# Current through capacitor   Whatsapp  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,
$$i_{C_{ins}} = \lim_{t \to 0} C{\Delta v_C \over \Delta t} = C {dv \over dt}$$
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].

## Do you want to say or ask something?

Only 250 characters are allowed. Remaining: 250