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�v
4
v 3
v2
0
v C (V)
1
t2 t3 �t
2
t1 3 4
5 6 7 8
FIG. 10.59
Example 10.13.
b. From 2 ms to 5 ms, the voltage remains constant at 4 V; the change
in voltage �v � 0. The change in time �t � 3 ms, and
iCav � C� �vC
0
� � C �� � 0
�t
�t
c. From 5 ms to 11 ms, the voltage decreases from 4 V to 0 V. The change
in voltage �v is, therefore, 4 V � 0 � 4V(with a negative sign since
the voltage is decreasing with time). The change in time �t �
11 ms � 5ms�6 ms, and
iCav � C� �vC
� ��(2 � 10
�t
�6 F) � �
��1.33 � 10 �3 4 V
���3
6 � 10 s
A ��1.33 mA
d. From 11 ms on, the voltage remains constant at 0 and �v � 0, so
iCav � 0. The waveform for the average current for the impressed
voltage is as shown in Fig. 10.60.
4
0
–1.33
i C (mA)
1 2 3 4
5 6 7 8
FIG. 10.60
The resulting current i C for the applied voltage of Fig. 10.59.
Note in Example 10.13 that, in general, the steeper the slope, the
greater the current, and when the voltage fails to change, the current is
zero. In addition, the average value is the same as the instantaneous
value at any point along the slope over which the average value was
found. For example, if the interval �t is reduced from 0 → t 1 to t 2 � t 3,
as noted in Fig. 10.59, �v/�t is still the same. In fact, no matter how
small the interval �t, the slope will be the same, and therefore the current
i Cav will be the same. If we consider the limit as �t → 0, the slope
will still remain the same, and therefore i Cav � i Cinst at any instant of
9 10 11 12
9 10 11 12
THE CURRENT i C ⏐⏐⏐ 409
t (ms)
t (ms)