Resistor-Capacitor (RC) Circuits

2025-10-26 – Dennis Evangelista

Describe the equivalent capacitance of multiple capacitors.

A collection of capacitors in a circuit may be analyzed as though it was a cingle capacitor with an equivalent capacitance \(C_{eq}\).

The inverse of the equivalent capacitance of a set of capacitors connected in series is equal to the sum of the inverses of the individual capacitances. \[\begin{equation} \dfrac{1}{C_{eq,s}} = \sum_i \dfrac{1}{C_i} \end{equation}\]

The equivalent capacitance of a set of capacitors in series is less than the capacitance of the smallest capacitor.

The equivalent capacitance of a set of capacitors in parallel is the sum of the individual capacitances. \[\begin{equation} C_{eq,p} = \sum_i C_i \end{equation}\]

As a result of conservation of charge, each of the capacitors in series must have the same magnitude of charge on each plate.

RC circuits

Describe the behavior of a circuit containing combinations of resistors and capacitors

The charge on a capacitor or the current in a resistor in an RC circuit can be described by a fundamental differential equation derived from Kirchoff’s voltage law. \[\begin{equation} \mathcal{E} = \dfrac{dq}{dt} R + \dfrac{q}{C} \end{equation}\]

The time constant (\(\tau\)) is a significant feature of an RC circuit.

The time constant of an RC circuit is a measure of how quickly the capacitor will charge or discharge and is defined as \[\begin{equation} \tau = R C \end{equation}\] For a charging capacitor, the time constant represents the time required for the capacitor’s charge to increase from zero to approximately 63% of its final asymptotic value. For a discharging capacitor, the time constant represents the time required for the capacitor’s charge to decrease from fully charged to approximately 37% of its initial value.

The potential difference across a capacitor and the current in the branch of the circuit containing the capacitor each change over time as the capacitor charges and discharges, but both will reach a steady state after a long time interval. Immediately after being placed in a circuit, an uncharged capacitor acts like a wire, and charge can easily flow to or from the plates of the capacitor. As a capacitor charges, changes to the potential difference across the capacitor affect the charge on the plates of the capacitor, the current in the circuit branch in which the capacitor is located, and the electric potential energy stored in the capacitor. The potential difference across a capactior, the current in the circuit branch in which the capacitor is located, and the electric potential energy stored in the capacitor all change with respect to time and asymptotically approach steady state conditions. After a long time, a charging capacitor approaches a state of being fully charged, reaching a maximum potential difference at which there is zero current in the circuit branch in which the capacitor is located.

Immediately after a charged capacitor begins discharging, the amount of charge on the capacitor and the energy stored in the capacitor begin to decrease. As a capacitor discharges, the amount of charge on the capacitor, the potential difference across the capacitor, and the current in the circuit branch in which the capacitor is located all decrease until a steady state is reached. After either charging or discharging for times much greater than the time constant, the capacitor and the relevant circuit branch may be modeled using steady-state conditions.

Solving an RC circuit

To expand our thinking about circuits, consider the following circuit:

(vin) at (2,2) ; (gndl) at (2,0) ; (vout) at (6,2) ; (gndr) at (6,0) ;

(0,2) to [V=v_i] (0,0); (0,2) to [short, i=i] (vin); (vin) to [R=R] (4,2); (4,2) to [C=C] (4,0); (4,0) to (0,0);

(4,2) to (vout); (4,0) to (gndr); (vout) to [open, v=v_o] (gndr);

RC circuit example

We wish to find the relationship between \(v_o\) and \(v_i\). First, we recognize that the resistor and capacitor are in series, thus the current in each is \(i\). As with any of the purely resistive circuits we have solved, we can apply Kirchoff’s Voltage Law, in addition to the device constitutive relations (Ohm’s Law for the resistor and \(i=C dv/dt\) for the capacitor).

Applying KVL gives us \(v_o\) in terms of \(v_i\): \[\begin{equation} v_o = v_i - i R \end{equation}\] The current is found using the capacitor: \[\begin{equation} i = C \frac{dv_o}{dt} \end{equation}\] Substituting the current \(i\) gives: \[\begin{equation} v_o = v_i - R C \frac{dv_o}{dt} \end{equation}\] Rearranging gives the following: \[\begin{equation} \frac{dv_o}{dt} + \frac{1}{RC} v_o = \frac{1}{RC} v_i \label{eq:RC} \end{equation}\] Equation \[eq:RC\] is a linear, constant coefficient, ordinary differential equation, which means it is reasonable to solve it. The actual solution of equation \[eq:RC\] will require knowing the input \(v_i\) and will require you to recall your differential equations class. Here I’ve typed out the solution; doing the math will take away the fear of it!

You will see this again before ES202 is over. First, we’ll look at two ways to solve for a solution to a step input (as you might see if you walked into Maury 115, built the circuit on a protoboard, and turned it on while watching \(v_o\) on an oscilloscope). Then we’ll look at the solution to a sinusoidal input.

Solution for step input from differential equations class

Consider \(v_i\) given by a step input \(u(t)\): \[\begin{equation} v_i(t) = V_i u(t) = \begin{cases} 0 & t \leq 0 \\ V_i & t > 0 \end{cases} \end{equation}\]

The initial condition for the circuit before it was ever energized is \(v_o=0\). In your differential equations class, you learned to break the solution into two pieces, a homogeneous solution that satisfies the initial condition and a particular solution that satisfies the input \(v_i\).

Let’s do the particular solution (or zero state response, ZSR) first, just by looking at the differential equation in equation \[eq:RC\]: \[\begin{equation} v_{o,p} = V_i \end{equation}\] Substituting this in for \(v_o\) will confirm this satisfies equation \[eq:RC\].

We now need to add a homogeneous solution (zero input response, ZIR) that ensures the total solution matches the initial condition. The ZIR satisfies the following: \[\begin{equation} RC\frac{dv_{o,h}}{dt} + v_{o,h} = 0 \end{equation}\]

You may recognize that it is separable: \[\begin{align} \frac{dv_{o,h}}{dt} &= - \frac{1}{RC} v_{o,h} \\ \frac{1}{v_{o,h}} dv_{o,h} &= - \frac{1}{RC} dt \\ \ln{v_{o,h}} + C &= - \frac{t}{RC} \\ v_{o,h} &= A e^{-t/RC} \end{align}\]

Or you may just remember the form of the solution is an exponential and write the solution by inspection: \[\begin{equation} v_{o,h} = A e^{-t/RC} \end{equation}\]

The total solution is then the ZSR plus the ZIR (particular solution plus homogeneous solution): \[\begin{equation} v_{o} = V_i + A e^{-t/RC} \end{equation}\] Apply the initial condition. Since \(v_o(t=0) = 0\), \(A = -V_i\), and the solution is \[\begin{equation} v_{o} = V_i - V_i e^{-t/RC} \end{equation}\] Substituting this in equation \[eq:RC\] shows it is correct for \(v_i = V_i u(t)\).

Solution for step input by inspection

This sort of circuit shows up enough in practical applications that it is useful to build some intuition for it by thinking about what the capacitor looks like just at \(t=0^{+}\) versus as \(t\rightarrow\infty\). In the instant just after the switch is turned on (\(t=0^{+}\)), the capacitor looks like a short, thus \(v_o(0^{+}) = 0\). After a long time, the capacitor becomes fully charged and looks like an open, thus \(v_o(\infty)\rightarrow v_i\). We also remember that the solution involves an exponential decay with time constant \(\tau=RC\). Putting these together, we can write a solution by inspection: \[\begin{equation} v_o(t) = V_i (1 - e^{-t/RC}) \end{equation}\]

Solution for sinusoidal input

Let’s briefly consider the particular solution for a sinusoidal input: \[\begin{equation} v_i = V_i \cos{\omega t} \end{equation}\] for the differential equation: \[\begin{equation*} \frac{dv_o}{dt} + \frac{1}{RC} v_o = \frac{1}{RC} v_i \end{equation*}\]

You may remember that the candidate solutions for this case are of the form: \[\begin{equation} v_o = A \cos{\omega t + \phi} \end{equation}\] You may need to remind yourself that there are some trigonometric identities you learned in high school that may help if you need to expand \(v_o\) or its derivative \(dv_o/dt\): \[\begin{equation} v_o = A \cos{\phi}\cos{\omega t} - A \sin{\phi}\sin{\omega t} \end{equation}\] \[\begin{align} \frac{dv_o}{dt} &= -A\omega\sin{\omega t + \phi} \\ &= -A\omega \sin{\omega t}\cos{\phi} - A \omega \cos{\omega t}\sin{\phi} \\ \end{align}\] Substitution gives: \[\begin{equation} - A\omega\cos{\phi}\sin{\omega t} - A\omega\sin{\phi}\cos{\omega t} + \frac{1}{RC} [A \cos{\phi}\cos{\omega t} - A \sin{\phi}\sin{\omega t}] = \frac{V_i}{RC}\cos{\omega t} \end{equation}\]

To solve this, collect the coefficients of \(\cos{\omega t}\) and \(\sin{\omega t}\) to get two equations with two unknowns \(A\) and \(\phi\): \[\begin{equation} -A \omega \cos{\phi} - \frac{A}{RC}\sin{\phi} = 0 \label{eq:prephase} \end{equation}\] \[\begin{equation} -A \omega \sin{\phi} + \frac{A}{RC}\cos{\phi} = \frac{V_i}{RC} \label{eq:premag} \end{equation}\]

From equation \[eq:prephase\] we get: \[\begin{equation} \phi = \arctan{-RC\omega} \end{equation}\] Solving for \(A\) gives: \[\begin{equation} A = \frac{V_i}{\cos{\phi} - \omega R C \sin{\phi}} \end{equation}\] The steady state response is then: \[\begin{equation} v_o = \frac{V_i}{\cos{\phi} - \omega R C \sin{\phi}} \cos{\omega t + \phi} \end{equation}\]

There are actually easier ways to solve for this result using complex numbers or using Laplace Transforms; you will encounter them in future courses.

Circuits with multiple sources

You may encounter circuits with multiple independent voltage or current sources. For circuits which include only linear things (i.e. for “linear systems”), there is a technique for dealing with multiple sources called superposition. For each independent source:

Short out all other voltage sources
Open all other current sources
Solve using the single isolated source

The total response is the sum (superposition) of the results for each isolated independent source.

Here’s a trivial case (circuit a) in which we might wish to find \(i\):

(0,2) to [V=v₁] (0,0); (0,2) to [R=R, i=i] (2,2); (0,0) to (2,0); (2,2) to [V=v₂] (2,0);

= b

(0,2) to [V=v₁] (0,0); (0,2) to [R=R, i=i₁] (2,2); (0,0) to (2,0); (2,0) to [short] (2,2);

c

(0,0) to [short] (0,2); (0,2) to [R=R, i=i₂] (2,2); (0,0) to (2,0); (2,2) to [V=v₂] (2,0);

Trivial example for superposition. (a) is a circuit with multiple sources. In (b) and (c) we solve for each independent source with the others shorted or opened as necessary.

In circuit b, shorting out \(v_2\) gives \(i_1=v_1/R\). Shorting out \(v_1\) in circuit c gives \(i_2=-v_2/R\). By superposition, \(i=v_1/R - v_2/R\), which is what we would expect by applying KVL and solving directly. For this trivial example, it seems more complicated, but superposition can be useful in circuits where there are many more sources and where the answer may not be as readily apparent.

A few additional circuit problems

whose utility will become apparent after Spring Break. In each circuit, let us solve for the output voltage \(v_o\) such that the current \(i\) is zero.

Circuit 1

Consider the circuit shown in figure 1.

(gnd) at (0,0) ; (v-) at (3,2) ; (gnd1) at (3,1) ; (gnd2) at (6,1) ;

(0,2) to [V=v_i] (gnd); (0,2) to [R=R₁] (v-); (v-) to [short,i=i] (gnd1); (v-) to [short] (3,3); (3,3) to [R=R₂] (6,3); (6,3) to [V=v₀] (gnd2);

Figure 1: Circuit 1. To find v₀ such than i = 0, write KCL at node a. This circuit will be seen again when we consider operational amplifiers (op amps) in an inverting amplifier configuration.

We wish to solve for output voltage \(v_o\) such that \(i=0\), for reasons that will become apparent during week 10. To do so, we write KCL at node a: \[\begin{equation} \frac{v_i - 0}{R_1} + \frac{v_o - 0}{R_2} = i = 0 \end{equation}\] Rearranging gives: \[\begin{equation} v_o = - \frac{R_2}{R_1} v_i \end{equation}\]

Note the negative sign so that the sign of \(v_o\) is the opposite of \(v_i\), i.e. it is inverting. In this example, using superposition would also work but doesn’t make things much simpler. The sign of \(v_o\) may also cause some confusion. If you like to think of the current in \(R_2\) as moving to the right (out of a), you would write it as \((0-v_o)/R\). If you prefer to think of it moving to the left (into a), you would write it as \(v_o/R\). Both ways will give the correct answer.

Circuit 2

As another example, consider the circuit shown in figure 2.

(gnd) at (0,0) ; (v-) at (3,2) ; (gnd1) at (3,0) ; (gnd2) at (6,1) ;

(0,2) to [V=v_i] (gnd); (0,2) to [short,i=i] (v-); (v-) to [R=R₁] (gnd1); (v-) to [short] (3,3); (3,3) to [R=R₂] (6,3); (6,3) to [V=v₀] (gnd2);

Figure 2: Circuit 2. To find v₀ such than i = 0, we again write KCL at node a. This circuit will be seen again when we consider operational amplifiers (op amps) in a non-inverting amplifier configuration.

Again we wish to solve for output voltage \(v_o\) such that \(i=0\). In this case, KCL at node a is:: \[\begin{equation} \frac{v_i - 0}{R_1} = \frac{v_o - v_i}{R_2} \end{equation}\] Rearranging gives: \[\begin{equation} v_o = \left( 1 + \frac{R_2}{R_1} \right) v_i \end{equation}\] The result looks similar to circuit 1, however, the sign of \(v_o\) is the same as the sign of \(v_i\), i.e. it is non-inverting.

There are many ways to solve circuits. Alternatively, you might derive the same result by looking at the series combination of \(R_1\) and \(R_2\) as a voltage divider: \[\begin{equation} v_a = v_i = \frac{R_1}{R_1+R_2} v_o \end{equation}\] \[\begin{equation} R_1 v_o = (R_1 + R_2) v_i \end{equation}\] \[\begin{equation} v_o = \left( 1 + \frac{R_2}{R_1} \right) v_i \end{equation}\]

Circuit 3

Now let us consider a modification to circuit 1 shown in figure 3. In this example, we have added an extra input voltage but the rest of the circuit remains unchanged from circuit 1.

(gnd) at (0,0) ; (gnd3) at (-2,0) ; (v-) at (3,2) ; (gnd1) at (3,1) ; (gnd2) at (6,1) ;

(0,2) to [V=v_i] (gnd); (0,2) to [R=R₁] (2,2); (2,2) to [short] (v-);

(-2,3) to [V=v₂] (gnd3); (-2,3) to [short] (0,3); (0,3) to [R=R₁] (2,3); (2,3) to [short] (2,2);

(v-) to [short,i=i] (gnd1); (v-) to [short] (3,3); (3,3) to [R=R₂] (6,3); (6,3) to [V=v₀] (gnd2);

Figure 3: Circuit 3 is a modification of circuit 1 with an additional input voltage v₂. This circuit will be seen again when we examine the inverting summing amplifier configuration.

Again, we wish to solve for output voltage \(v_o\) such that \(i=0\). Again, we write KCL at node a: \[\begin{equation} \frac{v_2-0}{R_1} + \frac{v_1 - 0}{R_1} + \frac{v_o - 0}{R_2} = i = 0 \end{equation}\] Rearranging gives: \[\begin{equation} v_o = - \frac{R_2}{R_1} v_1 - \frac{R_2}{R_1} v_2 \end{equation}\] \[\begin{equation} v_o = - \frac{R_2}{R_1} ( v_1 + v_2 ) \end{equation}\] The result here has an inverting gain (\(-R_2/R_1\)) and also sums the inputs \(v_1+v_2\).

Circuit 4

Finally, consider the circuit shown in figure 4.

(gnd) at (0,0) ; (v-) at (3,2) ; (gnd1) at (3,1) ; (gnd2) at (6,1) ;

(0,2) to [V=v_i] (gnd); (0,2) to [R=R] (v-); (v-) to [short,i=i] (gnd1); (v-) to [short] (3,3); (3,3) to [C=C] (6,3); (6,3) to [V=v₀] (gnd2);

Figure 4: Circuit 4. To find v₀ such than i = 0, write KCL at node a. This circuit will be seen again when we consider integrators.

We wish to solve for output voltage \(v_o\) such that \(i=0\), for reasons that will become apparent during week 10. To do so, we write KCL at node a: \[\begin{equation} \frac{v_i}{R_1} + C \frac{d v_o}{dt} = 0 \end{equation}\] Rearranging gives: \[\begin{equation} \frac{d v_o}{dt} = - \frac{1}{RC} v_i \end{equation}\] \[\begin{equation} v_o = - \frac{1}{RC} \int_{-\infty}^t v_i(\tau) d\tau \end{equation}\] Here the output is the integral of the input.