ) )
and the time to reach 10 V is required. Substituting into Eq. (10.24), we
have
20 V
t � (2 ms)loge��� 20 V � 10 V�
� (2 ms)loge2 � (2 ms)(0.693)
� 1.386 ms
Using Fig. 10.30, we find at (1 � e �t/t F
IN key on calculator
) � vC /E � 0.5 that t � 0.7t �
0.7(2 ms) � 1.4 ms, which is relatively close to the above.
Mathcad
It is time to see how Mathcad can be applied to the transient analysis
described in this chapter. For the first equation described in Section
10.10,
vC � 20(1 � e �t/(2�10�3 )
)
the value of t must be defined before the expression is written, or the
value can simply be inserted in the equation. The former approach is
INSTANTANEOUS VALUES ⏐⏐⏐ 403