Theory

Electrical oscillations and alternating-current circuits

How the charge oscillates in an LC circuit and how the oscillations are damped in an RLC circuit, how resistors, capacitors and inductors respond to a sinusoidal voltage, how phasors and impedance give the current and the phase shift, what resonance and the power factor are, and how a transformer changes an alternating voltage.

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LC circuit and electrical oscillations
Series RLC circuit and damped oscillations

LC circuit and electrical oscillations

The circuits are assumed to be in the quasi-static regime, that is, at every instant the current has the same value at all points of a branch; this hypothesis is satisfied when the time taken by an electromagnetic signal to cross the circuit is negligible compared with the period of the varying quantities. The wires have negligible resistance, except where a resistor is indicated.

1. Oscillations of the LC circuit

Consider a capacitor of capacitance $C$ with initial charge $Q_{0}$ , connected at time $t = 0$ to an ideal inductor of inductance $L$ , in a circuit of negligible resistance. Let $q (t)$ be the charge of the initially positive plate and let $i = \frac{d q}{d t}$ be the current, with the positive direction being the one that carries charge onto that plate. Then

q (t) = Q_{0} cos (ω_{0} t), i (t) = - ω_{0} Q_{0} sin (ω_{0} t), ω_{0} = \frac{1}{L C} .

The charge oscillates in harmonic motion with natural angular frequency $ω_{0}$ , period $T = 2 π L C$ and frequency $f_{0} = \frac{1}{2 π L C}$ , and the current has amplitude $I_{0} = ω_{0} Q_{0}$ . The $L C$ circuit is the simplest electrical oscillator and is used to generate signals of fixed frequency. With $L = 10 mH$ and $C = 1.0 μ F$ we have $ω_{0} = \frac{1}{1 0 ^{- 2} \cdot 1 0 ^{- 6}} rad/s = 1.0 \cdot 1 0^{4} rad/s$ and $f_{0} \approx 1.6 kHz$ .

L C

circuit: the positive direction of

i

is the one in which the current carries charge onto the plate with charge

q

, so

i = \frac{d q}{d t}

Proof

We traverse the loop in the positive direction of the current. The potential drop across the inductor is $L \frac{d i}{d t}$ , and across the capacitor, from the plate with charge $q$ to the other, it is $\frac{q}{C}$ . There are no sources in the loop, so by the loop rule the sum of the two drops is zero; with $i = \frac{d q}{d t}$ we obtain

L \frac{d ^{2} q}{d t ^{2}} + \frac{q}{C} = 0 ⟹ \frac{d ^{2} q}{d t ^{2}} + ω_{0}^{2} q = 0, ω_{0}^{2} = \frac{1}{L C} .

This is the equation of harmonic motion, therefore $q (t) = A cos (ω_{0} t + φ)$ . At the initial instant the charge is $Q_{0}$ and the current is zero, because before the circuit is closed no current flows through the inductor and the current in an inductor varies continuously. The initial conditions form the system

{A cos φ = Q_{0} - ω_{0} A sin φ = 0

which with $A > 0$ gives $φ = 0$ and $A = Q_{0}$ . Differentiating $q (t)$ we obtain $i (t)$ .

2. Conservation of energy in the LC circuit

In the $L C$ circuit the sum of the electric energy of the capacitor and the magnetic energy of the inductor is constant:

U = \frac{q ^{2}}{2 C} + \frac{1}{2} L i^{2} = \frac{Q _{0}^{2}}{2 C} .

The energy passes from the electric field of the capacitor to the magnetic field of the inductor and back, twice per period. When the capacitor is discharged all the energy is in the inductor, and the current is maximum in magnitude; equating the two expressions of the energy we obtain $\frac{Q _{0}^{2}}{2 C} = \frac{1}{2} L I_{0}^{2}$ , that is, $I_{0} = \frac{Q _{0}}{L C} = ω_{0} Q_{0}$ . In the circuit with $L = 10 mH$ and $C = 1.0 μ F$ , if the capacitor is charged to $10 V$ , we have $Q_{0} = 10 μ C$ , $U = 50 μ J$ and $I_{0} = 0.10 A$ .

q(t)

i(t)

Charge

q (t) = Q_{0} cos (ω_{0} t)

and current

i (t) = - ω_{0} Q_{0} sin (ω_{0} t)

, measured in units of

Q_{0}

and of

ω_{0} Q_{0}

, with time in units of

\frac{1}{ω _{0}}

: the current is maximum in magnitude when the capacitor is discharged

Proof

Differentiating $U$ with respect to time and using $i = \frac{d q}{d t}$ we obtain

\frac{d U}{d t} = \frac{q}{C} \frac{d q}{d t} + L i \frac{d i}{d t} = i (\frac{q}{C} + L \frac{d i}{d t}) = 0,

because the expression $\frac{q}{C} + L \frac{d i}{d t}$ is zero by the loop rule. Hence $U$ is constant and equal to the initial value $\frac{Q _{0}^{2}}{2 C}$ , computed when the current is zero.

Analogy with the mass-spring oscillator

The equation $L \overset{q}{¨} + \frac{q}{C} = 0$ has the same form as the equation $m \overset{x}{¨} + k x = 0$ of the mass-spring oscillator, so every result on one translates into the other with the following correspondence.

Mass-spring oscillator	$L C$ circuit
position $x$	charge $q$
velocity $v = \overset{x}{˙}$	current $i = \overset{q}{˙}$
mass $m$	inductance $L$
spring constant $k$	reciprocal of the capacitance $\frac{1}{C}$
kinetic energy $\frac{1}{2} m v^{2}$	magnetic energy $\frac{1}{2} L i^{2}$
potential energy $\frac{1}{2} k x^{2}$	electric energy $\frac{q ^{2}}{2 C}$
damping coefficient $b$	resistance $R$

For example, the angular frequency $\frac{k}{m}$ of the oscillator becomes $\frac{1}{L C}$ . Inductance plays the role of inertia, because it opposes changes in the current as mass opposes changes in velocity. See the Physics 1 topic on oscillations and harmonic motion.

Series RLC circuit and damped oscillations

Damped oscillations of the RLC circuit

Consider a capacitor of capacitance $C$ with initial charge $Q_{0}$ that discharges through an inductor of inductance $L$ and a resistor of resistance $R$ , connected in series. The charge satisfies the equation

L \frac{d ^{2} q}{d t ^{2}} + R \frac{d q}{d t} + \frac{q}{C} = 0, i.e. \overset{q}{¨} + 2 γ \overset{q}{˙} + ω_{0}^{2} q = 0, γ = \frac{R}{2 L}, ω_{0} = \frac{1}{L C} .

The behavior depends on the comparison between $R$ and the critical resistance $R_{c} = 2 \frac{L}{C}$ , which corresponds to $γ = ω_{0}$ .

If $R < R_{c}$ (underdamping), the charge oscillates with decreasing amplitude, $q (t) = A e^{- γ t} cos (ω_{s} t + φ)$ , with angular frequency $ω_{s} = ω_{0}^{2} - γ^{2}$ smaller than $ω_{0}$ .
If $R = R_{c}$ (critical damping), the charge tends to zero without oscillating; among the regimes without oscillations it is the one in which the charge vanishes most rapidly.
If $R > R_{c}$ (overdamping), the charge tends to zero without oscillating, more slowly than in the critical case.

The resistor plays in the circuit the role of viscous friction in the mechanical oscillator. With $L = 10 mH$ , $C = 1.0 μ F$ and $R = 2.0 Ω$ we have $R_{c} = 200 Ω$ , so the damping is weak, with $γ = 100 s^{- 1}$ and $ω_{s} \approx ω_{0} = 1.0 \cdot 1 0^{4} rad/s$ .

q(t)

\pm Q_0 e^{-\gamma t}

Discharge of an underdamped

R L C

circuit: the charge oscillates with angular frequency

ω_{s}

and its amplitude decreases as

Q_{0} e^{- γ t}

, with

γ = \frac{R}{2 L}

Proof

We traverse the loop in the positive direction of the current, with $i = \frac{d q}{d t}$ . The potential drop across the resistor is $R i$ by Ohm’s law, and the drops across the capacitor and the inductor are those of the $L C$ circuit. By the loop rule the sum of the three drops is zero, so $L \overset{q}{¨} + R \overset{q}{˙} + \frac{q}{C} = 0$ , and dividing by $L$ we obtain the second form. The characteristic equation $λ^{2} + 2 γ λ + ω_{0}^{2} = 0$ has discriminant $4 (γ^{2} - ω_{0}^{2})$ . If $γ < ω_{0}$ the roots are complex conjugates with real part $- γ$ , therefore the charge is an oscillation of angular frequency $ω_{s}$ multiplied by $e^{- γ t}$ . If $γ \geq ω_{0}$ the roots are real and negative, since $γ^{2} - ω_{0}^{2} < γ$ , so the solutions are combinations of decreasing exponentials, possibly multiplied by $t$ , which do not oscillate. Finally $γ < ω_{0}$ is equivalent to $\frac{R}{2 L} < \frac{1}{L C}$ , that is, to $R < 2 \frac{L}{C}$ .

Energy dissipated in the resistor

Differentiating the energy $U = \frac{q ^{2}}{2 C} + \frac{1}{2} L i^{2}$ and using the equation of the circuit we obtain

\frac{d U}{d t} = i (\frac{q}{C} + L \frac{d i}{d t}) = - R i^{2},

so the energy decreases at the rate $R i^{2}$ , which is the power dissipated by the Joule effect in the resistor. With weak damping and $γ$ much smaller than $ω_{0}$ , the energy is that of an $L C$ circuit with maximum charge $Q_{0} e^{- γ t}$ , that is,

U (t) \approx \frac{Q _{0}^{2}}{2 C} e^{- 2 γ t} = U_{0} e^{- \frac{R}{L} t} .

The energy halves after the time $\frac{L ln 2}{R}$ ; in the circuit with $R = 2.0 Ω$ and $L = 10 mH$ this time is about $3.5 ms$ , equal to about five and a half periods.

The theory continues with 7 more sections (alternating voltages and currents, resistor capacitor and inductor in ac, phasors…), plus all worked examples, the solutions and the PDF.

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