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 $V = Q/C$.
Since capacitance is a measure of the rate at which a capacitor will store charge on its plates,
for a particular change in voltage across the capacitor, the greater the value of capacitance, the greater will be the resulting capacitive current.
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.
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,
$$i_C = C V_m w \cos wt = wCV_m \cos wt$$
$$i_C = I_m \cos wt = I_m\sin (wt+90)$$
$$\bbox[10px,border:1px solid grey]{ I_m = wCV_m } \tag{1}$$
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
for a capacitor, $i_C$ leads $v_C$ by 90 degrees, or $v_C$ lags $i_C$ by 90 degree.
If a phase angle is included in the sinusoidal expression for $v_C$, such as
$$v_C = V_m \sin (wt+\theta)$$
$$i_C = wCV_m \sin (wt + \theta + 90)$$
$$ opposition = {cause \over effect}$$
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:
$$X_C = {1 \over wC} = {1 \over (400) ( 1\times 10^{-6})}=2500Ω$$
$$I_m = {V_m \over X_C} = {30V \over 2500Ω}=0.012A = 12mA$$
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.

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