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

13.2 Elementary integration 435
Engineeringapplication13.2
Voltageacrossacapacitor
Recall from Engineering application 10.2 that the current, i, through a capacitor
depends upon time,t, and isgiven by
i =C dv
dt
where v is the voltage across the capacitor andC is the capacitance of the capacitor.
Derive an expression for v.
Solution
If
i =C dv then
dt
Therefore,
∫ i
v =
C dt = 1 ∫
C
i dt
dv
dt
= i C
usinglinearity
Notethatwhereasthecapacitance,C,isconstant,thecurrent,i,isnotandsoitcannot
betakenthroughtheintegralsign.Inordertoperformtheintegrationweneedtoknow
i as a function oft.
13.2.2 Electronicintegrators
Often there is a requirement in engineering to integrate electronic signals. Various circuits
are available to carry out this task. One of the simplest circuits is shown in
Figure 13.3. The input voltage is v i
, the output voltage is v o
, the voltage drop across
theresistoris v R
andthecurrentflowinginthecircuitisi.ApplyingKirchhoff’svoltage
lawyields
v i
=v R
+v o
(13.1)
For the resistor with resistance,R,
v R
=iR (13.2)
For the capacitorwith capacitance,C,
i =C dv o
dt
Combining Equations (13.1) to(13.3) yields
(13.3)
v i
=RC dv o
+ v
dt o
(13.4)
In general, v i
will be a time-varying signal consisting of a range of frequencies. For
the case where v i
is sinusoidal we can specify a property of the capacitor known as the
capacitive reactance,X c
, given by
X c
= 1
2πfC