082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

16.3 Improper integrals 485
i
C 1 C 2
y 1
y
y 2
Figure16.1
Two capacitors
connected in series.
This can be written as a definite integral to give the voltage expression across the
capacitor atageneral pointintime,t. The expression is
v = 1 C
∫ t
−∞
idt
Now consider the situation depicted in Figure 16.1. Writing an equation for each of
the capacitors gives
v 1
= 1 C 1
∫ t
−∞
idt
v 2
= 1 C 2
∫ t
By Kirchhoff’s voltage law, v = v 1
+ v 2
and so
v = 1 ∫ t
idt+ 1 ∫ t
( 1
idt = + 1 ) ∫ t
idt
C 1 −∞ C 2 −∞ C 1
C 2 −∞
−∞
idt
Therefore,thetwocapacitorscanbereplacedbyanequivalentcapacitance,C,given
by
so that
1
C = 1 C 1
+ 1 C 2
= C 1 +C 2
C 1
C 2
C = C 1 C 2
C 1
+C 2
The result is easily generalised for more than two capacitors. For example, for three
capacitors wehave
1
C = 1 + 1 + 1 C 1
C 2
C 3
so that
C
C = 1
C 2
C 3
C 2
C 3
+C 1
C 3
+C 1
C 2
You may wish toprove this result.
∫ ∞
2
Example16.7 Evaluate
3 2t+1 − 1 t dt.
Solution
∫ ∞
3
2
2t+1 − 1 t dt =[ln|2t +1| −ln |t|]∞ 3
[
] = ln
2t+1
∞ [
∣ t ∣ = ln
∣ 2 + 1 3
t ∣
[ (
= lim ln 2 + 1 )]
−ln 7
t→∞ t 3
] ∞
3
=ln2−ln 7 3 = ln 6 7 = −0.1542