Chapter 5 · Section 5.6

5.6 Electric Circuits

Everything we learned about springs has an exact twin in electrical engineering. A circuit containing a resistor, an inductor and a capacitor in series obeys the same second-order linear equation as a damped mass on a spring. The charge on the capacitor plays the role of the displacement, the current plays the role of the velocity, and the voltage source plays the role of the external force. Once you see the dictionary, the whole of Sections 5.2–5.5 translates into circuit language for free. This is not a coincidence of notation: both systems store energy in two forms and dissipate it in a third.

Circuit elements and their voltage drops

A current $i (t)$ (in amperes, A) flows around a closed loop. It is the rate at which charge passes a point, so if $q (t)$ (in coulombs, C) is the charge on the capacitor,

i = \frac{d q}{d t} = q^{'} .

A source such as a generator or battery supplies an electromotive force (emf, voltage) $E (t)$ , in volts (V). Each element produces a voltage drop across itself, given by an experimental law:

Element	Constant (unit)	Voltage drop	Stores / does
Resistor	resistance $R$ (ohm, Ω)	$E_{R} = R i$ (Ohm's law)	dissipates energy as heat
Inductor	inductance $L$ (henry, H)	$E_{L} = L \frac{d i}{d t}$	stores energy $\frac{1}{2} L i^{2}$ in its magnetic field
Capacitor	capacitance $C$ (farad, F)	$E_{C} = \frac{q}{C}$	stores energy $\frac{1}{2} q^{2} / C$ in its electric field

The inductor resists changes of current, much as a mass resists changes of velocity. The capacitor pushes back in proportion to the charge it holds, much as a spring pushes back in proportion to its stretch. The resistor turns energy into heat, like a damper.

Kirchhoff's voltage law and the RLC equation

Kirchhoff's voltage law

Around any closed loop, the sum of the voltage drops across the elements equals the impressed emf.

For a resistor, an inductor and a capacitor in series with a source (Figure 5.6.1) the law gives

L \frac{d i}{d t} + R i + \frac{1}{C} q = E (t) .

This contains two unknowns, $q$ and $i$ , but they are linked by $i = q^{'}$ . Substituting $i = q^{'}$ and $i^{'} = q^{''}$ leaves one second-order equation for the charge. Alternatively, differentiating once and using $q^{'} = i$ gives an equation for the current.

Figure 5.6.1. A series RLC circuit. With the current

i

counted positive in the direction of the arrow and

q

the charge on the capacitor plate the current flows into,

i = q^{'}

Theorem 5.6.1 (Series RLC circuit)

The charge $q (t)$ on the capacitor of a series RLC circuit satisfies

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

and, if $E$ is differentiable, the current $i (t) = q^{'} (t)$ satisfies

L i^{''} + R i^{'} + \frac{1}{C} i = E^{'} (t) .

Initial conditions are $q (0) = q_{0}$ (initial charge) and $q^{'} (0) = i (0) = i_{0}$ (initial current).

Two simple special cases are first-order equations, which you met in Chapter 2:

RL circuit (no capacitor): $L i^{'} + R i = E (t)$ .
RC circuit (no inductor): $R q^{'} + \frac{1}{C} q = E (t)$ .

Common mistake: the current equation needs

E^{'}

In the equation for $i$ the right-hand side is the derivative of the emf. For a battery ( $E$ constant) it is zero, even though the battery is driving the circuit. Also, the second initial condition for $i$ is $i^{'} (0)$ , which you must get from the circuit law at $t = 0$ : $L i^{'} (0) = E (0) - R i (0) - q (0) / C$ . Working with $q$ avoids both traps.

The electrical–mechanical analogy

Put the two equations side by side:

m x^{''} + c x^{'} + k x = F (t) ⟷ L q^{''} + R q^{'} + \frac{1}{C} q = E (t) .

Mechanical system	Electrical system
displacement $x$	charge $q$
velocity $v = x^{'}$	current $i = q^{'}$
mass $m$	inductance $L$
damping constant $c$	resistance $R$
spring constant $k$	reciprocal capacitance $1/ C$
external force $F (t)$	emf $E (t)$
kinetic energy $\frac{1}{2} m v^{2}$	magnetic energy $\frac{1}{2} L i^{2}$
spring energy $\frac{1}{2} k x^{2}$	electric energy $\frac{1}{2} q^{2} / C$

Every result of Sections 5.2–5.5 now translates directly:

Natural frequency: $ω_{0} = k / m$ becomes $ω_{0} = \frac{1}{L C}$ .
Decay rate: $β = c / (2 m)$ becomes $β = R / (2 L)$ .
Damping classes: $c^{2}$ versus $4 mk$ becomes $R^{2}$ versus $4 L / C$ . The circuit is overdamped if $R > 2 L / C$ , critically damped if $R = 2 L / C$ , underdamped (the charge oscillates) if $R < 2 L / C$ .
Transient and steady state: if $R > 0$ the solution of the homogeneous equation dies out, and with a periodic emf only the steady-state charge and current remain.

Historically the analogy was used the other way round: engineers built cheap electrical circuits to simulate expensive mechanical systems, adjusting a resistor instead of rebuilding a damper.

Steady-state current and impedance

For an alternating emf $E (t) = E_{0} cos ω t$ we usually care about the current, the quantity an ammeter shows. Start from the steady-state charge, given by Theorem 5.4.2 with $m \to L$ , $c \to R$ , $k \to 1/ C$ :

q_{ss} = a cos ω t + b sin ω t, a = \frac{E _{0} ( 1/ C - L ω ^{2} )}{Δ}, b = \frac{E _{0} R ω}{Δ}, Δ = (\frac{1}{C} - L ω^{2})^{2} + R^{2} ω^{2} .

Differentiate. $i_{ss} = q_{ss}^{'} = ωb cos ω t - ωa sin ω t$ .
Introduce the reactance. Let $X = ω L - \frac{1}{ω C}$ (in ohms). Then $\frac{1}{C} - L ω^{2} = - ω X$ , and so $Δ = ω^{2} X^{2} + R^{2} ω^{2} = ω^{2} (R^{2} + X^{2}) .$
Simplify the coefficients. $ωb = \frac{E _{0} R ω ^{2}}{ω ^{2} ( R ^{2} + X ^{2} )} = \frac{E _{0} R}{R ^{2} + X ^{2}}, - ωa = \frac{E _{0} ω ^{2} X}{ω ^{2} ( R ^{2} + X ^{2} )} = \frac{E _{0} X}{R ^{2} + X ^{2}} .$
Amplitude–phase form. With $Z = R^{2} + X^{2}$ , $i_{ss} = \frac{E _{0}}{Z ^{2}} (R cos ω t + X sin ω t) = \frac{E _{0}}{Z} cos (ω t - ψ), cos ψ = \frac{R}{Z}, sin ψ = \frac{X}{Z} .$

Theorem 5.6.2 (Steady-state current)

Let $R, L, C > 0$ . For $E (t) = E_{0} cos ω t$ the steady-state current in a series RLC circuit is

i_{ss} (t) = \frac{E _{0}}{Z} cos (ω t - ψ),

where

X = ω L - \frac{1}{ω C} (reactance), Z = R^{2} + X^{2} (impedance), tan ψ = \frac{X}{R}, - \frac{π}{2} < ψ < \frac{π}{2} .

Since $cos ψ = R / Z > 0$ , the angle $ψ$ is always in $(- π /2, π /2)$ , so here $ψ = arctan (X / R)$ is safe. Its sign has a meaning:

$X > 0$ (inductive, high frequency): $ψ > 0$ , the current lags the emf;
$X < 0$ (capacitive, low frequency): $ψ < 0$ , the current leads the emf;
$X = 0$ : the current is in phase with the emf.

Current resonance: tuning a circuit

The current amplitude $E_{0} / Z$ is largest when $Z$ is smallest, i.e. when $X = 0$ :

ω L = \frac{1}{ω C} ⟺ ω = ω_{0} = \frac{1}{L C} .

At this frequency $Z = R$ , the current amplitude is $E_{0} / R$ , the voltage drops across $L$ and $C$ cancel exactly, and the circuit behaves as if it were a plain resistor. Note the contrast with Section 5.5: the charge amplitude (the analogue of displacement) peaks slightly below $ω_{0}$ , at $ω_{r} = 1/ (L C) - R^{2} / (2 L^{2})$ , but the current amplitude (the analogue of velocity) peaks exactly at $ω_{0}$ , whatever $R$ is. A small $R$ makes the peak tall and narrow (Figure 5.6.4): the circuit responds strongly to one frequency and weakly to the others. That is how a radio selects one station. Turning the tuning knob changes $C$ (or $L$ ), and so moves $ω_{0}$ onto the desired frequency.

Method in a nutshell: series RLC circuits

Convert units (mH, μF, pF to H and F) and write $L q^{''} + R q^{'} + q / C = E (t)$ with $q (0) = q_{0}$ , $q^{'} (0) = i_{0}$ .
Classify with $R^{2}$ versus $4 L / C$ ; compute $ω_{0} = 1/ L C$ and $β = R / (2 L)$ .
Solve exactly as for the mass–spring system: $q = q_{c} + q_{p}$ , then fit the initial conditions.
Get the current by differentiating: $i = q^{'}$ .
For the steady-state current with $E = E_{0} cos ω t$ , skip the charge: compute $X = ω L - 1/ (ω C)$ , $Z = R^{2} + X^{2}$ , and write $i_{ss} = (E_{0} / Z) cos (ω t - ψ)$ with $ψ = arctan (X / R)$ .

The circuit simulator below solves $L q^{''} + R q^{'} + q / C = E_{0} cos ω t$ (with $ω = 0$ meaning a constant emf $E_{0}$ ) and also shows the current. It starts with the circuit of Example 5.6.3. Set $E_{0} = 0$ and $q_{0} = 0.08$ to reproduce Example 5.6.2, or vary $ω$ and watch the steady-state current peak at $ω_{0} = 1/ L C$ .

Worked examples

Example 5.6.1 (RL circuit)

A 12 V battery is connected at $t = 0$ to a 6 Ω resistor and a 0.5 H inductor in series. Find the current if $i (0) = 0$ . How long does the current take to reach 95% of its final value?

Solution

Equation. No capacitor: $0.5 i^{'} + 6 i = 12$ , i.e. $i^{'} + 12 i = 24$ .
Solve. This is first-order linear with constant coefficients. The particular solution is the constant $i_{p} = 2$ ; the homogeneous solution is $C_{1} e^{- 12 t}$ . So $i = 2 + C_{1} e^{- 12 t}$ , and $i (0) = 0$ gives $C_{1} = - 2$ : $i (t) = 2 (1 - e^{- 12 t}) A .$
Interpretation. The current rises to $E / R = 2$ A. The inductor delays it; the time constant is $L / R = 1/12$ s.
95% time. $1 - e^{- 12 t} = 0.95$ gives $e^{- 12 t} = 0.05$ , so $t = ln 20/12 \approx 0.25$ s, about three time constants.

Example 5.6.2 (Free discharge of an RLC circuit)

A circuit has $L = 0.25$ H, $R = 5$ Ω and $C = 0.008$ F, and no source. The capacitor is charged to 10 V and the switch is closed at $t = 0$ (so $i (0) = 0$ ). Find $q (t)$ and $i (t)$ .

Solution

Initial charge. $q_{0} = C E_{C} = 0.008 \times 10 = 0.08$ C.
Classify. $R^{2} = 25$ and $4 L / C = 1/0.008 = 125$ . Since $25 < 125$ , the circuit is underdamped: the charge will oscillate.
Equation. $0.25 q^{''} + 5 q^{'} + 125 q = 0$ , i.e. $q^{''} + 20 q^{'} + 500 q = 0$ .
Roots. $r = - 10 \pm 100 - 500 = - 10 \pm 20 i$ , so $q = e^{- 10 t} (c_{1} cos 20 t + c_{2} sin 20 t) .$
Initial conditions. $q (0) = c_{1} = 0.08$ . $q^{'} (0) = - 10 c_{1} + 20 c_{2} = 0$ gives $c_{2} = 0.04$ : $q (t) = e^{- 10 t} (0.08 cos 20 t + 0.04 sin 20 t) C .$
Current. Differentiate with the product rule: $i = q^{'} = e^{- 10 t} [(- 0.8 + 0.8) cos 20 t + (- 0.4 - 1.6) sin 20 t] = - 2 e^{- 10 t} sin 20 t A .$ The negative sign means the current first flows opposite to the arrow in Figure 5.6.1: the capacitor discharges. The charge then overshoots (the capacitor charges with the opposite sign) and the oscillation decays like $e^{- 10 t}$ (Figure 5.6.2).

Figure 5.6.2. Example 5.6.2: the charge (in units of

1 0^{- 2}

C) and the current during a free, underdamped discharge. The current is zero exactly when the charge is at an extreme, just as the velocity of a mass is zero at its turning points.

Example 5.6.3 (Driven RLC circuit)

The circuit of Example 5.6.2 is connected, with the capacitor initially uncharged and no current, to an emf $E (t) = 100 cos 10 t$ V. Find $q (t)$ and $i (t)$ , and write the steady-state current in amplitude–phase form. Check the amplitude with Theorem 5.6.2.

Solution

Equation. $0.25 q^{''} + 5 q^{'} + 125 q = 100 cos 10 t$ , i.e. $q^{''} + 20 q^{'} + 500 q = 400 cos 10 t$ , with $q (0) = 0$ , $q^{'} (0) = 0$ .
Complementary solution. From Example 5.6.2: $q_{c} = e^{- 10 t} (c_{1} cos 20 t + c_{2} sin 20 t)$ .
Particular solution. Try $q_{p} = a cos 10 t + b sin 10 t$ . With $q_{p}^{''} = - 100 q_{p}$ : $cos 10 t : sin 10 t : (500 - 100) a + 200 b = 400, - 200 a + (500 - 100) b = 0.$ The second gives $a = 2 b$ ; then $800 b + 200 b = 400$ , so $b = 0.4$ and $a = 0.8$ .
Initial conditions. $q (0) = c_{1} + 0.8 = 0$ , so $c_{1} = - 0.8$ . Since $q_{p}^{'} (0) = 10 b = 4$ , $q^{'} (0) = - 10 c_{1} + 20 c_{2} + 4 = 8 + 20 c_{2} + 4 = 0$ , so $c_{2} = - 0.6$ .
Charge. $q (t) = - e^{- 10 t} (0.8 cos 20 t + 0.6 sin 20 t) + 0.8 cos 10 t + 0.4 sin 10 t C .$
Current. Differentiating, $i (t) = transient e^{- 10 t} (22 sin 20 t - 4 cos 20 t) + i_{ss} 4 cos 10 t - 8 sin 10 t A .$
Steady-state current. Amplitude $4^{2} + 8^{2} = 80 = 45 \approx 8.94$ A. Since $4 cos 10 t - 8 sin 10 t = 45 (\frac{1}{5} cos 10 t - \frac{2}{5} sin 10 t)$ , the angle $ψ$ has $cos ψ = 1/ 5$ , $sin ψ = - 2/ 5$ , so $ψ = - arctan 2 \approx - 1.107$ : $i_{ss} = 45 cos (10 t + 1.107) A .$
Check with impedance. $X = 10 (0.25) - \frac{1}{10 ( 0.008 )} = 2.5 - 12.5 = - 10$ Ω, $Z = 25 + 100 = 55$ Ω, $E_{0} / Z = 100/ (55) = 45$ . ✓ Since $X < 0$ the circuit is capacitive at this frequency, and the current leads the emf by $1.107$ rad, as found. (Here $ω = 10$ is well below $ω_{0} = 1/ 0.002 \approx 22.4$ rad/s.)

Figure 5.6.3. Example 5.6.3: the current (solid) starts at zero and merges with the steady-state current (dashed) once the transient, decaying like

e^{- 10 t}

, has died out.

Example 5.6.4 (Tuning a receiver)

The input stage of a receiver is a series circuit with $L = 0.5$ mH and resistance $R = 10$ Ω. (a) What capacitance tunes it to a station broadcasting at $f = 1$ MHz? (b) An emf of amplitude 2 mV at 1 MHz is induced in it. What is the current amplitude? (c) Compare with the current produced by an equally strong emf from a station at 1.1 MHz.

Solution

(a) Resonance condition. We need $ω_{0} = 1/ L C = 2 π f$ , so $C = \frac{1}{( 2 π f ) ^{2} L} = \frac{1}{( 2 π \times 1 0 ^{6} ) ^{2} \times 5 \times 1 0 ^{- 4}} \approx 5.07 \times 1 0^{- 11} F \approx 50.7 pF .$
(b) At resonance $Z = R$ , so the current amplitude is $E_{0} / R = 2 \times 1 0^{- 3} /10 = 2 \times 1 0^{- 4}$ A $= 0.2$ mA.
(c) At 1.1 MHz. $ω = 2 π (1.1 \times 1 0^{6}) \approx 6.912 \times 1 0^{6}$ rad/s, so $X = ω L - \frac{1}{ω C} \approx 3456 - 2856 \approx 600 Ω, Z = 1 0^{2} + 60 0^{2} \approx 600 Ω.$ The current amplitude is $2 \times 1 0^{- 3} /600 \approx 3.3 \times 1 0^{- 6}$ A, about $60$ times smaller than for the tuned station. The circuit strongly favours the frequency it is tuned to.

Figure 5.6.4. Steady-state current amplitude for

L = 0.25

C = 0.008

E_{0} = 100

V and three resistances. Every curve peaks exactly at

ω_{0} = 1/ L C \approx 22.4

rad/s, with height

E_{0} / R

; a smaller resistance gives a sharper peak.

Example 5.6.5 (Using the analogy)

Describe a mass–spring system whose displacement is described by exactly the same function as the charge in Example 5.6.2, and use Section 5.3 to predict the quasi-period and logarithmic decrement of the discharge.

Solution

Translate. $m \leftrightarrow L = 0.25$ , $c \leftrightarrow R = 5$ , $k \leftrightarrow 1/ C = 125$ : a 0.25 kg mass on a spring with $k = 125$ N/m and a damper with $c = 5$ N·s/m.
Initial conditions. $q_{0} = 0.08$ C, $i_{0} = 0$ become $x_{0} = 0.08$ m, $v_{0} = 0$ : pulled 8 cm down and released from rest. Then $x (t) = e^{- 10 t} (0.08 cos 20 t + 0.04 sin 20 t)$ m, the same function.
Predictions. $β = R / (2 L) = 10$ s $^{- 1}$ and $ω_{d} = 20$ rad/s, so $T_{d} = 2 π /20 \approx 0.314$ s and $δ = β T_{d} = π \approx 3.14$ . Each oscillation reduces the charge amplitude by the factor $e^{- π} \approx 0.043$ , which is why the oscillation in Figure 5.6.2 is barely visible after one cycle.

Exercises

An RL circuit has $L = 2$ H, $R = 10$ Ω and a 20 V battery switched on at $t = 0$ with $i (0) = 0$ . Find $i (t)$ .
Answer
$i = 2 (1 - e^{- 5 t})$ A
An RC circuit has $R = 1000$ Ω, $C = 100$ μF and a 9 V battery switched on at $t = 0$ ; the capacitor is initially uncharged. Find $q (t)$ and the time constant.
Answer
$q = 9 \times 1 0^{- 4} (1 - e^{- 10 t})$ C; $RC = 0.1$ s
An LC circuit ( $R = 0$ ) has $L = 0.5$ H, $C = 0.02$ F, $q (0) = 0.002$ C and $i (0) = 0$ . Find $q (t)$ , $i (t)$ and the frequency of oscillation in hertz.
Answer
$q = 0.002 cos 10 t$ , $i = - 0.02 sin 10 t$ ; $f_{0} = 10/ (2 π) \approx 1.59$ Hz
Classify the free behaviour of a circuit with $L = 1$ H, $R = 40$ Ω, $C = 1 0^{- 3}$ F, and find $β$ and the quasi-frequency if it oscillates.
Answer
$R^{2} = 1600 < 4 L / C = 4000$ : underdamped; $β = 20 s^{- 1}$ , $ω_{d} = 1000 - 400 = 600 \approx 24.5$ rad/s
Find the critical resistance of a circuit with $L = 0.2$ H and $C = 50$ μF.
Answer
$R_{cr} = 2 L / C = 24000 \approx 126.5$ Ω
A circuit with $L = 1$ H, $R = 50$ Ω, $C = 2.5$ mF and no source has $q (0) = 0.01$ C, $i (0) = 0$ . Find $q (t)$ and $i (t)$ .
Answer
$q = \frac{1}{75} e^{- 10 t} - \frac{1}{300} e^{- 40 t}$ ; $i = - \frac{2}{15} (e^{- 10 t} - e^{- 40 t})$
Find the steady-state current in a circuit with $L = 0.5$ H, $R = 30$ Ω, $C = 1$ mF and $E = 50 cos 20 t$ V. Does it lead or lag the emf?
Answer
$X = - 40$ Ω, $Z = 50$ Ω; $i_{ss} = cos (20 t + 0.927)$ A; it leads by $0.927$ rad
A circuit with $L = 0.4$ H and $R = 20$ Ω is driven by $E = 10 cos 50 t$ V. Which capacitance gives the largest steady-state current, and what is that current amplitude?
Answer
$C = 1/ (ω^{2} L) = 1 0^{- 3}$ F; amplitude $E_{0} / R = 0.5$ A
Give the mechanical analogue of the circuit $L = 0.5$ H, $R = 4$ Ω, $C = 0.02$ F, and classify both.
Answer
$m = 0.5$ kg, $c = 4$ N·s/m, $k = 50$ N/m; $c^{2} = 16 < 4 mk = 100$ : underdamped
A circuit with $L = 0.2$ H, $R = 2$ Ω, $C = 0.04$ F is connected at $t = 0$ to a 10 V battery, with $q (0) = 0$ and $i (0) = 0$ . Find $q (t)$ and $i (t)$ .
Answer
$q = 0.4 - e^{- 5 t} (0.4 cos 10 t + 0.2 sin 10 t)$ C; $i = 5 e^{- 5 t} sin 10 t$ A
Derive the current equation $L i^{''} + R i^{'} + i / C = E^{'} (t)$ from Kirchhoff's law, and explain why for a battery the current tends to zero as $t \to \infty$ when $R > 0$ , while the charge does not.
Answer
Differentiate $L i^{'} + R i + q / C = E$ and use $q^{'} = i$ . For constant $E$ the current equation is homogeneous, so $i \to 0$ ; the charge tends to $CE$ , the particular solution of the charge equation.