---
title: "Sinusoidal Steady State and Impedance"
book: "University Physics — Year 1"
subject: physics
language: en
chapter: 8
exercises: 12
source: https://one-course.com/books/physics/3/en/chapter/8-sinusoidal-steady-state-and-impedance
---

# Chapter 8 — Sinusoidal Steady State and Impedance

Turn the dial of an old radio and, out of the hundred stations that all reach the antenna at once, one comes through: a coil and a variable capacitor have been tuned so that exactly one frequency makes them resonate. The mains that hums in every wall is a $50\,\mathrm{Hz}$ sinusoid; the factory that runs big motors on it pays a penalty if its current lags its [voltage](https://one-course.com/books/physics/3/en/chapter/6-dc-circuits-kirchhoffs-laws-and-theorems#def-b1-dc-circuits-voltage) too much. Sinusoids are the natural language of [linear circuits](https://one-course.com/books/physics/3/en/chapter/6-dc-circuits-kirchhoffs-laws-and-theorems#def-b1-dc-circuits-linear) — they come out as they go in, only scaled and shifted — and this chapter gives them their algebra: [complex amplitudes](#def-b1-sinusoidal-impedance-complex), [impedances](#def-b1-sinusoidal-impedance-impedance), the resonance of the [RLC circuit](https://one-course.com/books/physics/3/en/chapter/7-transient-regimes-first-and-second-order#prop-b1-transient-regimes-rlc), and the power that sinusoidal currents actually deliver.

![An oscilloscope showing a sinusoid and a square wave: the sinusoid is the signal this chapter lives on, the square wave the one the next chapter decomposes into sinusoids.](https://one-course.com/images/onecourse/chapters/physics-3/b1-sinusoidal-impedance/img-1d22c9099832.jpg)

*An oscilloscope showing a sinusoid and a square wave: the sinusoid is the [signal](https://one-course.com/books/physics/3/en/chapter/5-signal-propagation-travelling-and-standing-waves#def-b1-wave-propagation-signal) this chapter lives on, the square wave the one the next chapter decomposes into sinusoids.*

## 8.1 The sinusoidal steady state

**Proposition 8.1 (Forced sinusoidal regime).**

In a [linear circuit](https://one-course.com/books/physics/3/en/chapter/6-dc-circuits-kirchhoffs-laws-and-theorems#def-b1-dc-circuits-linear) driven by a sinusoidal source of [angular frequency](https://one-course.com/books/physics/3/en/chapter/5-signal-propagation-travelling-and-standing-waves#def-b1-wave-propagation-signal) $\omega$, every current and [voltage](https://one-course.com/books/physics/3/en/chapter/6-dc-circuits-kirchhoffs-laws-and-theorems#def-b1-dc-circuits-voltage) is, once the transients have died out, a sinusoid of the same [angular frequency](https://one-course.com/books/physics/3/en/chapter/5-signal-propagation-travelling-and-standing-waves#def-b1-wave-propagation-signal) $\omega$, each with its own [amplitude](https://one-course.com/books/physics/3/en/chapter/5-signal-propagation-travelling-and-standing-waves#def-b1-wave-propagation-signal) and phase: $x(t) = X\cos(\omega t + \varphi)$.

**Proof.** The circuit’s equations are linear differential equations with constant coefficients; their general solution is a particular solution plus the solutions of the homogeneous equation, which are the transients of [Chapter 7](https://one-course.com/books/physics/3/en/chapter/7-transient-regimes-first-and-second-order#ch-b1-transient-regimes) and decay (every real circuit has some resistance). A sinusoidal particular solution exists because differentiating a sinusoid of frequency $\omega$ gives a sinusoid of the same frequency: the equations can be satisfied term by term — the complex method below constructs it explicitly. ∎

**Definition 8.2 (Complex amplitude).**

To the sinusoid $x(t) = X\cos(\omega t + \varphi)$ one associates the complex [signal](https://one-course.com/books/physics/3/en/chapter/5-signal-propagation-travelling-and-standing-waves#def-b1-wave-propagation-signal) $\underline{x}(t) = X\eu^{j(\omega t + \varphi)}$, so that $x = \Rea(\underline{x})$, and the *complex [amplitude](https://one-course.com/books/physics/3/en/chapter/5-signal-propagation-travelling-and-standing-waves#def-b1-wave-propagation-signal)*

$$
\underline{X} = X\eu^{j\varphi}, \qquad \underline{x}(t) = \underline{X}\,\eu^{j\omega t},
$$

which carries the [amplitude](https://one-course.com/books/physics/3/en/chapter/5-signal-propagation-travelling-and-standing-waves#def-b1-wave-propagation-signal) $X = \abs{\underline{X}}$ and the phase $\varphi = \arg\underline{X}$. Here $j$ denotes the imaginary unit, $j^2 = -1$, the letter $i$ being reserved for currents. Drawn as a vector of length $X$ at angle $\varphi$, $\underline{X}$ is a *phasor* (Fresnel vector).

**Proposition 8.3 (Rules of the complex method).**

1. Linear operations commute with taking the real part: the [complex amplitude](#def-b1-sinusoidal-impedance-complex) of a sum is the sum of the [complex amplitudes](#def-b1-sinusoidal-impedance-complex) .
2. Differentiating with respect to time multiplies the [complex amplitude](#def-b1-sinusoidal-impedance-complex) by $j\omega$ ; integrating divides it by $j\omega$ .
3. [Kirchhoff’s laws](https://one-course.com/books/physics/3/en/chapter/6-dc-circuits-kirchhoffs-laws-and-theorems#thm-b1-dc-circuits-kirchhoff) , being linear, hold for [complex amplitudes](#def-b1-sinusoidal-impedance-complex) .

**Proof.** $\Rea(\underline a + \underline b) = \Rea\underline a + \Rea\underline b$; $\dd(\underline X\eu^{j\omega t})/\dd t = j\omega\underline X\eu^{j\omega t}$, and $\dd x/\dd t = \Rea(\dd\underline x/\dd t)$ since differentiation is real-linear. Sums of currents at a node and of [voltages](https://one-course.com/books/physics/3/en/chapter/6-dc-circuits-kirchhoffs-laws-and-theorems#def-b1-dc-circuits-voltage) around a loop are linear. ∎

**Example 8.4 (Adding two sinusoids).**

$3\cos\omega t + 4\sin\omega t$: the [complex amplitudes](#def-b1-sinusoidal-impedance-complex) are $3$ and $4\eu^{-j\pi/2} = -4j$ (since $\sin\omega t = \cos(\omega t - \pi/2)$); their sum $3 - 4j$ has modulus $5$ and argument $-0.927\ \mathrm{rad}$: $5\cos(\omega t - 0.927)$. No trigonometric identity needed.

## 8.2 Impedance

**Definition 8.5 (Impedance, admittance).**

For a [linear dipole](https://one-course.com/books/physics/3/en/chapter/6-dc-circuits-kirchhoffs-laws-and-theorems#def-b1-dc-circuits-linear) in the [sinusoidal steady state](#prop-b1-sinusoidal-impedance-forced) ([receiver convention](https://one-course.com/books/physics/3/en/chapter/6-dc-circuits-kirchhoffs-laws-and-theorems#thm-b1-dc-circuits-kirchhoff)), the *impedance* is the ratio of [complex amplitudes](#def-b1-sinusoidal-impedance-complex)

$$
\underline Z = \frac{\underline U}{\underline I} = Z\eu^{j\varphi}
\quad (\text{ohms}),
$$

whose modulus $Z = U/I$ relates the [amplitudes](https://one-course.com/books/physics/3/en/chapter/5-signal-propagation-travelling-and-standing-waves#def-b1-wave-propagation-signal) and whose argument $\varphi = \varphi_u - \varphi_i$ is the phase lead of the [voltage](https://one-course.com/books/physics/3/en/chapter/6-dc-circuits-kirchhoffs-laws-and-theorems#def-b1-dc-circuits-voltage) over the current. Its inverse $\underline Y = 1/\underline Z$ is the *admittance* (siemens). The real part of $\underline Z$ is its resistance, the imaginary part its *reactance*.

**Proposition 8.6 (Impedances of R, L, C).**

$$
\underline Z_R = R, \qquad \underline Z_L = jL\omega, \qquad
\underline Z_C = \frac{1}{jC\omega} = -\frac{j}{C\omega} .
$$

A [resistor](https://one-course.com/books/physics/3/en/chapter/6-dc-circuits-kirchhoffs-laws-and-theorems#def-b1-dc-circuits-resistor) keeps [voltage](https://one-course.com/books/physics/3/en/chapter/6-dc-circuits-kirchhoffs-laws-and-theorems#def-b1-dc-circuits-voltage) and current in phase; in a coil the [voltage](https://one-course.com/books/physics/3/en/chapter/6-dc-circuits-kirchhoffs-laws-and-theorems#def-b1-dc-circuits-voltage) leads the current by $\pi/2$; in a capacitor it lags by $\pi/2$. At low frequency a coil tends to a wire and a capacitor to an open circuit; at high frequency the reverse.

**Proof.** $u = Ri$; $u = L\,\dd i/\dd t \to \underline U = jL\omega\,\underline I$; $i = C\,\dd u/\dd t \to \underline I = jC\omega\,\underline U$. The arguments of $1$, $j$ and $-j$ are $0$, $\pi/2$, $-\pi/2$. ∎

**Proposition 8.7 (Association; dividers; theorems).**

[Impedances](#def-b1-sinusoidal-impedance-impedance) in series add; [admittances](#def-b1-sinusoidal-impedance-impedance) in parallel add; the [voltage](https://one-course.com/books/physics/3/en/chapter/6-dc-circuits-kirchhoffs-laws-and-theorems#def-b1-dc-circuits-voltage) and [current dividers](https://one-course.com/books/physics/3/en/chapter/6-dc-circuits-kirchhoffs-laws-and-theorems#prop-b1-dc-circuits-dividers), Thévenin’s and Norton’s theorems, superposition and Millman’s theorem of [Chapter 6](https://one-course.com/books/physics/3/en/chapter/6-dc-circuits-kirchhoffs-laws-and-theorems#ch-b1-dc-circuits) all hold with [complex amplitudes](#def-b1-sinusoidal-impedance-complex) and [impedances](#def-b1-sinusoidal-impedance-impedance) in place of real [voltages](https://one-course.com/books/physics/3/en/chapter/6-dc-circuits-kirchhoffs-laws-and-theorems#def-b1-dc-circuits-voltage), currents and resistances.

**Proof.** Every proof in [Chapter 6](https://one-course.com/books/physics/3/en/chapter/6-dc-circuits-kirchhoffs-laws-and-theorems#ch-b1-dc-circuits) used only [Kirchhoff’s laws](https://one-course.com/books/physics/3/en/chapter/6-dc-circuits-kirchhoffs-laws-and-theorems#thm-b1-dc-circuits-kirchhoff) and the linearity of $u = Ri$, both of which hold in complex form. ∎

**Example 8.8 (The RC divider).**

A [resistor](https://one-course.com/books/physics/3/en/chapter/6-dc-circuits-kirchhoffs-laws-and-theorems#def-b1-dc-circuits-resistor) $R$ and a capacitor $C$ in series across $\underline E$, the output taken across $C$: $\underline U_C = \underline E\,\dfrac{1/jC\omega}
{R + 1/jC\omega} = \dfrac{\underline E}{1 + jRC\omega}$. [Amplitude](https://one-course.com/books/physics/3/en/chapter/5-signal-propagation-travelling-and-standing-waves#def-b1-wave-propagation-signal) $E/\sqrt{1 + (RC\omega)^2}$, phase $-\arctan(RC\omega)$: full transmission at low frequency, $1/\sqrt2$ and $-45^\circ$ at $\omega = 1/RC$, a fall as $1/\omega$ beyond — a low-pass filter, the object of [Chapter 9](https://one-course.com/books/physics/3/en/chapter/9-filters-and-transfer-functions#ch-b1-filters-transfer-functions).

## 8.3 The series RLC circuit: resonance

**Theorem 8.9 (Current resonance).**

A [resistor](https://one-course.com/books/physics/3/en/chapter/6-dc-circuits-kirchhoffs-laws-and-theorems#def-b1-dc-circuits-resistor) $R$, a coil $L$ and a capacitor $C$ in series across $e(t) = E\cos\omega t$ carry the current of [complex amplitude](#def-b1-sinusoidal-impedance-complex)

$$
\underline I = \frac{E}{R + j\left(L\omega - \dfrac{1}{C\omega}\right)},
\qquad
I = \frac{E}{\sqrt{R^2 + \left(L\omega - \dfrac{1}{C\omega}\right)^2}},
\qquad
\varphi_i = -\arctan\frac{L\omega - 1/C\omega}{R} .
$$

The [amplitude](https://one-course.com/books/physics/3/en/chapter/5-signal-propagation-travelling-and-standing-waves#def-b1-wave-propagation-signal) is maximal, $I_{\max} = E/R$, with current and [voltage](https://one-course.com/books/physics/3/en/chapter/6-dc-circuits-kirchhoffs-laws-and-theorems#def-b1-dc-circuits-voltage) in phase, at the *resonance* $\omega_0 = 1/\sqrt{LC}$. With $Q = L\omega_0/R = 1/(RC\omega_0)$ and $x = \omega/\omega_0$,

$$
\frac{I}{I_{\max}} = \frac{1}{\sqrt{1 + Q^2\left(x - \dfrac1x\right)^2}} ,
$$

and the *bandwidth* — the interval of $\omega$ over which $I \geq I_{\max}/\sqrt2$ — is

$$
\Delta\omega = \omega_2 - \omega_1 = \frac{\omega_0}{Q} = \frac{R}{L} .
$$

**Proof.** Series [impedances](#def-b1-sinusoidal-impedance-impedance) add: $\underline Z = R + jL\omega + 1/jC\omega$, and $\underline I = \underline E/\underline Z$. The modulus is largest when the [reactance](#def-b1-sinusoidal-impedance-impedance) $L\omega - 1/C\omega$ vanishes, i.e. at $\omega_0$. Factor $L\omega - 1/C\omega = L\omega_0(x - 1/x)$ and divide by $R$: $L\omega_0/R = Q$. At the edges of the band, $Q(x - 1/x) = \pm 1$, i.e. $x^2 \mp x/Q - 1 = 0$, whose positive roots are $x_{1,2} = \mp\frac{1}{2Q}
+ \sqrt{1 + \frac{1}{4Q^2}}$; their difference is $1/Q$. ∎

![Series RLC: amplitude of the current (left) and its phase relative to the source (right) against / _0, for Q = 1, 3, 10. The peak narrows as Q grows (= _0/Q); below resonance the circuit is capacitive (current leads), above it inductive (current lags).](https://one-course.com/images/onecourse/chapters/physics-3/b1-sinusoidal-impedance/fig-74f88f6a89eb.svg)

![Series RLC: amplitude of the current (left) and its phase relative to the source (right) against / _0, for Q = 1, 3, 10. The peak narrows as Q grows (= _0/Q); below resonance the circuit is capacitive (current leads), above it inductive (current lags).](https://one-course.com/images/onecourse/chapters/physics-3/b1-sinusoidal-impedance/fig-9bef167671c0.svg)

*Series RLC: [amplitude](https://one-course.com/books/physics/3/en/chapter/5-signal-propagation-travelling-and-standing-waves#def-b1-wave-propagation-signal) of the current (left) and its phase relative to the source (right) against $\omega/\omega_0$, for $Q = 1, 3, 10$. The peak narrows as $Q$ grows ($\Delta\omega =
\omega_0/Q$); below resonance the circuit is capacitive (current leads), above it inductive (current lags).*

![Phasor diagram of the series RLC above resonance: U_R along I, U_L a quarter turn ahead, U_C a quarter turn behind; their sum E leads the current by . At resonance U_L and U_C cancel and E = U_R.](https://one-course.com/images/onecourse/chapters/physics-3/b1-sinusoidal-impedance/fig-b403cb43517d.svg)

*[Phasor](#def-b1-sinusoidal-impedance-complex) diagram of the series RLC above resonance: $\underline U_R$ along $\underline I$, $\underline U_L$ a quarter turn ahead, $\underline U_C$ a quarter turn behind; their sum $\underline E$ leads the current by $\varphi$. At resonance $\underline U_L$ and $\underline U_C$ cancel and $\underline E = \underline U_R$.*

**Proposition 8.10 (Voltage magnification).**

At resonance the [voltages](https://one-course.com/books/physics/3/en/chapter/6-dc-circuits-kirchhoffs-laws-and-theorems#def-b1-dc-circuits-voltage) across the coil and the capacitor have the same [amplitude](https://one-course.com/books/physics/3/en/chapter/5-signal-propagation-travelling-and-standing-waves#def-b1-wave-propagation-signal) $QE$ and opposite phases: for $Q \gg 1$ the capacitor carries a [voltage](https://one-course.com/books/physics/3/en/chapter/6-dc-circuits-kirchhoffs-laws-and-theorems#def-b1-dc-circuits-voltage) $Q$ times larger than the source’s. The [amplitude](https://one-course.com/books/physics/3/en/chapter/5-signal-propagation-travelling-and-standing-waves#def-b1-wave-propagation-signal) $U_C(\omega)$ itself peaks, for $Q > 1/\sqrt2$, at $\omega_r =
\omega_0\sqrt{1 - 1/2Q^2}$, slightly below $\omega_0$, with $U_{C,\max} = QE/\sqrt{1 - 1/4Q^2}$.

**Proof.** At $\omega_0$, $\underline U_C = \underline I/jC\omega_0 = -jQE$ and $\underline U_L = jL\omega_0\underline I = jQE$ since $I = E/R$. In general $U_C = I/C\omega = E/\sqrt{(1 - x^2)^2 + x^2/Q^2}$ (multiply numerator and denominator by $1/C\omega_0$); the denominator’s square is $(1 - x^2)^2 + x^2/Q^2$, a quadratic in $x^2$ minimized at $x^2 =
1 - 1/2Q^2$ when that is positive, with minimum $1/Q^2 - 1/4Q^4$. ∎

**Example 8.11 (Tuning a radio).**

A coil $L = 200\,\text{µ}\mathrm{H}$ and a variable capacitor: $C =
127\,\mathrm{pF}$ tunes $f_0 = 1.00\,\mathrm{MHz}$. With the coil’s wire resistance $R = 12.6\,\Omega$, $L\omega_0 = 1257\,\Omega$ and $Q = 100$: bandwidth $\Delta f = f_0/Q = 10\,\mathrm{kHz}$ — the width of an AM channel — and the $10\,\text{µ}\mathrm{V}$ induced by a distant transmitter become $QE = 1\,\mathrm{mV}$ across the capacitor, free of charge. The weekend problem builds this receiver.

**Remark 8.12 (Why QQQ again).**

The $Q$ of this chapter is the $Q$ of [Chapter 7](https://one-course.com/books/physics/3/en/chapter/7-transient-regimes-first-and-second-order#ch-b1-transient-regimes): the circuit that rings for about $Q$ oscillations when struck is the one that resonates over a band $\omega_0/Q$ wide when driven. Both express one ratio: $2\pi$ times the energy stored over the energy dissipated per period ([Exercise 8.12](#exo-b1-sinusoidal-impedance-12)). A sharp resonance and a long ringing are the same property, and so is the slow response: a filter of bandwidth $\Delta\omega$ takes a time of order $1/\Delta\omega$ to settle.

## 8.4 Power in the sinusoidal regime

**Definition 8.13 (RMS value).**

The *root-mean-square* (rms, or effective) value of a periodic [signal](https://one-course.com/books/physics/3/en/chapter/5-signal-propagation-travelling-and-standing-waves#def-b1-wave-propagation-signal) $x(t)$ is $X_{\mathrm{rms}} = \sqrt{\langle x^2\rangle}$, the square root of the time average of $x^2$ over a period. For a sinusoid of [amplitude](https://one-course.com/books/physics/3/en/chapter/5-signal-propagation-travelling-and-standing-waves#def-b1-wave-propagation-signal) $X$, $X_{\mathrm{rms}} = X/\sqrt2$. The $230\,\mathrm{V}$ of the mains is an rms value; its [amplitude](https://one-course.com/books/physics/3/en/chapter/5-signal-propagation-travelling-and-standing-waves#def-b1-wave-propagation-signal) is $325\,\mathrm{V}$.

**Theorem 8.14 (Average power).**

A dipole with $u = U\cos\omega t$ and $i = I\cos(\omega t - \varphi)$ ([receiver convention](https://one-course.com/books/physics/3/en/chapter/6-dc-circuits-kirchhoffs-laws-and-theorems#thm-b1-dc-circuits-kirchhoff)) receives the instantaneous power $p = ui$, of average

$$
P = \tfrac12\,UI\cos\varphi = U_{\mathrm{rms}}I_{\mathrm{rms}}\cos\varphi
= \tfrac12\Rea\!\big(\underline U\,\underline I^{\,*}\big) = \tfrac12\Rea(\underline Z)\,I^2,
$$

where $\cos\varphi$ is the *[power factor](#thm-b1-sinusoidal-impedance-power)*. A [resistor](https://one-course.com/books/physics/3/en/chapter/6-dc-circuits-kirchhoffs-laws-and-theorems#def-b1-dc-circuits-resistor) receives $\tfrac12 RI^2 = RI_{\mathrm{rms}}^2$; an ideal coil or capacitor ($\varphi = \pm\pi/2$) receives no [average power](#thm-b1-sinusoidal-impedance-power) — it stores energy during half a period and returns it during the other half.

**Proof.** $p = UI\cos\omega t\cos(\omega t - \varphi) = \tfrac12 UI[\cos\varphi +
\cos(2\omega t - \varphi)]$; the second term averages to zero. With $\underline U = U$ and $\underline I = I\eu^{-j\varphi}$, $\underline U\,\underline I^* = UI\eu^{j\varphi}$, whose real part is $UI\cos\varphi$; and $\underline U = \underline Z\,\underline I$ gives $\underline U\,\underline I^* = \underline Z I^2$. ∎

![Instantaneous power received by a dipole with a 60 phase lag between current and voltage: it pulses at 2, dips below zero (energy returned to the source), and averages to 1/2 UI — here half of what it would be in phase.](https://one-course.com/images/onecourse/chapters/physics-3/b1-sinusoidal-impedance/fig-a6ed2eb7e694.svg)

*Instantaneous power received by a dipole with a $60^\circ$ phase lag between current and [voltage](https://one-course.com/books/physics/3/en/chapter/6-dc-circuits-kirchhoffs-laws-and-theorems#def-b1-dc-circuits-voltage): it pulses at $2\omega$, dips below zero (energy returned to the source), and averages to $\tfrac12 UI\cos\varphi$ — here half of what it would be in phase.*

**Example 8.15 (Power factor correction).**

A workshop motor on $230\,\mathrm{V}$ rms draws $2.0\,\mathrm{kW}$ with $\cos\varphi = 0.70$ (inductive): $I_{\mathrm{rms}} = 2000/(230 \times
0.70) = 12.4\,\mathrm{A}$, against $8.7\,\mathrm{A}$ if the [power factor](#thm-b1-sinusoidal-impedance-power) were $1$ — the extra current heats the utility’s cables for nothing, hence the penalty. A capacitor in parallel supplies the coil’s quarter-period energy swings locally: the *[reactive power](#thm-b1-sinusoidal-impedance-power)* $P\tan\varphi =
2.0\,\mathrm{kvar}$ needs $C = P\tan\varphi/(\omega U_{\mathrm{rms}}^2)
\approx 120\,\text{µ}\mathrm{F}$ at $50\,\mathrm{Hz}$, and the line current drops to $8.7\,\mathrm{A}$.

## 8.5 Exercises

**Exercise 8.1 ★.**

Give the [complex amplitude](#def-b1-sinusoidal-impedance-complex) of $u = 5\cos(\omega t + \pi/3)$ (V), in polar and Cartesian form. Using [complex amplitudes](#def-b1-sinusoidal-impedance-complex), write $3\cos\omega t
+ 4\sin\omega t$ as a single cosine.

**Solution of Exercise 8.1.**

$\underline U = 5\eu^{j\pi/3} = 2.5 + 4.33j$ (V). $3\cos\omega t +
4\sin\omega t$: $3 + 4\eu^{-j\pi/2} = 3 - 4j$, modulus $5$, argument $-\arctan(4/3) = -0.927\,\mathrm{rad}$: $5\cos(\omega t - 0.927)$.

**Exercise 8.2 ★.**

Compute the [impedance](#def-b1-sinusoidal-impedance-impedance) of $R = 100\,\Omega$, $L = 0.10\,\mathrm{H}$ and $C = 10\,\text{µ}\mathrm{F}$ at $50\,\mathrm{Hz}$ and at $1.0\,\mathrm{kHz}$. Give the modulus and phase of the [impedance](#def-b1-sinusoidal-impedance-impedance) of $R$ and $L$ in series at $50\,\mathrm{Hz}$.

**Solution of Exercise 8.2.**

$R$: $100\,\Omega$ at both. $L\omega$: $31.4\,\Omega$ ($50\,\mathrm{Hz}$), $628\,\Omega$ ($1\,\mathrm{kHz}$). $1/C\omega$: $318\,\Omega$, $15.9\,\Omega$. Series $RL$ at $50\,\mathrm{Hz}$: $\abs{\underline Z} = \sqrt{100^2 + 31.4^2} =
105\,\Omega$, phase $\arctan(0.314) = 17^\circ$.

**Exercise 8.3 ★.**

The mains is $230\,\mathrm{V}$ rms: what is its [amplitude](https://one-course.com/books/physics/3/en/chapter/5-signal-propagation-travelling-and-standing-waves#def-b1-wave-propagation-signal)? A $2.0\,\mathrm{kW}$ heater: rms and peak current. Show that the [average power](#thm-b1-sinusoidal-impedance-power) in a [resistor](https://one-course.com/books/physics/3/en/chapter/6-dc-circuits-kirchhoffs-laws-and-theorems#def-b1-dc-circuits-resistor) is $RI_{\mathrm{rms}}^2$.

**Solution of Exercise 8.3.**

$U = 230\sqrt2 = 325\,\mathrm{V}$. $I_{\mathrm{rms}} = 2000/230 = 8.7\,\mathrm{A}$, peak $12.3\,\mathrm{A}$. $\langle Ri^2\rangle = R\langle i^2\rangle =
RI_{\mathrm{rms}}^2$ by definition of the [rms value](#def-b1-sinusoidal-impedance-rms).

**Exercise 8.4 ★.**

Series RLC: $L = 10\,\mathrm{mH}$, $C = 100\,\mathrm{nF}$, $R = 100\,\Omega$, $E = 1.0\,\mathrm{V}$. Give $f_0$, $Q$, the bandwidth in hertz and the current [amplitude](https://one-course.com/books/physics/3/en/chapter/5-signal-propagation-travelling-and-standing-waves#def-b1-wave-propagation-signal) at resonance.

**Solution of Exercise 8.4.**

$f_0 = 1/(2\pi\sqrt{LC}) = 5.03\,\mathrm{kHz}$; $Q = \sqrt{L/C}/R = 3.16$; $\Delta f = f_0/Q = 1.6\,\mathrm{kHz}$; $I_{\max} = E/R = 10\,\mathrm{mA}$.

**Exercise 8.5 ★★.**

RC divider, output across $C$: $R = 1.0\,\mathrm{k}\Omega$, $C = 159\,\mathrm{nF}$. At what frequency is the output $1/\sqrt2$ of the input? Give the output [amplitude](https://one-course.com/books/physics/3/en/chapter/5-signal-propagation-travelling-and-standing-waves#def-b1-wave-propagation-signal) ratio and phase at $1.0\,\mathrm{kHz}$ and at $10\,\mathrm{kHz}$.

**Solution of Exercise 8.5.**

$1/\sqrt2$ at $\omega = 1/RC$: $f_c = 1/(2\pi RC) = 1.0\,\mathrm{kHz}$. At $1.0\,\mathrm{kHz}$: ratio $0.71$, phase $-45^\circ$. At $10\,\mathrm{kHz}$: $RC\omega = 10$, ratio $1/\sqrt{101} = 0.10$, phase $-84^\circ$.

**Exercise 8.6 ★★.**

At some frequency a series RLC has $R = 100\,\Omega$, $L\omega =
250\,\Omega$, $1/C\omega = 100\,\Omega$, and $E = 1.0\,\mathrm{V}$. Compute $\underline Z$, the current [amplitude](https://one-course.com/books/physics/3/en/chapter/5-signal-propagation-travelling-and-standing-waves#def-b1-wave-propagation-signal) and phase, and the three [voltage](https://one-course.com/books/physics/3/en/chapter/6-dc-circuits-kirchhoffs-laws-and-theorems#def-b1-dc-circuits-voltage) [amplitudes](https://one-course.com/books/physics/3/en/chapter/5-signal-propagation-travelling-and-standing-waves#def-b1-wave-propagation-signal); draw the [phasor](#def-b1-sinusoidal-impedance-complex) diagram. Is the circuit above or below resonance?

**Solution of Exercise 8.6.**

$\underline Z = 100 + j(250 - 100) = 100 + 150j$: $Z = 180\,\Omega$, $\arg = 56^\circ$. $I = 1.0/180 = 5.6\,\mathrm{mA}$, lagging the source by $56^\circ$. $U_R = 0.56\,\mathrm{V}$, $U_L = 1.39\,\mathrm{V}$, $U_C = 0.56\,\mathrm{V}$; [phasors](#def-b1-sinusoidal-impedance-complex): $U_R$ along $I$, $U_L$ up, $U_C$ down, resultant $1.0\,\mathrm{V}$. $L\omega > 1/C\omega$: above resonance (inductive).

**Exercise 8.7 ★★.**

Radio: $L = 200\,\text{µ}\mathrm{H}$, $C$ adjustable from $50\,$ to $500\,\mathrm{pF}$. Find the frequency range, the capacitance for $1.00\,\mathrm{MHz}$, and, with $Q = 100$, the bandwidth and the attenuation ([amplitude](https://one-course.com/books/physics/3/en/chapter/5-signal-propagation-travelling-and-standing-waves#def-b1-wave-propagation-signal) ratio) of a station $9\,\mathrm{kHz}$ away. An antenna emf of $10\,\text{µ}\mathrm{V}$: [voltage](https://one-course.com/books/physics/3/en/chapter/6-dc-circuits-kirchhoffs-laws-and-theorems#def-b1-dc-circuits-voltage) across $C$ at resonance?

**Solution of Exercise 8.7.**

$f_0 = 1/(2\pi\sqrt{LC})$: $1.59\,\mathrm{MHz}$ ($50\,\mathrm{pF}$) to $0.50\,\mathrm{MHz}$ ($500\,\mathrm{pF}$). At $1.00\,\mathrm{MHz}$: $C = 1/(4\pi^2f_0^2L) = 127\,\mathrm{pF}$. $\Delta f = 10\,\mathrm{kHz}$. Station at $1.009\,\mathrm{MHz}$: $x = 1.009$, $Q(x - 1/x) = 1.8$, ratio $1/\sqrt{1 + 3.2} = 0.49$. Capacitor [voltage](https://one-course.com/books/physics/3/en/chapter/6-dc-circuits-kirchhoffs-laws-and-theorems#def-b1-dc-circuits-voltage) $QE = 1.0\,\mathrm{mV}$.

**Exercise 8.8 ★★.**

Derive the two half-power frequencies of the series RLC and show that their difference is $\omega_0/Q$ and their product $\omega_0^2$.

**Solution of Exercise 8.8.**

$Q(x - 1/x) = \pm1 \Leftrightarrow x^2 \mp x/Q - 1 = 0$; positive roots $x_{1,2} = \mp 1/2Q + \sqrt{1 + 1/4Q^2}$. Difference $1/Q$, so $\Delta\omega = \omega_0/Q$; product $(1 + 1/4Q^2) - 1/4Q^2 = 1$, so $\omega_1\omega_2 = \omega_0^2$ (the band is symmetric on a log scale).

**Exercise 8.9 ★★.**

A motor on $230\,\mathrm{V}$ rms, $50\,\mathrm{Hz}$, absorbs $2.0\,\mathrm{kW}$ with $\cos\varphi = 0.70$. Compute the rms current, the [reactive power](#thm-b1-sinusoidal-impedance-power), the capacitance in parallel that brings the [power factor](#thm-b1-sinusoidal-impedance-power) to $1$, and the new current. The supply line has $0.50\,\Omega$: Joule losses before and after.

**Solution of Exercise 8.9.**

$I = P/(U\cos\varphi) = 2000/(230 \times 0.70) = 12.4\,\mathrm{A}$. [Reactive power](#thm-b1-sinusoidal-impedance-power) $P\tan\varphi = 2000 \times 1.02 = 2.0\,\mathrm{kvar}$. $C = 2040/(314
\times 230^2) = 123\,\text{µ}\mathrm{F}$. New current $2000/230 = 8.7\,\mathrm{A}$. Losses $0.50 \times 12.4^2 = 77\,\mathrm{W}$ before, $0.50 \times 8.7^2 =
38\,\mathrm{W}$ after.

**Exercise 8.10 ★★★.**

Parallel RLC fed by a [current source](https://one-course.com/books/physics/3/en/chapter/6-dc-circuits-kirchhoffs-laws-and-theorems#def-b1-dc-circuits-sources): write the [admittance](#def-b1-sinusoidal-impedance-impedance), show that the [voltage](https://one-course.com/books/physics/3/en/chapter/6-dc-circuits-kirchhoffs-laws-and-theorems#def-b1-dc-circuits-voltage) is maximal at $\omega_0 = 1/\sqrt{LC}$ and equals $RI$ there, and that the bandwidth is $\Delta\omega = 1/RC$, i.e. $Q =
R\sqrt{C/L}$. Numbers: $R = 10\,\mathrm{k}\Omega$, $L = 1.0\,\mathrm{mH}$, $C = 1.0\,\mathrm{nF}$.

**Solution of Exercise 8.10.**

$\underline Y = 1/R + j(C\omega - 1/L\omega)$; $\underline U = \underline I/
\underline Y$ is maximal when the susceptance vanishes, at $\omega_0$, where $U = RI$. Half-power: $\abs{C\omega - 1/L\omega} = 1/R$, i.e. $RC\omega_0(x - 1/x) = \pm1$: $\Delta\omega = \omega_0/(RC\omega_0) = 1/RC$ and $Q = RC\omega_0 = R\sqrt{C/L}$. Numbers: $\omega_0 = 1.0 \times 10^{6}\,\mathrm{rad}/\mathrm{s}$ ($f_0 = 159\,\mathrm{kHz}$), $Q = 10^4\sqrt{10^{-6}} = 10$, $\Delta f =
16\,\mathrm{kHz}$.

**Exercise 8.11 ★★★.**

Show that the capacitor [voltage](https://one-course.com/books/physics/3/en/chapter/6-dc-circuits-kirchhoffs-laws-and-theorems#def-b1-dc-circuits-voltage) of a series RLC is $U_C = E/\sqrt{(1 -
x^2)^2 + x^2/Q^2}$, that it peaks at $x_r = \sqrt{1 - 1/2Q^2}$ if $Q > 1/\sqrt2$, and compute $x_r$ and $U_{C,\max}/E$ for $Q = 3.16$. What happens for $Q < 1/\sqrt2$?

**Solution of Exercise 8.11.**

$U_C = I/C\omega = E/[C\omega\sqrt{R^2 + (L\omega - 1/C\omega)^2}]$; multiply inside by $C\omega$: $\sqrt{(RC\omega)^2 + (LC\omega^2 - 1)^2} =
\sqrt{x^2/Q^2 + (1 - x^2)^2}$. With $s = x^2$: $(1 - s)^2 + s/Q^2$ is minimal at $s = 1 - 1/2Q^2$ (if positive), minimum $1/Q^2 - 1/4Q^4$: $U_{C,\max} = QE/\sqrt{1 - 1/4Q^2}$. $Q = 3.16$: $x_r = 0.975$, $U_{C,\max} = 3.2E$. For $Q < 1/\sqrt2$ the minimum is at $s = 0$: $U_C$ decreases monotonically from $E$, no peak.

**Exercise 8.12 ★★★.**

At resonance of a series RLC, show that the total stored energy $\tfrac12 Li^2 + \tfrac12 Cu_C^2$ is constant and equal to $\tfrac12 LI^2$, compute the energy dissipated in $R$ per period, and prove that $2\pi\,(\text{stored})/(\text{dissipated per period}) = Q$.

**Solution of Exercise 8.12.**

At resonance $i = I\cos\omega_0 t$ and $u_C = (I/C\omega_0)\sin\omega_0 t$, so, using $1/C\omega_0^2 = L$,

$$
\tfrac12 Li^2 + \tfrac12 Cu_C^2 = \tfrac12 LI^2\cos^2\omega_0 t
+ \tfrac12 LI^2\sin^2\omega_0 t = \tfrac12 LI^2 .
$$

Dissipated per period: $\tfrac12 RI^2T$. Ratio $\times 2\pi$: $2\pi L/(RT)
= L\omega_0/R = Q$.

![Tuning a valve radio: turning the knob changes a capacitor, the capacitor sets the resonance frequency of an LC circuit, and the station whose frequency matches is the one heard — the weekend problem.](https://one-course.com/images/onecourse/chapters/physics-3/b1-sinusoidal-impedance/img-9d0e0e3a0a3d.jpg)

*Tuning a valve radio: turning the knob changes a capacitor, the capacitor sets the resonance frequency of an $LC$ circuit, and the station whose frequency matches is the one heard — the weekend problem.*

## 8.6 Problem: The resonant radio receiver

**Problem 8.1.**

Weekend problem — a coil, a variable capacitor and an antenna: how a series resonance picks one station out of a hundred, multiplies its faint signal by a hundred, and why it cannot be made arbitrarily selective

The antenna is modeled by an ideal sinusoidal emf $e(t) = E\cos\omega t$ in series with a coil $L = 200\,\text{µ}\mathrm{H}$, whose wire has resistance $R = 12.6\,\Omega$, and a variable capacitor $C$. The receiver reads the [voltage](https://one-course.com/books/physics/3/en/chapter/6-dc-circuits-kirchhoffs-laws-and-theorems#def-b1-dc-circuits-voltage) across $C$. Later (Part IV) the antenna’s own resistance $R_a = 50\,\Omega$ is taken into account.

**Part I — The resonant current.**

1. Write the complex [impedance](#def-b1-sinusoidal-impedance-impedance) of the series circuit.
2. Give the [complex amplitude](#def-b1-sinusoidal-impedance-complex) of the current, then its [amplitude](https://one-course.com/books/physics/3/en/chapter/5-signal-propagation-travelling-and-standing-waves#def-b1-wave-propagation-signal) $I(\omega)$ and its phase $\varphi_i(\omega)$ relative to $e$ .
3. Show that $I$ is maximal at $\omega_0 = 1/\sqrt{LC}$ ; give $I_{\max}$ and the phase then.
4. What capacitance tunes $f_0 = 1.000\,\mathrm{MHz}$ ? Compute $L\omega_0$ .
5. Define $Q = L\omega_0/R$ and compute it.
6. Show that $I/I_{\max} = 1/\sqrt{1 + Q^2(x - 1/x)^2}$ with $x = \omega/\omega_0$ .
7. State the sign of $\varphi_i$ below and above resonance and interpret (capacitive or inductive behavior).

**Part II — Selectivity.**

8. Write the condition defining the two frequencies at which $I = I_{\max}/\sqrt2$ .
9. Solve it and show that $\Delta\omega = \omega_2 - \omega_1 =  \omega_0/Q$ .
10. Compute the bandwidth $\Delta f$ of this receiver.
11. A second station broadcasts at $1.010\,\mathrm{MHz}$ , a third at $1.020\,\mathrm{MHz}$ : compute the [amplitude](https://one-course.com/books/physics/3/en/chapter/5-signal-propagation-travelling-and-standing-waves#def-b1-wave-propagation-signal) ratio $I/I_{\max}$ for each, and the corresponding power ratio in decibels.
12. AM stations are spaced $9\,$ to $10\,\mathrm{kHz}$ apart. Is the receiver selective enough? What would improve it?
13. The capacitor can be set between $50\,$ and $500\,\mathrm{pF}$ : what frequency band does the receiver cover?

**Part III — [Voltage](https://one-course.com/books/physics/3/en/chapter/6-dc-circuits-kirchhoffs-laws-and-theorems#def-b1-dc-circuits-voltage) magnification.**

14. Express the [complex amplitude](#def-b1-sinusoidal-impedance-complex) of the capacitor [voltage](https://one-course.com/books/physics/3/en/chapter/6-dc-circuits-kirchhoffs-laws-and-theorems#def-b1-dc-circuits-voltage) and show that at resonance its [amplitude](https://one-course.com/books/physics/3/en/chapter/5-signal-propagation-travelling-and-standing-waves#def-b1-wave-propagation-signal) is $QE$ .
15. The antenna emf is $E = 10\,\text{µ}\mathrm{V}$ : compute $I_{\max}$ and the capacitor [voltage](https://one-course.com/books/physics/3/en/chapter/6-dc-circuits-kirchhoffs-laws-and-theorems#def-b1-dc-circuits-voltage) at resonance. Comment.
16. Show that the coil [voltage](https://one-course.com/books/physics/3/en/chapter/6-dc-circuits-kirchhoffs-laws-and-theorems#def-b1-dc-circuits-voltage) has the same [amplitude](https://one-course.com/books/physics/3/en/chapter/5-signal-propagation-travelling-and-standing-waves#def-b1-wave-propagation-signal) and is in phase opposition with the capacitor [voltage](https://one-course.com/books/physics/3/en/chapter/6-dc-circuits-kirchhoffs-laws-and-theorems#def-b1-dc-circuits-voltage) at resonance: what does the source “see”?
17. Compute the energy stored in the circuit at resonance and the energy dissipated per period; check that their ratio times $2\pi$ is $Q$ .
18. From [Chapter 7](https://one-course.com/books/physics/3/en/chapter/7-transient-regimes-first-and-second-order#ch-b1-transient-regimes) , what is the [time constant](https://one-course.com/books/physics/3/en/chapter/7-transient-regimes-first-and-second-order#def-b1-transient-regimes-firstorder) $\tau = 2Q/\omega_0$ of this circuit? Why does a very high $Q$ eventually distort speech and music (bandwidth of audio: about $5\,\mathrm{kHz}$ for AM)?
19. If the coil wire resistance doubled (thinner wire), how would $Q$ , the bandwidth and the capacitor [voltage](https://one-course.com/books/physics/3/en/chapter/6-dc-circuits-kirchhoffs-laws-and-theorems#def-b1-dc-circuits-voltage) change?

**Part IV — The antenna’s resistance, and power.**

20. With $R_a = 50\,\Omega$ in series, recompute $Q$ , the bandwidth and the capacitor [voltage](https://one-course.com/books/physics/3/en/chapter/6-dc-circuits-kirchhoffs-laws-and-theorems#def-b1-dc-circuits-voltage) for $E = 10\,\text{µ}\mathrm{V}$ . Conclude.
21. Compute the [average power](#thm-b1-sinusoidal-impedance-power) delivered by the antenna emf at resonance and the fraction dissipated in $R_a$ .
22. The maximum-power theorem would ask for a load resistance equal to $R_a$ : why is that goal in conflict with selectivity?
23. Real receivers connect the antenna to a tap on the coil, or through a small coupling capacitor. Explain qualitatively what this achieves.
24. Using $P = \tfrac12 UI\cos\varphi$ , compute the [power factor](#thm-b1-sinusoidal-impedance-power) at the two half-power frequencies and explain the name “half-power”.
25. Summarize the design: what $Q$ buys, what it costs, and the value chosen for a $1\,\mathrm{MHz}$ receiver with $10\,\mathrm{kHz}$ channels.

**Solution of Problem 8.1.**

**1.** $\underline Z = R + j(L\omega - 1/C\omega)$.

**2.** $\underline I = E/\underline Z$; $I = E/\sqrt{R^2 + (L\omega -
1/C\omega)^2}$; $\varphi_i = -\arctan[(L\omega - 1/C\omega)/R]$.

**3.** The [reactance](#def-b1-sinusoidal-impedance-impedance) vanishes at $\omega_0 = 1/\sqrt{LC}$: $I_{\max}
= E/R$, $\varphi_i = 0$.

**4.** $C = 1/(4\pi^2f_0^2L) = 127\,\mathrm{pF}$; $L\omega_0 = 2\times10^{-4}
\times 6.28\times10^6 = 1257\,\Omega$.

**5.** $Q = 1257/12.6 = 100$.

**6.** $L\omega - 1/C\omega = L\omega_0(x - 1/x)$, divide by $R$.

**7.** $x < 1$: [reactance](#def-b1-sinusoidal-impedance-impedance) negative, $\varphi_i > 0$, the current leads (capacitive); $x > 1$: lags (inductive).

**8.** $Q(x - 1/x) = \pm1$.

**9.** $x^2 \mp x/Q - 1 = 0$, $x_{1,2} = \mp1/2Q + \sqrt{1 + 1/4Q^2}$, $x_2 - x_1 = 1/Q$: $\Delta\omega = \omega_0/Q$.

**10.** $\Delta f = 10\,\mathrm{kHz}$.

**11.** $1.010\,\mathrm{MHz}$: $Q(x - 1/x) = 100 \times 0.0199 = 2.0$, ratio $1/\sqrt{5.0} = 0.45$, power $0.20$: $-7\,\mathrm{dB}$. $1.020\,\mathrm{MHz}$: $3.96$, ratio $0.25$, power $0.06$: $-12\,\mathrm{dB}$.

**12.** Marginal: the neighbor is only $7\,\mathrm{dB}$ down. A higher $Q$ (lower $R$), or several tuned stages in cascade, sharpens the selection.

**13.** $0.50\,\mathrm{MHz}$ to $1.59\,\mathrm{MHz}$: the AM band.

**14.** $\underline U_C = \underline I/jC\omega$; at $\omega_0$, $U_C = I_{\max}/C\omega_0 = E/(RC\omega_0) = QE$.

**15.** $I_{\max} = 10^{-5}/12.6 = 0.79\,\text{µ}\mathrm{A}$; $U_C = QE =
1.0\,\mathrm{mV}$: a hundredfold [voltage](https://one-course.com/books/physics/3/en/chapter/6-dc-circuits-kirchhoffs-laws-and-theorems#def-b1-dc-circuits-voltage) gain with no amplifier.

**16.** $\underline U_L = jL\omega_0\underline I = jQE$, $\underline U_C
= -jQE$: equal and opposite, they cancel; the source sees $R$ alone.

**17.** Stored $\tfrac12 LI_{\max}^2 = 6.2 \times 10^{-17}\,\mathrm{J}$; dissipated per period $\tfrac12 RI_{\max}^2T = 3.9 \times 10^{-18}\,\mathrm{J}$; $2\pi \times 6.2/0.39
\approx 100 = Q$.

**18.** $\tau = 2Q/\omega_0 = 32\,\text{µ}\mathrm{s}$. The circuit cannot follow changes faster than $\sim\tau$: a bandwidth $\Delta f$ passes only modulations slower than $\Delta f$; with $Q = 1000$, $\Delta f =
1\,\mathrm{kHz}$ and the treble of the program would be lost.

**19.** $Q$ halves to $50$, $\Delta f$ doubles to $20\,\mathrm{kHz}$, $U_C$ halves to $0.5\,\mathrm{mV}$.

**20.** $R_{\text{tot}} = 62.6\,\Omega$: $Q = 1257/62.6 = 20$, $\Delta f = 50\,\mathrm{kHz}$, $U_C = 20 \times 10\,\text{µ}\mathrm{V} = 0.2\,\mathrm{mV}$: sensitivity and selectivity both collapse.

**21.** $P = E^2/(2R_{\text{tot}}) = 10^{-10}/125 = 8 \times 10^{-13}\,\mathrm{W}$; fraction in $R_a$: $50/62.6 = 80\%$.

**22.** Matching wants a load resistance equal to $R_a$, i.e. a large total resistance; selectivity wants the smallest possible total resistance ($Q = L\omega_0/R_{\text{tot}}$). One cannot have both with a direct connection.

**23.** A tap or a small coupling capacitor presents the antenna with only a fraction of the resonant circuit: the antenna’s resistance is “seen” by the circuit much reduced, keeping $Q$ high, while the antenna still delivers power near its optimum.

**24.** At the half-power frequencies the [reactance](#def-b1-sinusoidal-impedance-impedance) equals $\pm R$, so $\varphi = \pm45^\circ$ and $\cos\varphi = 1/\sqrt2$; with $I =
I_{\max}/\sqrt2$ too, $P = \tfrac12 EI\cos\varphi$ is half its resonant value.

**25.** $Q$ multiplies the [signal](https://one-course.com/books/physics/3/en/chapter/5-signal-propagation-travelling-and-standing-waves#def-b1-wave-propagation-signal) by $Q$ and narrows the band to $f_0/Q$, at the cost of a slower response ($\tau = 2Q/\omega_0$) and a fragile coupling to the antenna; for $10\,\mathrm{kHz}$ channels at $1\,\mathrm{MHz}$, $Q \approx 100$ is the number — and the antenna must be coupled lightly to keep it.