RC Circuit Analysis: Series & Parallel (Explained in Plain English)

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Key learnings:

RC Circuit Definition: An RC circuit is an electrical configuration consisting of a resistor and a capacitor used to filter signals or store energy.
Parallel RC Circuit Dynamics: In a parallel RC circuit, the voltage is uniform across all components, while the total current is the sum of individual currents through the resistor and capacitor.
Impedance and Phase Calculation: The impedance in an RC circuit helps determine how the voltage and current are phased, impacting the signal’s overall behavior.
Transfer Function Importance: The transfer function in RC circuits, whether series or parallel, provides a crucial link between input and output, influencing how signals are processed.
Time Constant Significance: The time constant of an RC circuit is fundamental in understanding how quickly the circuit responds to changes in voltage or current, essential for timing applications.

What is an RC Circuit?

An RC circuit, or resistor-capacitor network, is an electrical circuit containing the passive circuit components of a resistor (R) and capacitor (C). Depending on its connection and output point, the network can shape signals, set timing or store and release energy while driven by a voltage source or current source.

When current flows, the resistor dissipates energy as heat, while an ideal capacitor stores energy in its electric field and can return it to the circuit. The same distinction applies to the resistive and reactive parts of an RL circuit or RLC circuit.

An ideal LC circuit has no resistance and therefore no resistive loss. Real inductors, capacitors and connecting conductors have non-zero resistance and other losses, so practical circuits dissipate some energy.

Series RC Circuit

In a series RC circuit, a resistor with resistance R and a capacitor with capacitance C carry the same current. Their voltage drops have different phases under sinusoidal steady-state conditions.

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.

R-C Circuit Vector Diagram — **VECTOR DIAGRAM**

Because the current $'I'$ is common to both series components, the phasor diagram uses it as the reference.

$V_R = IR$ is in phase with current $'I'$ because an ideal resistor has voltage and current in phase.

$V_C=I X_C$ lags current $'I'$ by $90^0$ for an ideal capacitor. Their voltage and current are $90^0$ apart: voltage lags current by $90^0$ , or current leads 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 magnitude of 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 figure shows the voltage and impedance triangles. The complex impedance is R minus jXC; the displayed triangle gives its magnitude.

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 figure shows the sinusoidal voltage and current waveforms of the R-C series circuit.

R C Circuit Waveform — **VOLTAGE AND CURRENT WAVEFORM**

Power in an RC Series Circuit

Instantaneous power is the product of instantaneous voltage and current. In the derivation below, the uppercase symbols in the first line represent time-varying values before peak-value expressions are substituted.

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

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

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

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

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

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

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

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

Thus the instantaneous power consists of two parts.

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

2. A varying component with the minus sign shown in the derivation = $\frac{V_m I_m}{2} cos({2\omega}t+{\phi})$ , 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 resistor and a capacitor are connected across the same two nodes. The resistor has resistance $R$ , and the capacitor has capacitance $C$ .

Each branch of a parallel RC circuit has the same voltage. With branch-current directions defined away from the source node, the source current is the phasor or instantaneous sum of the resistor and capacitor currents.

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

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

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:

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

For the source-free natural response, applying KCL (Kirchhoff’s Current Law) with the displayed current directions gives

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

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

The resulting first-order differential equation describes the natural voltage decay of this source-free parallel R-C circuit. A driven circuit would include its input current in the KCL equation.

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

With zero initial conditions, the capacitor has impedance $\frac {1} {sC}$ in the Laplace domain. A non-zero initial capacitor voltage can be represented with an additional source $\frac {vC(0^-)} {s}$ , where $vC (0^-)$ is the voltage immediately before switching. The source polarity depends on the chosen reference directions.

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

A step response describes the circuit’s change from its initial condition after an input changes abruptly, such as when a switch connects a DC source.

For a linear circuit, the complete response is the sum of its zero-state response and zero-input response. The terms forced response and natural response are also used, though the forced response may refer to either the complete zero-state transient or only its steady-state part depending on convention.

The zero-state response is calculated with the input applied and all initial stored energy set to zero.

The natural, or zero-input, response is calculated with independent sources set to zero while retaining the initial capacitor voltage or inductor current.

Therefore, total response = forced response + natural response

What is an Initial Condition?

For an ideal inductor, finite voltage cannot change current instantaneously, so the current at $t=0^-$ equals the current at $t=0^+$ . The displayed equation is the special case in which that continuous current is zero:

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

For an ideal capacitor, finite current cannot change its voltage instantaneously, so the voltage at $t=0^-$ equals the voltage at $t=0^+$ . The extra equality to V in the displayed expression applies only when the initial capacitor voltage equals that named voltage:

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

Forced Response of Driven Series RC Circuit

Assume the capacitor is initially discharged. Switch K remains open for a long time and closes at $t=0$ to connect the source.

Force Response Of Driven Series R C Circuit

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

For the stated discharged condition, voltage continuity requires the capacitor voltage immediately before and after switching to equal zero. The protected equation below incorrectly equates the initial value with the source voltage and should not be used for this example.

(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 final equation is the capacitor-voltage solution for a zero-volt initial condition and a constant DC step Vs. The trailing V denotes volts rather than another multiplier.

The response contains the final steady-state response $V_S$ and a decaying transient. In the preceding expression, the transient term is subtracted from the steady-state value:

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 sides gives the next line. That protected line is missing time t in its first term; the correct term is negative t divided by RC.

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

(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,

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

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

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

Exponentiating both sides gives the following result. The final protected line contains one extra closing brace after the capacitor-voltage function, but its intended exponential decay is clear.

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

(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 describes the rate of exponential change; it is not the time needed to reach the final value exactly. An ideal first-order response approaches its final value asymptotically.

After one time constant, a rising response has completed about 63.2% of the change from its initial value to its final value. A decaying response has about 36.8% of its initial difference remaining.

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

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

Its SI unit is the second.

RC Circuit Frequency Response

Using Impedance method: The frequency response is the output-to-input ratio. In the circuit shown, the output is taken across the capacitor, so it is a first-order low-pass response:

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

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 capacitor

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

Substitute this in equation (10), we get,

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

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

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

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

This is the complex frequency response of the capacitor-output RC low-pass circuit. Taking the output across the resistor instead gives the high-pass response shown earlier.

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

R C Charging Circuit — **R-C CHARGING CIRCUIT**

The figure shows a capacitor (C) and resistor (R) in series with a DC source and switch K. The capacitor is initially uncharged. When K closes, capacitor voltage rises towards the source voltage while current falls towards zero. For an ideal linear capacitor, charge and voltage are related by 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.

Variation Of Voltage Vs Time — **Variation of Voltage Vs Time**

Variation Of Current Vs Time — **Variation of Current Vs Time**

In this ideal R-C charging circuit, capacitor voltage rises exponentially while current magnitude decays with the same time constant. Both approach their final values asymptotically: capacitor voltage approaches the source voltage, and current approaches zero.

RC Circuit Discharging

If a fully charged ideal capacitor is isolated, its charge and terminal voltage remain constant. A real capacitor gradually loses charge through leakage, dielectric loss and any connected measurement path.

If the source is replaced by a short circuit and the switch closes, the capacitor discharges through the resistor. This arrangement is an RC discharging circuit.

R C Discharging Circuit — **R-C DISCHARGING CIRCUIT**

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

The capacitor voltage decreases exponentially as it discharges through resistor R. For this simple circuit, charging and discharging use the same resistance and capacitance, so both have the same time constant:

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

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.

In the R-C discharging circuit, capacitor voltage and discharge-current magnitude both decay exponentially towards zero with the same time constant. The negative sign in the current equation indicates that current flows opposite to the selected charging-current reference direction.