An accurate RC series circuit with a source connected is shown in the following Fig.

The charging transient begins when a switch is closed, at a time t=0 sec. With the switch closed, the voltage equation is:

$E={{V}_{R}}+{{V}_{c}}$

$E=iR+{{V}_{c}}~~~\cdots ~~\left( 2 \right)$

At the instant, the switch is closed, the capacitor which is considered initially uncharged, has zero voltage (V_{c}=0), so equation (2) becomes;

$E=iR$

Or

$i={{I}_{o}}=\frac{E}{R}~~~~~~\cdots ~~~~\left( 3 \right)$

The initial charging current I_{o} is then limited by the resistance of the circuit and is determined by the simple application of ohm’s law. On the other hand, after sufficient time, the capacitor is fully charged and no current flows. Then, from eq. (2)

$E=0+{{V}_{c}}~~~~~\cdots \text{ }~~\left( 4 \right)$

Equation (3) and (4) state initial and final conditions for simple charging circuit and it becomes evident to obtain expressions for V_{c} and i as function of time throughout the transient period. So, we can write eq. (2) as

$E=iR+{{V}_{c}}~~~~~~~~~~~\cdots ~~~~~~~~~~~\left( 5 \right)$

Whereas;

$i=\frac{dq}{dt}$

And

${{V}_{C}}=\frac{q}{C}$

Putting both values in eq. (5) yields

$E=R\frac{dq}{dt}+\frac{q}{C}$

Dividing, both sides by “R”

$\frac{E}{R}=\frac{dq}{dt}+\frac{q}{RC}$

Multiplying both sides by dt

$\frac{E}{R}dt=\frac{dq}{dt}dt+\frac{q}{RC}dt$

$\left( \frac{E}{R}-\frac{q}{RC} \right)dt=dq$

_{$\frac{dq}{\left( \frac{E}{R}-\frac{q}{RC} \right)}=dt~~~~~~~~~~~~\cdots ~~~~~~~~~~~~\left( 6 \right)$ }

After integrating eq. (6), we have

$-RC\ln (\frac{E}{R}-\frac{q}{RC})=t+k~~~~~~~~~~~~~~\cdots ~~~~~~~~~~~\left( 7 \right)$

At t=0

\[q=0\]

$k=-RC\ln \frac{E}{R}$

So,

After putting the the value the of k into eq. (7) we come up with following expression;

$\frac{\ln (\frac{E}{R}-\frac{q}{RC})}{{}^{E}/{}_{R}}=-\frac{t}{RC}$

Taking exponent on both sides yield;

$E-\frac{q}{C}=E{{e}^{-\frac{t}{RC}}}$

$\frac{q}{C}=E-E{{e}^{-\frac{t}{RC}}}$

So final expression for voltage across a capacitor would be;

${{V}_{C}}=E\left( 1-{{e}^{-\frac{t}{RC}}} \right)~~~~~~~~\cdots ~~~~~~~\left( 8 \right)$

Now the voltage across the resistor is;

${{V}_{R}}=E-{{V}_{C}}~~~~~~~~~\cdots ~~~~~~~~~\left( 9 \right)$

Putting value of V_{C }from (8) into (9) yields

${{V}_{R}}=E-E+E{{e}^{-\frac{t}{RC}}}$

${{V}_{R}}=E{{e}^{-\frac{t}{RC}}}$

And

$i=\frac{{{V}_{R}}}{R}=\frac{E}{R}{{e}^{-\frac{t}{RC}}}$

Final expression for current would be,

$i={{I}_{o}}{{e}^{-\frac{t}{RC}}}~~~~~~~~\cdots ~~~~~~~\left( 10 \right)$

The factor of e^{-t/RC} in all the transient equations clearly indicates that current i_{c} and voltage V_{R} approach zero at the same exponential rate as V_{C} approaches E as shown in Fig.

The product RC appearing in the exponential term has the units of seconds.

$RC=\left( ohms \right)\left( farad \right)$

$=\left( \frac{volts}{ampere} \right)\left( \frac{coulombs}{volt} \right)$

$=\left( \frac{volts}{{}^{coulombs}/{}_{second}} \right)\left( \frac{coulombs}{volt} \right)$

$=second$

This product is defined as the time constant of the RC charging circuit and has the symbol 𝝉 (tau)

$\tau =RC$

Since the time constant affects the exponential term, it determines how fast the curve charges. The following figure shows relative effect of on the transient curve; for larger , the rate of change is less.

For t=τ =RC, the exponential factor becomes

${{\varepsilon }^{-1}}=0.368$

So that

${{V}_{C}}=E\left( 1-0.368 \right)=0.632E$

And

$i={{I}_{0}}\left( 0.368 \right)=0.368{{I}_{0}}$

That is; during one time constant, the voltage rises to 63.2 % of its final value and current drops to 36.8 % of its initial value.

**RC Series Circuit discharging Transient**

Discharging is the process of removing charge from a previously charged capacitor with a subsequent delay in capacitor voltage.

Consider the following figure:

If the switch is initially placed at position 1 the capacitor will charge towards the supply voltage E and after 5-time constants can be considered fully charged.

If the switch is at position 2, the capacitor is across the resistor so that charge leaks through the resistor. The Kirchhoff’s voltage law for discharged case is,

$0=iR+{{V}_{c}}~~~~\cdots ~~~\left( 11 \right)$

Substitute current i_{C} into above equation, we have

$RC\frac{d{{V}_{C}}}{dt}+{{V}_{c}}=0$

So, the final differential equation yields the expression for V_{C}

${{V}_{C}}={{V}_{0}}{{e}^{-\frac{t}{RC}}}$

Similarly, the discharging current can be obtained by solving eq. (11), which is

$i=-\frac{{{V}_{0}}}{R}{{e}^{-\frac{t}{RC}}}$

The minus sing indicates that current is opposite to that shown in above figure, which is, the discharging current has a direction opposite to that for the charging current. The V_{o} appearing in the discharging equations is the initial value of the capacitor voltage at the start of the discharge transient. The discharge equations are summarized by the following figure.

** ****Energy storage in a capacitor**

Capacitance stores energy in an electric field. This energy must come from a source. However, all the energy delivered by the source is not stored in the electric field. The energy stored by the capacitance is only one-half of the total energy delivered by the source. In equitation form, the average energy stored by a capacitance is;

${{W}_{E}}=\frac{1}{2}C{{V}^{2}}~~~~~\cdots ~~~~~\left( 12 \right)$

Where W_{E} represents the average energy stored in an electric field in Joules, C is the value of the capacitance in farads, V is an average value of the voltage across the capacitor in volts.

Eq. (12) is written in terms of average values. This is because it is the average value that is of primary concern, not an instantaneous value. If an AC voltage waveform is impressed across a capacitor, the average energy stored over one cycle is zero. The energy storage feature makes a capacitor useful device for the generation of large current for a very short time.

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