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

∫ 2
either of the integrals failstoconverge then
∫ b
−1
1
dx = ln |b| −ln |−1| = ln|b|
x
∫ b
1
lim dx = lim
b→0 − −1 x b→0 −(ln|b|)
This limitfailstoexist and so
∫ 2
−1
−1
1
dx diverges.
x
16.3 Improper integrals 487
1
dx diverges. Now,
x
Engineeringapplication16.2
Energystoredinacapacitor
Acapacitorprovidesausefulmeansofstoringenergy.Thisenergycanbedischarged
by connecting the capacitor in series with a load resistor and closing a switch. The
stored electrical energy is converted into heat energy as a result of electrical current
flowing through the resistor. Consider the circuit shown in Figure 16.2 which
consists of a capacitor,C F, connected in series with a resistor with valueRand
isolatedbymeansofaswitch,S.Wewishtocalculatetheamountofenergystoredin
the capacitor. The switch is closed att = 0 and a current,i, flows in the circuit. We
have already seen in Chapter 2 that for such a case the time-varying voltage across
the capacitordecaysexponentially and isgiven by
v =Ve −t/RC
So,usingOhm’s law
i = v R = Ve−t/RC
R
S
i
C
R
Figure16.2
The capacitor is dischargedbyclosingthe switch.
Now the effect of closing the switch is to allow the energy stored in the capacitor to
be dissipated in the resistor. Therefore, if the total energy dissipated in the resistor
is calculated then this will allow the energy stored in the capacitor to be obtained.
However, the energy dissipation rate, that is power dissipated, is not a constant for
theresistorbutdependsonthecurrentflowingthroughit.Thetotalenergydissipated,
E, isgiven by
E =
∫ ∞
0
P(t)dt
whereP(t) is the power dissipated in the resistor at timet. This equation has been
discussedinEngineeringapplication 13.3.Now,
P=i 2 R= RV2 e −2t/RC
R 2
= V2 e −2t/RC
R
➔