[Fig. 1] following the closing of the switch (to
position 1).
**Fig. 1: **Basic charging network
You will recall that the instant the switch is closed, electrons are
drawn from the top plate and deposited on the bottom plate by the battery, resulting in a net positive charge on the top plate and a negative charge on the bottom plate. The transfer of electrons is very rapid at first, slowing down as the potential across the capacitor approaches the
applied voltage of the battery. When the voltage across the capacitor
equals the battery voltage, the transfer of electrons will cease and the
plates will have a net charge determined by $Q = CV_C = CE$.
Plots of the changing current and voltage appear in [Fig. 2(a)] and
[2(b)], respectively. When the switch is closed at $t = 0 s$, the current
jumps to a value limited only by the resistance of the network and then
decays to zero as the plates are charged.
**(a)**
**(b)**
**Fig. 2: **Plots of the changing current and voltage: (a) ic during the charging phase. (b) vc during the charging phase
Note the rapid decay in current level, revealing that the amount of charge deposited on the plates per unit time is rapidly decaying also. Since the voltage across the plates is
directly related to the charge on the plates by $v_C = q/C$, the rapid rate
with which charge is initially deposited on the plates will result in a
rapid increase in $v_C$. Obviously, as the rate of flow of charge (I)
decreases, the rate of change in voltage will follow suit. Eventually, the
flow of charge will stop, the current I will be zero, and the voltage will
cease to change in magnitude-the charging phase has passed.
**(a)**
**(b)**
**Fig. 3: **A capacitor
following the charging phase: (a) Open-circuit equivalent. (b) Short-circuit equivalent
At this
point the capacitor takes on the characteristics of an open circuit: a voltage drop across the plates without a flow of charge "between" the plates. As demonstrated in [Fig. 3(a)], the voltage across the capacitor
is the source voltage since
and
Looking back at the instant the switch is closed, we can also surmise
that a capacitor behaves as a short circuit the moment the switch is
closed in a dc charging network, as shown in [Fig. 3(b)].
and the voltage
**Fig. 4: ** The $e^{-x}$ function (x ≥ 0)
Through the use of calculus, the following mathematical equation
for the charging current $i_C$ can be obtained:
The factor $e^{-t/RC}$ is an exponential function of the form $e^{-x}$, where
$x =-t/RC$ and $e = 2.71828 . . . .$ A plot of $e^{-x}$ for $x > 0$ appears in
Fig. 4. Exponentials are mathematical functions that all students
of electrical, electronic, or computer systems must become very familiar with. They will appear throughout the analysis to follow in this course, and in succeeding courses.
#### Time constants in charging phase

The factor RC in Eq. (1) is called the time constant of the system and has the units of time as follows:
Its symbol is the Greek letter $\tau$ (tau), and its unit of measure is the second. Thus,
If we substitute $t = RC$ into the exponential function $e^{-t/RC}$, we
obtain $e^{-t/\tau}$. In one time constant,
or the function equals 36.79% of its maximum value of 1. At $t = 2\tau$,
and the function has decayed to only 13.53% of its maximum value.
The rate of change of $i_C$ is therefore quite sensitive to the time constant
determined by the network parameters R and C. For this reason, the universal time constant chart of [Fig. 5] is provided to permit a more
accurate estimate of the value of the function $e^{-x}$ for specific time intervals related to the time constant.
**Fig. 5:** Universal time constant chart.
Returning to Eq. (1), we find that the multiplying factor E/R is
the maximum value that the current $i_C$ can attain, as shown in [Fig. 2]. Substituting t = 0 s into Eq. (1) yields
verifying our earlier conclusion.
For increasing values of t, the magnitude of $e^{-t/\tau}$, and therefore the
value of $i_C$, will decrease, as shown in [Fig. 6]. Since the magnitude
of $i_C$ is less than 1% of its maximum after five time constants, we will
assume the following for future analysis:
Since C is usually found in microfarads or picofarads, the time constant $\tau = RC$ will never be greater than a few seconds unless R is very large.
**Fig. 6:**$i_C$ versus t during the charging phase.
#### Capacitor voltage during charging phase

Let us now turn our attention to the charging voltage across the capacitor. By applying KCL to [Fig. 1], we get
Rearranging for $v_C$,
Now replacing $i_C$ from equation (1) in the above equation, we get
After cancelation,
Through mathematical analysis, the following equation for the voltage across the capacitor can be determined:
In addition, since E is the multiplying factor, we can conclude that, for all practical purposes, the voltage $v_C$ is E volts after five time constants of the charging phase. A plot of $v_C$ versus t is provided in [Fig. 7(a)].
**(a)**
**(b)**
**Fig. 7: **(a) vC versus t during the charging phase. (b) Effect of C on the charging phase.
If we keep R constant and reduce C, the product RC will decrease,
and the rise time of five time constants will decrease. The change in
transient behavior of the voltage $v_C$ is plotted in [Fig. 7(b)] for various
values of C. The product RC will always have some numerical value,
even though it may be very small in some cases. For this reason:
In fact, the capacitance of a network is also a measure of how much it
will oppose a change in voltage across the network. The larger the
capacitance, the larger the time constant, and the longer it takes to
charge up to its final value (curve of $C_3$ in [Fig. 7(b)]). A lesser capacitance would permit the voltage to build up more quickly since the time
constant is less (curve of $C_1$ in [Fig. 7(b)]).
#### Charging rate on the plates

The rate at which charge is deposited on the plates during the charging phase can be found by substituting the following for $v_C$ in Eq. (3):
indicating that the charging rate is very high during the first few time
constants and less than 1% after five time constants.
#### Properties of a capacitor

We should note the following important properties of a capacitor:

Previous sections have described how a capacitor acquires its charge. Let us now
extend this discussion to include the potentials and current developed
within the network of $$ i = i_C = i_R = 0 A$$

$$v_R = i_RR = (0)R Ω =0 V$$

$$i = i_C = i_R = E/R$$

$$\begin{split}
v_C &= E - V_R = E - i_RR\\
&= E - (E/R)R \\ &= E - E = 0 V \text{at t = 0 s}
\end{split}
$$

$$\bbox[10px,border:1px solid grey]{i_C = {E \over R} e^{-t/RC}} \tag{1}$$

$$RC = {V \over I} {Q \over V} = {Q \over I} = t$$

$$ \tau = RC \quad \text{(seconds, s)} \tag{2}$$

$$e^{-t/\tau} = e^{-\tau/\tau} = e^{-1} = 0.3679$$

$$e^{-t/\tau} = e^{-2\tau/\tau} = e^{-2} = 0.1353$$

$$i_C = {E \over R} e^{-t/RC} = {E \over R} e^{-0/RC}={E \over R} e^0={E \over R}$$

$$ E = i_C R + v_C$$

$$ v_C = E - i_CR$$

$$ v_C = E - {E \over R} e^{-t/RC} R$$

$$ v_C = E - E e^{-t/RC}$$

$$ \bbox[10px,border:1px solid grey]{v_C = E(1 - e^{-t/RC})} \tag{3} $$

$$ q = C v_C$$

$$ \bbox[10px,border:1px solid grey]{q = C E(1-e^{-t/RC})} \tag{4}$$

- when the voltage across a capacitor is not changing with time (i.e., dc voltage), the current through the capacitor is zero. Thus, However, if a battery (dc voltage) is connected across a capacitor, the capacitor charges.
- The voltage on the capacitor must be continuous. The capacitor resists an abrupt change in the voltage across it.
- The ideal capacitor does not dissipate energy. It takes power from the circuit when storing energy in its field and returns previously stored energy when delivering power to the circuit.
- A real, non-ideal capacitor has a parallel-model leakage resistance. The leakage resistance may be as high as $100 MΩ$ and can be neglected for most practical applications. For this reason, we will assume ideal capacitors in this book.

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