RC Circuit Analysis: Series, Parallel, Equations & Transfer Function

Electrical4u

03/13/2024 14:50:32

What is an RC Circuit?

An RC circuit (also known as an RC filter or RC network) stands for a resistor-capacitor circuit. An RC circuit is defined as an electrical circuit composed of the passive circuit components of a resistor (R) and capacitor (C), driven by a voltage source or current source.

Due to the presence of a resistor in the ideal form of the circuit, an RC circuit will consume energy, akin to an RL circuit or RLC circuit.

This is unlike the ideal form of an LC circuit, which will consume no energy due to the absence of a resistor. Although this is only in the ideal form of the circuit, and in practice, even an LC circuit will consume some energy because of the non-zero resistance of the components and connecting wires.

Series RC Circuit

In an RC series circuit, a pure resistor having resistance R in ohms and a pure capacitor of capacitance C in Farads are connected in series.

Here $I$ is the RMS value of the current in the circuit.

$V_R$ is the voltage across the resistor R.

$V_C$ is the voltage across the capacitor C.

$V$ is the RMS value of the supply voltage.

The figure shows a vector diagram of the series RC circuit.

Since in a series circuit current $'I'$ is the same so it is taken as a reference.

$V_R = IR$ is drawn in phase with current $'I'$ because in a pure resistor the voltage and current are in phase with each other.

$V_C=I X_C$ is drawn lagging with current $'I'$ by $90^0$ because in a pure capacitor voltage and current are $90^0$ out of each other i.e. voltage lags current by $90^0$ or current leads the voltage by $90^0$ .

Now $V$ is the vector sum of $V_R$ and $V_C$ .

$\begin{align*} \,\, therefore, \,\, V^2 = {V_R}^2 + {V_C}^2 \end{align*}$

$\begin{align*} \begin{split} V = {\sqrt{{V_R}^2 + {V_C}^2}} \ & = {\sqrt{{IR}^2 + {IX_C}^2}} \ & = I {\sqrt{{R}^2 + {X_C}^2}} \ & = IZ \ \end{split} \end{align*}$

The impedance of an R-C series circuit is

$\begin{align*} Z = {\sqrt{{R}^2 + {X_C}^2}} \end{align*}$

$\begin{align*} \,\, where, \,\, X_C = \frac{1}{{\omega}C} = \frac{1}{2{\pi}fC} \end{align*}$

The voltage and impedance triangle are shown in figure.

As seen, the vector $V$ lags $I$ by an angle ø where

$\begin{align*} tan{\phi} = \frac{IX_C}{IR} \end{align*}$

$\begin{align*} {\phi} =tan^-^1 \frac{X_C}{R} \end{align*}$

Thus in an R-C series circuit current $'I'$ leads the supply voltage $'V'$ by an angle

$\begin{align*} {\phi} =tan^-^1 \frac{X_C}{R} \end{align*}$

$\begin{align*} \,\, i.e. \,\ if \,\,V = V_m sin{\omega}t \end{align*}$

$\begin{align*} i = I_m sin({\omega}t + {\phi}) \end{align*}$

$\begin{align*} \,\, where, \,\, I_m = \frac{V_m}{Z} \end{align*}$

The voltage and current waveforms of the R-C series circuit are shown in fig.

Power in an RC Series Circuit

The instantaneous value of the power is the product of the instantaneous values of the voltage and current.

$\begin{align*} P = V I \end{align*}$

$\begin{align*} = (V_m sin{\omega}t) [I_m sin({\omega}t + {\phi})] \end{align*}$

$\begin{align*} = \frac{V_m I_m}{2} [2sin{\omega}t * sin({\omega}t + {\phi})] \end{align*}$

$\begin{align*} = \frac{V_m I_m}{2} [cos[{\omega}t-({\omega}t+{\phi})] - cos[{\omega}t+({\omega}t+{\phi})]] \end{align*}$

$\begin{align*} = \frac{V_m I_m}{2} [cos({-\phi}) - cos({2\omega}t+{\phi})] \end{align*}$

$\begin{align*} = \frac{V_m I_m}{2} [cos{\phi} - cos({2\omega}t+{\phi})] \end{align*}$

$\begin{align*} \,\, [where, \,\, cos ({-\phi}) = cos {\phi} \,\, because \,\, cos \,\, curve \,\, is \,\, symmetric] \,\, \end{align*}$

$\begin{align*} = \frac{V_m I_m}{2} cos{\phi} - \frac{V_m I_m}{2} cos({2\omega}t+{\phi}) \end{align*}$

Thus the instantaneous power consists of two parts.

1. A constant part = $\frac{V_m I_m}{2} cos{\phi}$

2. A varying component = $\frac{V_m I_m}{2} cos({2\omega}t+{\phi})$ which varies at twice the supply frequency.

The average value of the varying power component over a complete cycle is zero.

Thus the average power consumed in an RC series circuit over one cycle is

$\begin{align*} \begin{split} P = \frac{V_m I_m}{2} cos{\phi} \ & = \frac{V_m}{\sqrt{2}} \frac{I_m}{\sqrt{2}} cos{\phi} \ & = V I cos{\phi} \ \end{split} \end{align*}$

Where $V$ and $I$ are the RMS values of the applied voltage and current in the circuit.

Power Factor in an RC Series Circuit

Consider the figure showing the power and impedance triangles.

$\begin{align*} \begin{split} \,\, (power \,\, factor) \,\, cos{\phi} = \frac{P \,\, (active \,\, power)\,\,} {S \,\, (apparent \,\, power)\,\,} \ & = \frac{R} {Z} \ & = \frac{R} {\sqrt{{R}^2 +{X_C}^2}} \ \end{split} \end{align*}$

Parallel RC Circuit

In a parallel R-C circuit a pure resistor having resistance $R$ in ohms and a pure capacitor of capacitance $C$ in Farads are connected in parallel.

Voltage drops in a parallel RC circuit are the same hence the applied voltage is equal to the voltage across the resistor and voltage across the capacitor. Current in a parallel R-C circuit is the sum of the current through the resistor and capacitor.

$\begin{align*} V = V_R = V_C \end{align*}$

$\begin{align*} I = I_R + I_C \end{align*}$

For the resistor, current through it given by ohm’s law:

$\begin{align*} I_R = \frac {V_i_n} {R} \end{align*}$

The voltage-current relationship for the capacitor is:

$\begin{align*} I_C = C \frac {dV_i_n} {dt} \end{align*}$

Applying KCL (Kirchhoff’s Current Law) to parallel R-C circuit

$\begin{align*} I_R + I_C = 0 \end{align*}$

$\begin{align*} \frac{v} {R} +C \frac {dV} {dt} = 0 \end{align*}$

Above equation is the first-order differential equation of an R-C circuit.

Transfer Function of the Parallel RC Circuit:

$\begin{align*} H(s) = \frac {V_o_u_t} {I_i_n} = \frac {R}{1+RCs} \end{align*}$

RC Circuit Equations

Capacitor C behaves as a $\frac {1} {sC}$ in the frequency domain with a voltage source of $\frac {vC(0^-)} {s}$ in series with it where $vC (0^-)$ is the initial voltage across the capacitor.

Impedance: The complex impedance, $Z_C$ of a capacitor C is

$\begin{align*} Z_C = \frac {1} {sC} \end{align*}$

$\begin{align*} \,\, Where, \,\, s = j{\omega} \end{align*}$

$\,\,1.\,\, j$ represents the imaginary part $j^2 = -1$

$\,\,2.\,\, \omega$ represents sinusoidal angular frequency (radians per second)

$\begin{align*} Z_C = \frac{1}{j\omega C} = \frac{j}{j2\omega C} = -\frac{j}{\omega C} \end{align*}$

Current: The current is same everywhere in series R-C circuit.

$\begin{align*} I(s) = \frac{V_i_n(s)}{R+\frac{1}{Cs}} = {\frac{Cs}{1+RCs}}V_i_n(s) \end{align*}$

Voltage: By applying the voltage divider rule, the voltage across the capacitor is:

$\begin{align*} \begin{split} V_C(s) = \frac {\frac{1}{Cs}}{{R+\frac{1}{Cs}}} V_i_n(s) \ & = \frac {\frac{1}{Cs}}{{\frac{1+RCs}{Cs}}} V_i_n(s) \ & = \frac{1}{1+RCs}V_i_n(s) \ \end{split} \end{align*}$

and the voltage across the resistor is:

$\begin{align*} \begin{split} V_R(s) = \frac{R}{R+\frac{1}{Cs}} V_i_n(s) \ & = \frac{R}{\frac{1+RCs}{Cs}} V_i_n(s) \ &= \frac{RCs}{1+RCs}V_i_n(s) \ \end{split} \end{align*}$

RC Circuit Current

The current is the same everywhere in the series R-C circuit.

$\begin{align*} I(s) = \frac{V_i_n(s)}{R+\frac{1}{Cs}} = {\frac{Cs}{1+RCs}}V_i_n(s) \end{align*}$

Transfer Function of RC Circuit

The transfer function from the input voltage to the voltage across the capacitor is

$\begin{align*} H_C(s) = \frac{V_C(s)}{V_i_n(s)} = \frac{1}{1+RCs} \end{align*}$

Similarly, the transfer function from the input voltage to the voltage across the resistor is

$\begin{align*} H_R(s) = \frac{V_R(s)}{V_i_n(s)} = \frac{RCs}{1+RCs} \end{align*}$

Step Response of RC Circuit

When something changes in a circuit, as a switch closes, the voltage and current also change and adjust to the new conditions. If the change is an abrupt step the response is called the step response.

The total response of a circuit is equal to the forced response plus natural response. These responses can be combined using the principle of superposition.

The forced response is one in which source of supply is turned on but with the initial conditions (internally stored energy) assumed to be zero.

The natural response is one in which source of supply is turned off but the circuit does including the initial conditions (initial voltage on capacitors and current in inductors). The natural response is also called the zero input response because the source of supply is turned off.

Therefore, total response = forced response + natural response

What is an Initial Condition?

In the case of an inductor, the current through it cannot be changed instantaneously. That means the current through inductor at instant $t=0^-$ will remain same just after transition at instant $t=0^+$ . i.e.,

$\begin{align*} i (0^-) = I_0 = 0 = i (0^+) \end{align*}$

In the case of a capacitor, the voltage across the capacitor cannot be changed instantaneously. That means the voltage across the capacitor at instant $t=0^-$ will remain the same just after transition at instant $t=0^+$ . i.e.,

$\begin{align*} V_C (0^-) = V_0 = V = V_C (0^+) \end{align*}$

Forced Response of Driven Series RC Circuit

Let us assume that the capacitor is initially fully discharged and switch (K) is kept open for a very long time and it is closed at $t=0$ .

Force Response Of Driven Series R C Circuit

At $t=0^-$ switch K is open

This is an initial condition hence we can write,

(1)

$\begin{equation*} V_C (0^-) = V_0 = V = V_C (0^+) \end{equation*}$

Because the voltage across the capacitor cannot change instantaneously.

For all $t\geq0$ switch K is closed.

Now the voltage source is introduced in the circuit. Hence applying KVL to the circuit, we get,

$\begin{align*} -R i(t) - V_c(t) + V_s =0 \end{align*}$

(2)

$\begin{equation*} R i(t) + V_c(t) = V_s \end{equation*}$

Now i(t) is the current through the capacitor and it can be expressed in terms of voltage across capacitor as

$\begin{align*} i (t) = i_c (t) = C \frac {dV_c(t)}{dt} \end{align*}$

Substitute this into equation (2), we get,

$\begin{align*} RC \frac {dV_c(t)}{dt} + V_c (t) = V_s \end{align*}$

$\begin{align*} RC \frac {dV_c(t)}{dt} = V_s - V_c (t) \end{align*}$

Separating variables, we get

$\begin{align*} \frac{dV_c(t)} {[V_s - V_c (t)]} = \frac {1} {RC} dt \end{align*}$

Integrating both the sides

$\begin{align*} \int \frac {dV_c(t)} {[V_s - V_c (t)]} = \int \frac {1} {RC} dt \end{align*}$

(3)

$\begin{equation*} -ln [V_s - V_c (t)] = \frac {t} {RC} + K^' \end{equation*}$

Where $K^'$ is the arbitrary constant

To find $K'$ : Using initial condition i.e. substituting equation (1) into equation (3), we get,

$\begin{align*} -ln [V_s - 0] = \frac {0} {RC} + K^' \end{align*}$

(4)

$\begin{equation*} {K^'} = -ln [V_s] \end{equation*}$

Substituting value of K’ in equation (3) we get,

$\begin{align*} -ln [V_s - V_c (t)] = \frac {t} {RC} - ln[V_s] \end{align*}$

$\begin{align*} -ln [V_s - V_c (t)] + ln[V_s] = \frac {t} {RC} \end{align*}$

$\begin{align*} ln [V_s - V_c (t)] - ln[V_s] = -\frac {t} {RC} ([ln[a] - ln[b] = ln \frac{a}{b}]) \end{align*}$

$\begin{align*} ln \frac {V_s - V_c (t)}{V_s} = -\frac {t} {RC} \end{align*}$

Taking antilog, we get,

$\begin{align*} \frac {V_s - V_c (t)}{V_s} = e^ {-\frac {t} {RC}} \end{align*}$

$\begin{align*} V_s - V_c (t) = V_s e^ {-\frac {t} {RC}} \end{align*}$

$\begin{align*} V_c (t) = V_s - V_s e^ {-\frac {t} {RC}} \end{align*}$

(5)

$\begin{equation*} V_c (t) = V_s (1 - e^ {-\frac {t} {RC}}) V \end{equation*}$

The above equation indicates the solution of a first-order differential equation of a series R-C circuit.

The above response is a combination of steady-state response i.e. $V_S$

and transient response i.e. $V_s * e^ {-\frac {t} {RC}}$

Natural Response of Source Free Series RC Circuit

The source free response is the discharge of a capacitor through a resistor in series with it.

For all $t>=0^+$ switch K is closed

Applying KVL to the above circuit, we get,

$\begin{align*} -R i(t) - V_c(t) = 0 \end{align*}$

(6)

$\begin{equation*} R i(t) = - V_c(t) \end{equation*}$

$\begin{align*} \,\, Now \,\, i(t) = i_c (t) = C \frac {dV_c(t)} {dt} \end{align*}$

Substitute this value of current into equation (6), we get,

$\begin{align*} R C \frac {dV_c(t)} {dt} = - V_c (t) \end{align*}$

Separating variables, we get

$\begin{align*} \frac {dV_c(t)} {V_c(t)} = - \frac {1} {R C} dt \end{align*}$

Integrating both the sides

$\begin{align*} \int \frac {dV_c(t)} {V_c(t)} = \int - \frac {1} {R C} dt \end{align*}$

(7)

$\begin{equation*} ln [{V_c(t)}] = - \frac {1} {R C} + K^' \end{equation*}$

Where $K^'$ is arbitrary constant

To find $K^'$ : Using initial condition i.e. substituting equation (1) into equation (7), we get,

$\begin{align*} ln [V_0] = - \frac {0} {RC} + K^' \end{align*}$

(8)

$\begin{equation*} {K^'} = ln [V_0] \end{equation*}$

Substituting the value of $K^'$ in equation (7) we get,

$\begin{align*} ln [V_c (t)] = - \frac {t} {RC} + ln[V_0] \end{align*}$

$\begin{align*} ln [V_c (t)] - ln[V_0] = -\frac {t} {RC} \end{align*}$

$\begin{align*} ln \frac {V_c (t)} {V_0} = -\frac {t} {RC} \end{align*}$

Taking antilog, we get,

$\begin{align*} \frac {V_c (t)} {V_0} = e^{-\frac {t} {RC}} \end{align*}$

(9)

$\begin{equation*} V_c (t)} = V_0 e^{-\frac {t} {RC}} \end{equation*}$

The above equation indicates the natural response of the series RC circuit.

Now, total response = forced response + natural response

$\begin{align*} V_c (t) = V_s (1 - e^{-\frac {t} {RC}})+ V_0 e^{-\frac {t} {RC}} \end{align*}$

$\begin{align*} V_c (t) = V_s - V_s e^{-\frac {t} {RC}}+ V_0 e^{-\frac {t} {RC}} \end{align*}$

$\begin{align*} V_c (t) = V_s + (V_0 - V_s) e^{-\frac {t} {RC}} \end{align*}$

Where, $V_S$ is the step voltage.

$V_0$ is the initial voltage on the capacitor.

Time Constant of RC Circuit

The time constant of an R-C circuit can be defined as the time during which the voltage across the capacitor would reach its final steady-state value.

One time constant is the time required for the voltage to rises 0.632 times steady-state value or time required for the current to decay 0.368 times the steady-state value.

The time constant of the R-C circuit is the product of resistance and capacitance.

$\begin{align*} \tau = R C \end{align*}$

Its unit is second.

RC Circuit Frequency Response

Using Impedance method: General equation for frequency response system is

$\begin{align*} H (\omega) = \frac {Y(\omega)} {X(\omega)} = \frac {V_o_u_t} {V_i_n} \end{align*}$

Now apply potential divider rule to the above circuit

(10)

$\begin{equation*} V_o_u_t = V_i_n \frac {Z_c} {Z_c + R} \end{equation*}$

Where, $Z_C$ = Impedance of capacito

$\begin{align*} Z_c = \frac {1} {j\omega C} \end{align*}$

Substitute this in equation (10), we get,

$\begin{align*} V_o_u_t = V_i_n \frac {\frac{1}{j\omega C}}{{\frac{1}{j\omega C} + R}} \end{align*}$

$\begin{align*} \frac {V_o_u_t} {V_i_n} =\frac {\frac{1}{j\omega C}}{\frac{1+j\omega RC}{j\omega C}} \end{align*}$

$\begin{align*} \frac {V_o_u_t} {V_i_n} = \frac {1} {1+j\omega R C} \end{align*}$

$\begin{align*} H (\omega) = \frac {V_o_u_t} {V_i_n} = \frac {1} {1+j\omega R C} \end{align*}$

The above response is a frequency response of an R-C circuit in complex form.

RC Circuit Differential Equation

RC Charging Circuit Differential Equation

Voltage across capacitor is given by

(11)

$\begin{equation*} V_c(t) = V - V e^{-\frac {t} {R C}} V \end{equation*}$

Now current through the capacitor is given by

$\begin{align*} i(t) = i_c(t) = C \frac {dV_c(t)}{dt} = C \frac {d}{dt} [V - V e^ {\frac{-t}{RC}}] \end{align*}$

$\begin{align*} i(t) = C [0 - V (\frac{-t}{RC})e^ {\frac{-t}{RC}}] \end{align*}$

$\begin{align*} i(t) = C [- V (\frac{-1}{R})e^ {\frac{-t}{RC}}] \end{align*}$

$\begin{align*} i(t) = \frac{V}{R}e^ {\frac{-t}{RC}} \end{align*}$

(12)

$\begin{equation*} i(t) = \frac{V}{R}e^ {\frac{-t}{\tau}} A \end{equation*}$

RC Discharging Circuit Differential Equation

The voltage across the capacitor is given by

(13)

$\begin{equation*} V_c(t) = V_0 e^{-\frac {t} {R C}} V \end{equation*}$

Now current through the capacitor is given by

$\begin{align*} i(t) = i_c(t) = C \frac {dV_c(t)}{dt} = C \frac {d}{dt} [V_0 e^ {\frac{-t}{RC}}] \end{align*}$

$\begin{align*} i(t) = C [V_0 (\frac{-t}{RC})e^ {\frac{-t}{RC}}] \end{align*}$

$\begin{align*} i(t) = C [V_0 (\frac{-1}{R})e^ {\frac{-t}{RC}}] \end{align*}$

$\begin{align*} i(t) = -\frac{V_0}{R}e^ {\frac{-t}{RC}} \end{align*}$

(14)

$\begin{equation*} i(t) = -\frac{V_0}{R}e^ {\frac{-t}{\tau}} A \end{equation*}$

RC Circuit Charging and Discharging

RC Circuit Charging

The figure shows the simple R-C circuit in which capacitor (C), in series with a resistor (R) that is connected to the DC voltage source via a mechanical switch (K). The capacitor is initially uncharged. When switch K is closed, the capacitor will gradually charge up through the resistor until the voltage across the capacitor becomes equal to the supply voltage source. The charge on the plates of the capacitor is given as Q = CV.

$\begin{align*} V_c(t) = V (1 - e^{-\frac {t} {R C}}) V \end{align*}$

From the above equation, it is clear that the capacitor voltage increases exponentially.

Where,

$V_C$ is the voltage across the capacitor
$V$ is the supply voltage.

RC is the time constant of the RC charging circuit. i.e. $\tau = R C$

Let us substitute different values of time t in equation (11) and (12),we get capacitor charging voltage, i.e.

$\begin{align*} t = \tau \,\, then \,\, V_c(t) = V - V * e^-^1 = (0.632) V \,\, (where, e = 2.718) \,\, \end{align*}$

$\begin{align*} t = 2\tau \,\, then \,\, V_c(t) = V - V * e^-^2 = (0.8646) V \end{align*}$

$\begin{align*} t = 4\tau \,\, then \,\, V_c(t) = V - V * e^-^4 = (0.9816) V \end{align*}$

$\begin{align*} t = 6\tau \,\, then \,\, V_c(t) = V - V * e^-^6 = (0.9975) V \end{align*}$

and capacitor charging current

$\begin{align*} t = \tau \,\, then \,\, i(t) = \frac {V}{R} * e^-^1 = \frac {V}{R}(0.368) A \,\, (where, e = 2.718) \,\, \end{align*}$

$\begin{align*} t = 2\tau \,\, then \,\, i(t) = \frac {V}{R} e^-^2 = \frac {V}{R}(0.1353) A \end{align*}$

$\begin{align*} t = 4\tau \,\, then \,\, i(t) = \frac {V}{R} e^-^4 = \frac {V}{R} (0.0183) A \end{align*}$

$\begin{align*} t = 6\tau \,\, then \,\, i(t) = \frac {V}{R} e^-^6 = \frac {V}{R}(0.0024) A \end{align*}$

The variation of voltage across the capacitor $V_C(t)$ and current through capacitor $i(t)$ as a function of time is shown in the figure.

Thus in R-C charging circuit if the voltage across the capacitor rises exponentially, the current through capacitor decays exponentially with the same rate. When the voltage across the capacitor reaches the steady-state value, the current decreases to zero value.

RC Circuit Discharging

If fully charged capacitor is now disconnected from the battery supply voltage, the stored energy in the capacitor during the charging process would stay indefinitely on its plates, keeping the voltage stored across its terminals at a constant value.

Now if the battery was replaced by a short circuit and when the switch is closed the capacitor will discharge through the resistor, now we have a circuit called RC discharging circuit.

$\begin{align*} V_c(t) = V_0 e^{\frac {-t}{RC}} V \end{align*}$

From the above equation, it is clear that the capacitor voltage decreases exponentially. That means in discharging the R-C circuit, the capacitor discharges through resistor R in series with it. Now the time constant of R-C charging circuit and R-C discharging circuit are same and is

$\begin{align*} \tau = R C \end{align*}$

Let us substitute different values of time t in equation (13) and (14),we get capacitor discharging voltage, i.e.

$\begin{align*} t = \tau \,\, then \,\, V_c(t) = V_0 * e^-^1 = V_0 (0.368) V \end{align*}$

$\begin{align*} t = 2\tau \,\, then \,\, V_c(t) = V_0 * e^-^2 = V_0 (0.1353) V \end{align*}$

$\begin{align*} t = 4\tau \,\, then \,\, V_c(t) = V_0 * e^-^4 = V_0 (0.0183) V \end{align*}$

$\begin{align*} t = 6\tau \,\, then \,\, V_c(t) = V_0 * e^-^6 = V_0 (0.0024) V \end{align*}$

The variation of voltage across the capacitor $V_C(t)$ as a function of time is shown in the figure.

Thus in the R-C Discharging circuit, similarly if the voltage across the capacitor decreases exponentially, the current through the capacitor rises exponentially with the same rate. When the voltage across the capacitor reaches zero value, the current reaches a steady-state value.

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Topics Circuit Theory RLC Circuit