RC Circuit

An RC circuit is made up of a resistor (R) and a capacitor (C) connected either in series or parallel. These circuits are widely used in electronics for tasks like filtering, timing, and processing signals. The way the circuit behaves is mainly determined by how the capacitor charges and discharges over time, which is influenced by the time constant. RC circuits play an essential role in analyzing how circuits respond to changes and are important in applications like high-pass and low-pass filters, oscillators, and many other electronic systems.

1.0Charging of A Capacitor

• In the following circuit, when key 1 is closed, the capacitor initiates to charge, and the charging process takes a finite amount of time. The charge on the capacitor at any given time tt is expressed as, $q=q_0\left[1-e^{-\left(\frac{t}{RC}\right)}\right]$                              

Here, $q_0$ represents the maximum charge the capacitor will reach as $t=\infty$. According to these equations, the charge on the capacitor increases exponentially over time.

• If $t=RC=\tau$,then 

$q=q_0\left[1-e^{-\left(\frac{RC}{RC}\right)}\right]=q_0\left[1-\frac{1}{e}\right]=$

$q=q_0(1-0.37)=0.63q_0=63\%\text{ of }{q}_0$

• Time t = RC is known as time constant.The time constant is the duration required for the charge on the capacitor plates to reach 63% of its maximum value.
• The voltage across the capacitor plates at any given moment is expressed as:

$V=V_0\left[1-e^{-\left(\frac{t}{RC}\right)}\right]Volt$

• The potential curve follows a similar pattern to that of the charge. During the charging process, a temporary electric current, known as the transient current, flows through the circuit for a brief period. The current at any given moment is given by,

$I=I_0\left[e^{-\left(\frac{t}{RC}\right)}\right]Ampere$

According to this equation, the current in the circuit decreases exponentially over time.

• If  $t=RC=\tau =\text{Time Constant}$

$I=I_0\left[e^{-\left(\frac{RC}{RC}\right)}\right]=\frac{I_0}{e}=0.37I_0=37\%\text{ of }{I}_0$

The time constant is the duration required for the current in the circuit to drop to 37% of its maximum value.

2.0 Derivation of Capacitor Charging Formulas

• Assuming the capacitor is initially uncharged, let the charge on the capacitor at any time be q. Applying Kirchhoff's voltage law

$\epsilon -iR-\frac{q}{C}=0\Rightarrow iR=\frac{\epsilon C-q}{C}$

$i=\frac{\epsilon C-q}{CR}\Rightarrow \frac{dq}{dt}=\frac{\epsilon C-q}{CR}$

$\frac{\epsilon C-q}{CR}dq=dt$

${\int }_0^q\frac{dq}{\epsilon C-q}={\int }_0^t\frac{dt}{RC}\Rightarrow -\ln (\epsilon C-q)+\ln \epsilon C=\frac{t}{RC}$

$\ln \frac{\epsilon C}{\epsilon C-q}=\frac{t}{RC};\epsilon C-q=\epsilon C.e^{-\frac{t}{RC}}$

$q=\epsilon C\left(1-e^{-\frac{t}{RC}}\right)$

$RC=\text{time constant of the RC series circuit.}$

After one time constant

$q=\epsilon C\left(1-\frac{1}{e}\right)=\epsilon C(1-0.37)=0.63\epsilon C$

Current at any time t 

$i=\frac{dq}{dt}=\epsilon C\left(e^{-\frac{t}{RC}}\left(-\frac{1}{RC}\right)\right)=\frac{\epsilon }{R}{e}^{-\frac{t}{RC}}$

Voltage across capacitor after one-time constant $V=0.63\epsilon$

$Q=CV;V_c=\epsilon \left(1-e^{-\frac{t}{RC}}\right)$

3.0Voltage Across the Resistor

$V_R=iR=\epsilon e^{-\frac{t}{RC}}$

By energy conservation,

$\text{Heat dissipated}=\text{work done by battery}-\Delta U_{capacitor}$

$\text{Heat dissipated}=C\epsilon (\epsilon )-\left(\frac{1}{2}C{\epsilon }^2-0\right)=\frac{1}{2}C{\epsilon }^2$

Alternatively,

$\text{Heat}=H={\int }_0^{\infty }{i}^{2}Rdt$

$R{\left[\frac{e^{-\frac{2t}{RC}}}{-\frac{2}{RC}}\right]}_0^{\infty }$

$H=\frac{{\epsilon }^{2}RC}{2R}{\left[e^{-\frac{2t}{RC}}\right]}_0^{\infty }=\frac{{\epsilon }^{2}C}{2}$

Note:

In the figure, the time constant of  (2) is greater than that of  (1).

4.0Discharging of a Capacitor

• In the circuit shown below, when key 1 is opened and key 2 is closed, the capacitor begins to discharge.

$\text{(Initial charge on capacitor }=q_0)$

• The charge on the capacitor at any given time t is expressed by,

$q=q_0{e}^{-\left(\frac{t}{RC}\right)}$

The charge decreases exponentially over time.

$q_0=\text{initial charge of capacitor}$

• $\text{ If }t=RC=\tau =\text{time constant}$, then $q=\frac{q_0}{e}=0.37q_0=37\%\text{ of }{q}_0$
•  The time constant is the duration required for the charge on the capacitor plates to decrease to 37% during the discharge process.
• The dimensions of RC are those of time, represented as $[M^0{L}^0{T}^1]$, while the dimensions of 1/RC correspond to frequency, represented as $[M^0{L}^0{T}^{-1}]$.
• The voltage across the capacitor plates at any given time t is expressed as:

$V=V_0{e}^{-\left(\frac{t}{RC}\right)}Volt$

• The instantaneous transient current is given by,

$I=-I_0{e}^{-\left(\frac{t}{RC}\right)}Ampere$

The current in the circuit reduces exponentially, but its direction is opposite to that of the charging current. (The negative sign simply indicates that the current streams in the opposite direction compared to the charging current.)

5.0Derivation of Discharging Circuit Equation

Applying K.V.L.

$+\frac{q}{C}-iR=0$

$i=\frac{q}{CR},-\frac{dq}{dt}=\frac{q}{CR}$

${\int }_q^Q\frac{dq}{q}={\int }_0^t\frac{dt}{CR}$

$-\ln \frac{q}{Q}=+\frac{t}{RC}$

$q=Q.e^{-\frac{t}{RC}}$

$i=-\frac{dq}{dt}=\frac{Q}{RC}{e}^{-\frac{t}{RC}}=i_0{e}^{-\frac{t}{RC}}$

$V=\frac{q}{C}=\frac{Q}{C}{e}^{-\frac{t}{RC}}=V_0{e}^{-\frac{t}{RC}}$

6.0Thevenin's Theorem(Time Constant Calculation in Complex Circuits)

Thevenin's Theorem:

• To calculate time constant and write charge as function of time directly in those circuits where only one capacitor is present.
• It states that any linear circuit containing any amount of emfs source and resistance can be replaced by a single voltage source called Thevenin's voltage $\left(V_{Th}\right)$
• and single series resistance connected to load which is termed as Thevenin's resistance $\left(R_{Th}\right)$.

• In RC circuits, use Thevenin’s theorem to simplify the circuit and calculate the time constant and charge on the capacitor without analyzing the entire circuit.
• Open the capacitor's terminal and replace voltage sources with their internal resistances.
• The equivalent resistance between the terminals is the effective resistance $\left(R_{eff}\right)$.
• The time constant is $\tau =R_{eff}C$
• If the capacitor is initially uncharged, the charge on the capacitor at time $q=Q_{st}\left(1-e^{-\frac{t}{\tau }}\right)$ where $Q_{st}$ is the steady-state charge on the capacitor.

Illustration Thevenin's Theorem

Let's try to understand the application of this method by illustrating following example

Solution:

Step-1.Effective resistance will be equivalent of the combination between terminals A and B

$R_{eff}=\frac{3R}{2}$

Step-2.Time Constant, $\tau =R_{eff}C\Rightarrow \tau =\frac{3RC}{2}$

Step-3.The given circuit can be redrawn as

$q=Q_{st}\left(1-e^{-\frac{t}{\tau }}\right)$

By substituting the time constant and steady-state charge, we obtain the expression for the charge as a function of time.

$q=\frac{\epsilon C}{2}\left(1-e^{-\frac{2t}{3RC}}\right)$

7.0Charging of complex RC circuits

• Complex RC circuits may involve multiple resistors and capacitors. The charge as a function of time can be calculated using Kirchhoff's Voltage Law (KVL) and Kirchhoff's Current Law (KCL). This process is demonstrated using KVL in the following simple case, where the capacitor is initially uncharged, and we aim to find the charge on the capacitor as a function of time.

Applying KVL in loop ABCDA

$\epsilon -iR-(i-i_1)R=0$

$\epsilon -2iR+i_{1}R=0......(1)$

Applying KVL in loop ABCEFDA

$\epsilon -iR-i_{1}R-\frac{q}{C}=0$

By using equation (1)

$\frac{2\epsilon -\epsilon -i_{1}R-2i_{1}R}{2}=\frac{q}{C}\Rightarrow \epsilon C-3i_{1}RC=2q$

$\epsilon C-2q=3\frac{dq}{dt}\cdot RC\Rightarrow {\int }_0^q\frac{dq}{\epsilon C-2q}={\int }_0^t\frac{dt}{3RC}$

$-\frac{1}{2}\ln \frac{\epsilon C-2q}{\epsilon C}=\frac{t}{3RC}\Rightarrow q=\frac{\epsilon C}{2}\left(1-e^{-\frac{2t}{3RC}}\right)$

Illustration:In circuits with multiple capacitors, without using the equivalent formula, we will determine the charge on each capacitor and the current in all branches as a function of time.

Applying KVL in ABDEA

$\epsilon -iR=\frac{q}{2C}$

$i=\frac{\epsilon }{R}-\frac{q}{2CR}=\frac{2C\epsilon -q}{2CR}$

$\frac{dq}{2C\epsilon -q}=\frac{dt}{2CR}$

${\int }_0^q\frac{dq}{(2C\epsilon -q)}=\frac{t}{2CR}$

$\frac{2C\epsilon -q}{2C\epsilon }=e^{-\frac{t}{2RC}}$

$q=2C\epsilon \left(1-e^{-\frac{t}{2RC}}\right)$

$q_1=\frac{q}{2}=\epsilon C\left(1-e^{-\frac{t}{2RC}}\right)\Rightarrow i_1=\frac{\epsilon }{2R}{e}^{-\frac{t}{2RC}}$

$q_2=\frac{q}{2}=\epsilon C\left(1-e^{-\frac{t}{2RC}}\right)\Rightarrow i_2=\frac{\epsilon }{2R}{e}^{-\frac{t}{2RC}}$

8.0Solved Examples: RC Circuits

Q-1.Determine the current in the circuit and the charge on the capacitor, which is initially uncharged, in the following scenarios: (a) Immediately after the switch is closed. (b) After a long period, once the switch has been closed.

Solution:

For just after closing the switch:

potential difference across capacitor = 0

$Q_c=0\therefore i=5A$

After a long time at steady state current, i=0 and potential difference across capacitor = 10 V       

$Q_c=3\times 10=30C$

Q-2.Find out current $I_1,I_2,I_3$ charge on capacitor and $\frac{dQ}{dt}$ of capacitor in the

circuit which is at first uncharged in the following situations.

(a) Just after the switch is closed

(b) After a long time when switch is closed.

Solution:

(a).Initially the capacitor is uncharged so its behaviour is like a conductor

 Apply Kirchhoff’s $1^{st}$ law at point E 

$\frac{x-\epsilon }{R}+\frac{x-0}{R}+\frac{x-0}{R}=0\Rightarrow \frac{3x}{R}=\frac{\epsilon }{R}$

$x=\frac{\epsilon }{3};Q_C=0$

$I_1=\frac{\frac{\epsilon }{3}+\epsilon }{R}=\frac{2\epsilon }{3R},I_2=\frac{dQ}{dt}=\frac{\epsilon }{3R}\text{ and }{I}_3=\frac{\epsilon }{3R}$

$\text{(B). At }t=\infty \text{ (finally)}$

The capacitor is fully charged, so no current will flow through it.

$I_2=0,I_1=I_3=\frac{\epsilon }{2R}$

$V_E-V_B=V_D-V_C=\left(\frac{\epsilon }{2R}\right)R=\frac{\epsilon }{2}\Rightarrow Q_c=\frac{\epsilon C}{2},\frac{dQ}{dt}=I_2=0$

Example-3.At t=0, switch is closed, if initially $C_1$ is uncharged and $C_2$ is charged to a potential difference 2 then find out following

$(Given{C}_1=C_2=C)$.

(a) Charge on $C_1$ and $C_2$ as a function of time.

(b) Find out current in the circuit as a function of time.

(c) Also plot the graphs for the relations derived in part (a).

Solution :Now applying KVL

$\epsilon -\frac{q}{C}-iR-\frac{q-2\epsilon C}{C}=0$

$\epsilon -\frac{q}{C}-\frac{q}{C}+2iR=0$

$3\epsilon =\frac{2q}{C}+iR\Rightarrow 3\epsilon -iR=\frac{2q}{C}$

$3\epsilon C-iRC=2q\Rightarrow \frac{dq}{dt}RC=3\epsilon C-2q$

${\int }_0^q\frac{dq}{3\epsilon C-2q}={\int }_0^t\frac{dt}{RC}\Rightarrow -\frac{1}{2}\ln \left(\frac{3\epsilon C-2q}{3\epsilon C}\right)=\frac{t}{RC}$

$\ln \left(\frac{3\epsilon C-2q}{3\epsilon C}\right)=-\frac{2t}{RC}\Rightarrow 3\epsilon C-2q=3\epsilon C{e}^{-\frac{2t}{RC}}$

$3\epsilon C\left(1-e^{-\frac{2t}{RC}}\right)=2q\Rightarrow q=\frac{3}{2}\epsilon C\left(1-e^{-\frac{2t}{RC}}\right)\text{ (charge on C, as function of time)}$

$i=\frac{dq}{dt}=\frac{3\epsilon }{R}{e}^{-\frac{2t}{RC}}$

Charge on $C_2$ as function of time

$q^{\prime }=2\epsilon C-q$

$=2\epsilon C-\frac{3}{2}\epsilon C+\frac{3}{2}\epsilon C{e}^{-\frac{2t}{RC}}=\frac{\epsilon C}{2}+\frac{3}{2}\epsilon C{e}^{-\frac{2t}{RC}}$

$=\frac{\epsilon C}{2}\left[1+3e^{-\frac{2t}{RC}}\right]$

Example-4. A $5.0\mu F$ capacitor having a charge of $20\mu C$ is discharged through a wire of resistance $5.0\Omega$. Find the heat dissipated in the wire between $25to50\mu s$ after the connections are made. $(Given:e^{-2}=0.135)$

Solution:

$i_0=\frac{20}{5}\times \frac{1}{4}=\frac{4}{5}A$

$i=i_0{e}^{-\frac{t}{RC}}$

$H=\int i^{2}Rdt={\int }_{25\mu s}^{50\mu s}\frac{16}{5}{e}^{-\frac{2t}{25\times 10^{-6}}}dt$

$=\frac{16}{5}\left(-\frac{25}{2}\times 10^{-6}\right){\left[e^{-\frac{2t}{25\times 10^{-6}}}\right]}_{25\mu s}^{50\mu s}$

$=40\times 10^{-6}\left[\frac{e^2-1}{e^4}\right]=4.7\mu J$

Example-5.Find the time constant given circuit if

$R_1=4\Omega ,R_2=12\Omega ,C_1=3\mu F,C_2=6\mu F$

Solution: Given circuit can be reduced to

$C=\frac{C_1{C}_2}{C_1+C_2}=\frac{3\times 6}{3+6}=2\mu F,R=\frac{R_1{R}_2}{R_1+R_2}=\frac{4\times 12}{4+12}=3\Omega$

$\text{Time constant}=RC=(3)(2)=6\mu s$.