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example: 3.5-6
A periodic current waveform is applied to a capacitor of value 0.1 F from t 
=
0 s as in Fig. 3.5-8. The 
voltage across the capacitor is found to vary periodically between 1 V and 5 V. (i) What is the full-
cycle average value of the applied current waveform? (ii) What is the half-cycle average of the applied 
current waveform? (iii) What was the initial voltage in the capacitor? (iv) Find I
p
.
i
S
(
t
) (A)
I
p
0.5 
I
p
t
(s)
1 2 3 4 5 6 7 8 9
i
S
(
t
)
i
C
(
t
)
v
C
(
t
)
C = 
0.1 
µ
F
+
–
Fig. 3.5-8 
Circuit and waveform for Example 3.5-6

---

Series Connection of Capacitors 
3.41
Solution
(i) The voltage is stated to be periodic. Therefore, it must either be a pure alternating waveform or 
such an alternating waveform plus a DC offset. Differentiation of a pure alternating waveform 
gives another pure alternating waveform. Differentiation of a DC term can give only zero. 
Hence the derivative of capacitor voltage will not contain a DC term. Derivative of voltage 
multiplied by capacitance value is the current through the capacitor. Therefore, current through 
the capacitor will not have a DC value. But the DC content in a periodic waveform is nothing 
but its average over a cycle period. Therefore, this current waveform has a full-cycle average
of 0V.
(ii) Half-cycle average of alternating current 
=
Half-cycle area / half the time period. Half-cycle area 
of the alternating current is given by change in charge of capacitor between the maximum and 
minimum voltage values. It is (5
-
1)V 
× 
0.1F 
=
0.4C in this case. Therefore, the half-cycle area of 
current waveform is 0.4C and its half-cycle average value is 0.4C/4 s 
=
0.1A.
(iii) The capacitor voltage is periodic between 1V and 5V. There is no impulse content in the applied 
current. Hence its initial voltage must have been 1V.
(iv) The half-cycle area in terms of I
p
is 
=
(0.5 I
p
+ 
0.5 I
p
+ 
0.25 I
p
)
× 
2 
=
2.5 I
p
C. This must be equal 
to 0.4 C. Therefore, I
p
=
0.4/2.5 
=
0.16A.