Response of Capacitor to a Sinusoidal Voltage
Our investigation of the inductor revealed that the inductive voltage
across a coil opposes the instantaneous change in current through the
coil. For capacitive networks, the voltage across the capacitor is limited
by the rate at which charge can be deposited on, or released by, the
plates of the capacitor during the charging and discharging phases,
respectively. In other words, an instantaneous change in voltage across
a capacitor is opposed by the fact that there is an element of time
required to deposit charge on (or release charge from) the plates of a
capacitor, and $V = Q/C$.
In addition, the fundamental equation relating the voltage across a
capacitor to the current of a capacitor [$i = C(dv/dt)$] indicates that
for a particular capacitance, the greater the rate of change of voltage
across the capacitor, the greater the capacitive current.
Certainly, an increase in frequency corresponds to an increase in the
rate of change of voltage across the capacitor and to an increase in the
current of the capacitor.
The current of a capacitor is therefore directly related to the frequency (or, again more specifically, the angular velocity) and the capacitance of the capacitor. An increase in either quantity will result in an
increase in the current of the capacitor. For the basic configuration of
[Fig. 1], however, we are interested in determining the opposition of
the capacitor as related to the resistance of a resistor and $wL$ for the
inductor. Since an increase in current corresponds to a decrease in
opposition, and $i_C$ is proportional to w and C, the opposition of a capacitor is inversely related to $w (= 2\pi f )$ and C.
We will now verify, as we did for the inductor, some of the above
conclusions using a more mathematical approach.
For the capacitor of [Fig. 2],
applying differentiation,
where
Note that the peak value of iC is directly related to $w (= 2 \pi f )$ and C,
as predicted in the discussion above.
A plot of vC and iC in [Fig. 3] reveals that
If a phase angle is included in the sinusoidal expression for $v_C$, such
as
then
Applying
and substituting values, we obtain
which agrees with the results obtained above.
The quantity $1/wC$, called the reactance of a capacitor, is symbolically represented by $X_C$ and is measured in ohms; that is,
In an Ohm's law format, its magnitude can be determined from
Capacitive reactance is the opposition to the flow of charge, which
results in the continual interchange of energy between the source and
the electric field of the capacitor. Like the inductor, the capacitor does
not dissipate energy in any form (ignoring the effects of the leakage
resistance).
It is possible now to determine whether a network with one or more elements is predominantly capacitive or inductive by noting the phase relationship between the input voltage and current.
If the source current leads the applied voltage, the network is
predominantly capacitive, and if the applied voltage leads the source
current, it is predominantly inductive.
Since we now have an equation for the reactance of an inductor or
capacitor, we do not need to use derivatives or integration in the
examples to be considered. Simply applying Ohm's law, $I_m = E_m/X_L$
(or $X_C$), and keeping in mind the phase relationship between the voltage and current for each element, will be sufficient to complete the examples.
Fig. 1: Defining the parameters that determine the opposition of a capacitive element
to the flow of the charge.
Fig. 2: Investigating the sinusoidal response of a capacitive element.
Fig. 3: The current of a purely capacitive element
leads the voltage across the element by 90 degrees.
$$i_C = C {dv_C \over dt}$$
$$i_C = C {dV_m \sin wt \over dt}$$
$$i_C = C V_m w \cos wt = wCV_m \cos wt$$
$$i_C = I_m \cos wt = I_m\sin (wt+90)$$
$$\bbox[10px,border:1px solid grey]{ I_m = wCV_m } \tag{1}$$
$$v_C = V_m \sin (wt+\theta)$$
$$i_C = wCV_m \sin (wt + \theta + 90)$$
$$ opposition = {cause \over effect}$$
$$ opposition = {V_m \over I_m}= {V_m \over wCV_m} = {1 \over wC}$$
$$\bbox[10px,border:1px solid grey]{X_C = {1 \over wC}} \,(ohms) \tag{2}$$
$$\bbox[10px,border:1px solid grey]{X_C = {V_m \over I_m}} \,(ohms) \tag{3}$$
Example 1: The voltage across a $1 \mu \, F$ capacitor is provided
below. What is the sinusoidal expression for the current? Sketch the v
and i curves.
Solution:
Eq. 2:
and
and we know that for a capacitor i leads v by 90 degrees. Therefore,
The curves are sketched in [Fig. 4].
$$v = 30 \sin 400t$$
$$X_C = {1 \over wC} = {1 \over (400) ( 1\times 10^{-6})}=2500Ω$$
$$I_m = {V_m \over X_C} = {30V \over 2500Ω}=0.012A = 12mA$$
$$ \bbox[10px,border:1px solid grey]{i = 12\times10^{-3} \sin(400t + 90)}$$
Fig. 4: Example 1.
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