Encyclopedia of Electrical Engineering

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 $V = Q/C$.
**Fig. 1: **Defining the parameters that determine the opposition of a capacitive element
to the flow of the charge.
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.
**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.
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,
$$i_C = C {dv_C \over dt}$$
$$i_C = C {dV_m \sin wt \over dt}$$
applying differentiation,
$$i_C = C V_m w \cos wt = wCV_m \cos wt$$
$$i_C = I_m \cos wt = I_m\sin (wt+90)$$
where
$$\bbox[10px,border:1px solid grey]{ I_m = wCV_m } \tag{1}$$
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 vC, such
as
$$v_C = V_m \sin (wt+\theta)$$
then
$$i_C = wCV_m \sin (wt + \theta + 90)$$
Applying
$$ opposition = {cause \over effect}$$
and substituting values, we obtain
$$ opposition = {V_m \over I_m}= {V_m \over wCV_m} = {1 \over wC}$$
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,
$$\bbox[10px,border:1px solid grey]{X_C = {1 \over wC}} \,(ohms) \tag{2}$$
In an Ohm's law format, its magnitude can be determined from
$$\bbox[10px,border:1px solid grey]{X_C = {V_m \over I_m}} \,(ohms) \tag{3}$$
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.
**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.
$$v = 30 \sin 400t$$
**Solution: **
Eq. 2:
$$X_C = {1 \over wC} = {1 \over (400) ( 1\times 10^{-6})}=2500Ω$$
and
$$I_m = {V_m \over X_C} = {30V \over 2500Ω}=0.012A = 12mA$$
and we know that for a capacitor i leads v by 90 degrees. Therefore,
$$ \bbox[10px,border:1px solid grey]{i = 12\times10^{-3} \sin(400t + 90)}$$
The curves are sketched in Fig. 4.
**Fig. 4: **Example 1.

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.