RLC Circuits and Electromechanical Analogies, differential equations lesson

Charge a capacitor, then let it discharge through a coil and a resistor. Depending on the components, the charge can drain away steadily, or it can swing back and forth so that the two plates trade the sign of their charge again and again while the swings die out. That is the behavior of a mass on a damped spring, and it is no coincidence. This lesson derives the equation of the series circuit from the voltage drops around the loop, matches its coefficients with the spring model, and checks a free discharge through its charge, its current and its energy.

Key terms

Charge and current

Definition Throughout, $t$ is time in seconds and a prime means a derivative with respect to $t$ . Choose a positive direction around the loop, and call the capacitor plate that a current in this direction flows onto the reference plate. Let $q(t)$ be the charge on the reference plate in coulombs, so the other plate holds $-q(t)$ . The current $i(t)$ , in amperes, is the rate at which charge arrives at the reference plate,

i(t) = q'(t) = \frac{dq}{dt}.

Conversely, the charge is recovered from the current by

q(t) = q(t_0) + \int_{t_0}^{t} i(s)\,ds.

A negative current means that charge flows against the chosen direction. Reversing the chosen direction changes the signs of $q$ , $i$ and the source voltage together, and the circuit equation keeps the same form.

In plain words Charge says how much is stored on the capacitor, and current says how fast that amount changes. The sign of the current gives the direction of flow. It does not say whether the reference plate is positively charged.

Example Suppose the charge is $q(t) = 0.004\sin 50t$ coulombs. Then the current is $i(t) = 0.2\cos 50t$ amperes. At $t = 0.05$ seconds the charge is about $0.0024$ coulombs, still positive, while the current is about $-0.160$ amperes. The reference plate holds positive charge but is losing it.

Resistance, inductance and capacitance

Definition Measure the voltage drop across each component in the chosen direction of the current. The three ideal components obey

\begin{gathered} v_R = R\,i, \qquad v_L = L\,i', \\[0.4em] v_C = \frac{q}{C}, \end{gathered}

where $q$ is the charge on the capacitor plate that a positive current flows onto. The resistance $R$ is measured in ohms (volts per ampere), the inductance $L$ in henries (volt seconds per ampere) and the capacitance $C$ in farads (coulombs per volt). In this lesson they are constants with $R \ge 0$ , $L > 0$ and $C > 0$ . Real components depart from these laws, for example through leakage in a capacitor, saturation of an inductor’s core or the resistance of its wire, and then need a richer model.

In plain words The resistor’s drop is proportional to the current, the inductor’s drop to the rate of change of the current, and the capacitor’s drop to the charge it holds. A large inductance makes the current hard to change, and a large capacitance holds a lot of charge at a low voltage.

Example An inductor of $0.5$ henries whose current rises at $4$ amperes per second has a drop of $0.5 \cdot 4 = 2$ volts. The units confirm that each law produces volts. For the inductor,

\text{H} \cdot \frac{\text{A}}{\text{s}} = \frac{\text{V}\,\text{s}}{\text{A}} \cdot \frac{\text{A}}{\text{s}} = \text{V}.

For the resistor an ohm times an ampere is a volt, and for the capacitor coulombs divided by coulombs per volt leave volts.

Kirchhoff’s voltage law

Definition For a circuit of ideal components joined by ideal wires, Kirchhoff’s voltage law says that the signed voltage changes around any closed loop add to zero. In a single loop with a source of voltage $V(t)$ , counted positive when it pushes current in the chosen direction, this means that the drops across the components add up to the source voltage. For a loop with a resistor, an inductor and a capacitor,

v_L + v_R + v_C = V(t).

In plain words A trip once around the loop ends at the potential where it started, so whatever voltage the source supplies is used up across the components.

Example A loop has $R = 10$ ohms, $L = 0.5$ henries and $C = 0.002$ farads. At one instant the current is $0.4$ amperes and falling at $6$ amperes per second, and the reference plate holds $0.01$ coulombs. The drops are

\begin{aligned} v_R &= 10 \cdot 0.4 = 4, \\[0.4em] v_L &= 0.5 \cdot (-6) = -3, \\[0.4em] v_C &= \frac{0.01}{0.002} = 5 \end{aligned}

volts, so the source voltage at that instant is $4 - 3 + 5 = 6$ volts. The inductor’s drop is negative because its current is falling. At that instant the inductor gives up stored energy and helps push the current along.

Series RLC circuit

Definition A series RLC circuit is one loop containing a resistor, an inductor and a capacitor with constants $R \ge 0$ , $L > 0$ , $C > 0$ , driven by a source voltage $V(t)$ . Substituting the component laws and $i = q'$ into Kirchhoff’s voltage law gives the charge equation

L q'' + R q' + \frac{1}{C}\,q = V(t).

If $V$ is continuous on an interval $I$ containing $t_0$ , then for any initial charge $q(t_0) = q_0$ and initial current $i(t_0) = i_0$ the equation has exactly one solution on all of $I$ . Its characteristic equation is $Lr^2 + Rr + \frac{1}{C} = 0$ , and the natural angular frequency is

\omega_0 = \frac{1}{\sqrt{LC}}.

With $R > 0$ the free circuit, where $V = 0$ , is underdamped when $R^2 < 4L/C$ , critically damped when $R^2 = 4L/C$ and overdamped when $R^2 > 4L/C$ . In the underdamped case every free solution other than $q = 0$ oscillates with the damped angular frequency

\omega_d = \sqrt{\frac{1}{LC} - \frac{R^2}{4L^2}}

inside an envelope that decays like $e^{-Rt/(2L)}$ . With $R = 0$ every free solution other than $q = 0$ oscillates at $\omega_0$ with constant amplitude.

In plain words The circuit equation is a damped oscillator for charge. Once the source voltage is given, two initial facts, how much charge is on the capacitor and how much current flows, decide everything the circuit does afterward.

Example Take $L = 1$ henry, $R = 2$ ohms, $C = \frac{1}{5}$ farad and no source, with $q(0) = 1$ coulomb and $i(0) = 0$ . Since $\frac{1}{C} = 5$ , the equation is $q'' + 2q' + 5q = 0$ . Here $R^2 = 4$ is less than $4L/C = 20$ , so the circuit is underdamped. Since the letter $i$ already names the current, write $j$ for the imaginary unit. The roots of $r^2 + 2r + 5 = (r + 1)^2 + 4 = 0$ are $-1 \pm 2j$ , so $q = e^{-t}(A\cos 2t + B\sin 2t)$ . The charge condition gives $A = 1$ . By the product rule $i(0) = -A + 2B$ , and setting this to zero gives $B = \frac{1}{2}$ . Setting $B = 0$ would be a mistake, because the decaying exponential also contributes to the initial current. The solution is

\begin{aligned} q(t) &= e^{-t}\left(\cos 2t + \tfrac{1}{2}\sin 2t\right), \\[0.4em] i(t) &= -\tfrac{5}{2}\,e^{-t}\sin 2t. \end{aligned}

Differentiating again gives $q'' = e^{-t}\left(\frac{5}{2}\sin 2t - 5\cos 2t\right)$ , and in $q'' + 2q' + 5q$ the sine terms and the cosine terms both cancel. For $0 < t < \frac{\pi}{2}$ the current is negative, so charge leaves the reference plate. The charge first reaches zero at $t = \frac{1}{2}(\pi - \arctan 2) \approx 1.017$ seconds and is $-e^{-\pi/2} \approx -0.208$ coulombs at $t = \frac{\pi}{2}$ , where the current is zero and changes sign. Here $\omega_0 = \sqrt{5} \approx 2.236$ and $\omega_d = 2$ radians per second.

Electromechanical analogy

Definition The mass-spring-damper equation

m x'' + c x' + k x = F(t),

for a displacement $x(t)$ with mass $m > 0$ , damping $c \ge 0$ , stiffness $k > 0$ and applied force $F(t)$ , has the same form as the series RLC equation. Matching the two equations term by term gives an electromechanical analogy, called the force-voltage analogy because the applied force corresponds to the source voltage.

Mass-spring-damper	Series RLC circuit
displacement $x$	charge $q$
velocity $x'$	current $i = q'$
mass $m$	inductance $L$
damping $c$	resistance $R$
stiffness $k$	reciprocal capacitance $1/C$
applied force $F(t)$	source voltage $V(t)$
kinetic energy $\tfrac{1}{2}m(x')^2$	inductor energy $\tfrac{1}{2}Li^2$
spring energy $\tfrac{1}{2}kx^2$	capacitor energy $q^2/(2C)$

Under this matching the natural angular frequency $\sqrt{k/m}$ becomes $1/\sqrt{LC}$ and the discriminant $c^2 - 4mk$ becomes $R^2 - 4L/C$ , so every result about the spring equation becomes a result about the circuit after renaming the symbols. The analogy matches equations, not physical quantities. Each symbol keeps its own units, and a henry is not a kilogram.

In plain words Inductance acts as electrical inertia, resistance as electrical friction, and the capacitor as a spring with stiffness $1/C$ . A large capacitance is a soft spring, because a small voltage already holds a given charge.

Example The circuit with $L = 1$ henry, $R = 2$ ohms and $C = \frac{1}{5}$ farad matches a $1$ kilogram mass with damping $2$ newton seconds per meter and stiffness $5$ newtons per meter, the system of Free and Damped Mechanical Oscillations. Released at rest from $x(0) = 2$ meters, that mass moves as $x(t) = e^{-t}(2\cos 2t + \sin 2t)$ . Both problems have the same equation, and the mechanical initial data are twice the electrical ones, so by uniqueness the displacement in meters is exactly twice the charge in coulombs at every time. The mechanical energy at release is $10$ joules, four times the circuit’s $\frac{5}{2}$ joules, because energy is quadratic in the state.

Energy and passive components

Definition The energy stored in a series RLC circuit at time $t$ , in joules, is the magnetic energy of the inductor plus the electric energy of the capacitor,

E(t) = \tfrac{1}{2}L\,i^2 + \frac{q^2}{2C}.

Along every solution of the charge equation,

\begin{aligned} E'(t) &= i\left(L\,i' + \frac{q}{C}\right) \\[0.4em] &= V(t)\,i - R\,i^2. \end{aligned}

The term $V(t)\,i$ is the power supplied by the source, and $R\,i^2$ is the power the resistor turns into heat. A component is passive when it can never deliver more energy to the rest of the circuit than it has received. With $R \ge 0$ , $L > 0$ and $C > 0$ the three ideal components are passive. The resistor only absorbs energy, and the inductor and the capacitor only store energy and give it back. With the source disconnected, $E' = -R\,i^2 \le 0$ , so the stored energy never increases. If moreover $R > 0$ and the solution is not identically zero, the current vanishes only at isolated instants and the energy is strictly decreasing.

In plain words Energy can move back and forth between the capacitor and the inductor, while the resistor drains it away as heat. Without a source nothing puts energy back, which is why the swings shrink.

Example In the free discharge worked out above, with $L = 1$ henry, $R = 2$ ohms and $C = \frac{1}{5}$ farad, the initial current is zero, so all the initial energy sits in the capacitor,

E(0) = \frac{1^2}{2 \cdot \frac{1}{5}} = \frac{5}{2} \text{ joules}.

At $t = \frac{\pi}{4}$ the capacitor holds about $0.130$ joules and the inductor about $0.650$ joules, a total of $\frac{15}{4}e^{-\pi/2} \approx 0.780$ joules, so energy has moved into the inductor while some has been lost. The rate of loss is

E'(t) = -2\,i^2 = -\tfrac{25}{2}\,e^{-2t}\sin^2 2t,

which is zero only at the multiples of $\frac{\pi}{2}$ . Since $E(t) \to 0$ , integrating $2\,i^2$ from $0$ to $\infty$ gives exactly $\frac{5}{2}$ joules, and all of the stored energy ends up as heat in the resistor.

Check yourself

A series RLC circuit has $L = 2$ henries and $C = 0.02$ farads. Find its natural angular frequency and the resistance that makes it critically damped. Is it underdamped with $R = 10$ ohms, and if so, what is its damped angular frequency?

Show the answer

The natural angular frequency is $\omega_0 = 1/\sqrt{2 \cdot 0.02} = 1/\sqrt{0.04} = 5$ radians per second. Critical damping needs $R^2 = 4L/C = 400$ , so $R = 20$ ohms. With $R = 10$ ohms, $R^2 = 100 < 400$ , and the circuit is underdamped. Its decay rate is $R/(2L) = 2.5$ per second, and

\begin{aligned} \omega_d &= \sqrt{25 - 2.5^2} \\[0.4em] &= \frac{5\sqrt{3}}{2} \approx 4.33 \end{aligned}

radians per second.

A classmate translates the circuit with $L = 1$ henry, $R = 2$ ohms and $C = \frac{1}{5}$ farad into a spring by setting the stiffness equal to $C$ . From $x'' + 2x' + 0.2x = 0$ they conclude that the charge never oscillates. Find the mistake and the correct conclusion.

Show the answer

The capacitor contributes the term $q/C$ , so the stiffness corresponds to $1/C = 5$ , not to $C$ . The units agree with this, since a stiffness in newtons per meter matches volts per coulomb, the units of $1/C$ . The classmate’s equation has the roots $-1 \pm \sqrt{0.8}$ , about $-0.106$ and $-1.894$ , which are real and negative and would mean overdamping. The correct equation $q'' + 2q' + 5q = 0$ has the complex roots $-1 \pm 2j$ , where $j$ is the imaginary unit, and indeed $R^2 = 4 < 4L/C = 20$ . The circuit is underdamped, and its charge oscillates with damped angular frequency $2$ radians per second inside the envelope $e^{-t}$ .

A series circuit has $L = 1$ henry, $R = 6$ ohms and $C = \frac{1}{5}$ farad. At $t = 0$ the source is disconnected, the reference plate holds $1$ coulomb and no current flows. Find $q(t)$ and $i(t)$ for $t \ge 0$ , decide whether the charge ever changes sign, and check that the stored energy decreases.

Show the answer

The equation is $q'' + 6q' + 5q = 0$ . Since $R^2 = 36 > 4L/C = 20$ , the circuit is overdamped, and the characteristic polynomial factors as $(r + 1)(r + 5)$ . So $q = Ae^{-t} + Be^{-5t}$ with $A + B = 1$ and $i(0) = -A - 5B = 0$ , which give $A = \frac{5}{4}$ and $B = -\frac{1}{4}$ . Hence

\begin{aligned} q(t) &= \tfrac{1}{4}\left(5e^{-t} - e^{-5t}\right), \\[0.4em] i(t) &= -\tfrac{5}{4}\left(e^{-t} - e^{-5t}\right). \end{aligned}

For $t \ge 0$ we have $5e^{-t} > e^{-5t}$ , so the charge stays positive and never changes sign. For $t > 0$ we have $e^{-t} > e^{-5t}$ , so the current is negative and charge flows off the reference plate the whole time. Its size is largest at $t = \frac{1}{4}\ln 5 \approx 0.402$ seconds, where $i = -5^{-1/4} \approx -0.669$ amperes. The initial energy is $E(0) = \frac{5}{2}$ joules, all in the capacitor, and $E' = -6\,i^2 < 0$ for every $t > 0$ , so the energy strictly decreases toward zero.

RLC Circuits and Electromechanical Analogies

Key terms

Charge and current

Resistance, inductance and capacitance

Kirchhoff’s voltage law

Series RLC circuit

Electromechanical analogy

Energy and passive components

Check yourself

How this lesson connects

Builds on

Used later in