Encyclopedia of Electrical Engineering

We already know that after charging a capacitor, the charge q and voltage $v_C$ across capacitor's plates is formulated with the following equation:
$$ \bbox[10px,border:1px solid grey]{q = Cv} \tag{1}$$
By taking derivative w.r.t t of the above equation, we will get
$$ {dq \over dt} = C{dv \over dt} \tag{2}$$
where
$$ {dq \over dt} = i = i_C$$
The current $i_C$ associated with a capacitance C is related to the voltage
across the capacitor. Therefore eq. (2) becomes,
$$ \bbox[10px,border:1px solid grey]{i_C = C{dv_C \over dt}} \tag{3}$$
where $C{dv \over dt}$ is a measure of the change in $v_C$ in a vanishingly small
period of time. The function ${dv \over dt}$ is called the derivative of the voltage $v_C$ with respect to time t.
If the voltage fails to change at a particular instant, then
$$dv_C = 0$$
and
$$ i_C = C{dv \over dt} = 0 A$$
In other words, if the voltage across a capacitor fails to change with
time, the current $i_C$ associated with the capacitor is zero. To take this a
step further, the equation also states that the more rapid the change in
voltage across the capacitor, the greater the resulting current.
In an effort to develop a clearer understanding of Eq. (3), let us
calculate the average current associated with a capacitor for various
voltages impressed across the capacitor. The average current is defined
by the equation
$$ \bbox[10px,border:1px solid grey]{i_{Cav} = C{\Delta v_C \over \Delta t}} \tag{4}$$
where $\Delta$ indicates a finite (measurable) change in charge, voltage, or
time. The instantaneous current can be derived from Eq. (4) by letting $\Delta t$ become vanishingly small; that is,
$$ i_{Cins} = \lim_{t \to 0} C{\Delta v_C \over \Delta t} = C {dv \over dt}$$
**Example 1: **Find the waveform for the average current if the
voltage across a $2-\mu F$ capacitor is as shown in Fig. 1.
**Fig. 1: **For example 1.
**Solution: **

**a.** From 0 ms to 2 ms, the voltage increases linearly from 0 V to 4 V, the change in voltage
$$\Delta v = 4 V - 0 = 4 V $$
(with a positive sign since the voltage increases with time). The change in time
$$\Delta t =2 ms - 0 = 2 ms$$
and
$$ \begin{split}
i_{Cav} &= C{\Delta v_C \over \Delta t} \\
&= (2 \times 10^{-6} F) {4 \over (2 \times 10^{-3})}\\
&= (4 \times 10^{-3}) = 4mA
\end{split}
$$
**b.** From 2 ms to 5 ms, the voltage remains constant at 4 V; the change
in voltage $\Delta v = 0$. The change in time $\Delta t = 3 ms$, and
$$i_{Cav} = C{\Delta v_C \over \Delta t} = C{0 \over \Delta t} = 0 A$$
**c.** From 5 ms to 11 ms, the voltage decreases from 4 V to 0 V. The change
in voltage $\Delta v$ is, therefore, 4 V - 0 = 4 V (with a negative sign since the voltage is decreasing with time). The change in time $\Delta t =11 ms - 5 ms = 6 ms$, and
$$ \begin{split}
i_{Cav} &= C{\Delta v_C \over \Delta t} \\
&= -(2 \times 10^{-6} F) {4 \over (6 \times 10^{-3})}\\
&= -(1.33 \times 10^{-3}) = -1.33mA
\end{split}
$$
**d.** From 11 ms on, the voltage remains constant at 0 and $\Delta v = 0$, so
$i_{Cav} = 0$. The waveform for the average current for the impressed voltage is as shown in Fig. 2.
**Fig. 2: **The resulting current $i_C$ for the applied voltage of Fig. 1.
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