---
title: "RC and RL Circuits"
book: "High School Physics"
subject: physics
language: en
chapter: 30
exercises: 15
source: https://one-course.com/books/physics/2/en/chapter/30-rc-and-rl-circuits
---

# Chapter 30 — RC and RL Circuits

Press a camera’s shutter halfway: a [thin](https://one-course.com/books/physics/2/en/chapter/10-lenses-images-and-the-eye#def-g11-lenses-and-eye-lens) whine climbs for two seconds, a light turns green — only then will the flash fire, dumping in a millisecond what it spent two seconds gathering. This chapter meets the two components that give circuits a memory and a clock — [capacitor](#def-g12-rc-rl-circuits-capacitor) and [coil](https://one-course.com/books/physics/2/en/chapter/15-magnetism-and-magnetic-fields#def-g11-magnetic-fields-solenoid) — and physics’ first differential equations, checkable with this year’s derivatives.

## 30.1 The capacitor

**Definition 30.1 (Capacitor and capacitance).**

A *capacitor* is a pair of facing metal plates separated by a [thin](https://one-course.com/books/physics/2/en/chapter/10-lenses-images-and-the-eye#def-g11-lenses-and-eye-lens) insulator. Wired to a source, the plates take opposite charges $+q$ and $-q$ — an [electric field](https://one-course.com/books/physics/2/en/chapter/14-electric-and-gravitational-fields#def-g11-electric-gravitational-fields-efield) fills the gap ([Chapter 14](https://one-course.com/books/physics/2/en/chapter/14-electric-and-gravitational-fields#ch-g11-electric-gravitational-fields)) — and the [voltage](https://one-course.com/books/physics/2/en/chapter/12-electric-circuits-and-power#def-g11-circuits-and-power-voltage) $u$ across the plates is proportional to the stored charge:

$$
q = C\,u .
$$

The constant $C$ is the *capacitance*, in *farads* ($\mathrm{F}$): $1\,\mathrm{F} = 1\,\mathrm{C}/\mathrm{V}$.

**Example 30.2 (Orders of magnitude).**

One [farad](#def-g12-rc-rl-circuits-capacitor) is enormous — one [coulomb](https://one-course.com/books/physics/2/en/chapter/13-the-fundamental-interactions#def-g11-fundamental-interactions-charge) parked per [volt](https://one-course.com/books/physics/2/en/chapter/12-electric-circuits-and-power#def-g11-circuits-and-power-voltage). Ceramics of radio circuits: $\mathrm{pF}$ to $\mathrm{nF}$; electrolytics of [power](https://one-course.com/books/physics/2/en/chapter/17-work-of-a-force#def-g11-work-of-force-power) supplies and flashes: $\text{µ}\mathrm{F}$ to $\mathrm{mF}$; supercapacitors: thousands of [farads](#def-g12-rc-rl-circuits-capacitor).

**Proposition 30.3 (Current into a capacitor).**

The current arriving on a [capacitor](#def-g12-rc-rl-circuits-capacitor)’s plate is the rate at which its charge grows:

$$
i = \frac{dq}{dt} = C\,\frac{du}{dt} ,
$$

lowercase marking time-varying quantities ($U$, $I$ stay steady, [Chapter 12](https://one-course.com/books/physics/2/en/chapter/12-electric-circuits-and-power#ch-g11-circuits-and-power)).

**Proof.** Current is charge delivered per second ([Definition 12.1](https://one-course.com/books/physics/2/en/chapter/12-electric-circuits-and-power#def-g11-circuits-and-power-current)): over a shrinking interval, $i$ is the derivative of $q$; and $q = Cu$ with $C$ constant gives $C\,du/dt$. ∎

**Remark 30.4 (Two consequences).**

If $u$ is constant, $i = 0$: a charged [capacitor](#def-g12-rc-rl-circuits-capacitor) passes no steady current. And $u$ can never jump — a step would demand infinite current: the [capacitor](#def-g12-rc-rl-circuits-capacitor)’s [voltage](https://one-course.com/books/physics/2/en/chapter/12-electric-circuits-and-power#def-g11-circuits-and-power-voltage) is the circuit’s memory.

## 30.2 Charging a capacitor: the RC circuit

One loop — source $\mathcal{E}$ ([Definition 12.18](https://one-course.com/books/physics/2/en/chapter/12-electric-circuits-and-power#def-g11-circuits-and-power-emf)), switch, resistor, empty [capacitor](#def-g12-rc-rl-circuits-capacitor): close $K$ at $t = 0$.

![The RC charging circuit: closing K lets the source push charge through R onto the plates of C.](https://one-course.com/images/onecourse/chapters/physics-2/g12-rc-rl-circuits/fig-2f8ee4d9b2da.svg)

*The RC charging circuit: closing $K$ lets the source push charge through $R$ onto the plates of $C$.*

**Theorem 30.5 (Charging law of the RC circuit).**

After the switch closes, the [capacitor](#def-g12-rc-rl-circuits-capacitor) [voltage](https://one-course.com/books/physics/2/en/chapter/12-electric-circuits-and-power#def-g11-circuits-and-power-voltage) obeys, then follows from $u(0) = 0$:

$$
RC\,\frac{du}{dt} + u = \mathcal{E},
\qquad
u(t) = \mathcal{E}\left(1 - e^{-t/\tau}\right),
\qquad \tau = RC .
$$

**Proof.** [Voltages](https://one-course.com/books/physics/2/en/chapter/12-electric-circuits-and-power#def-g11-circuits-and-power-voltage) along the loop add ([Proposition 12.4](https://one-course.com/books/physics/2/en/chapter/12-electric-circuits-and-power#prop-g11-circuits-and-power-laws)): $\mathcal{E} = u_R + u = Ri + u$ at every instant; substitute $i = C\,du/dt$ ([Proposition 30.3](#prop-g12-rc-rl-circuits-cdudt)). Now *verify* the candidate: $du/dt = (\mathcal{E}/\tau)e^{-t/\tau}$, so $RC\,(\mathcal{E}/RC)\,e^{-t/\tau} + \mathcal{E}(1 - e^{-t/\tau}) =
\mathcal{E}$, true for all $t$; and $u(0) = \mathcal{E}(1-1) = 0$. That no *other* curve fits the equation and the start is proved in this year’s mathematics course on differential equations. ∎

**Definition 30.6 (Time constant).**

The *time constant* of an [RC circuit](#thm-g12-rc-rl-circuits-charging) is $\tau = RC$. [Units](https://one-course.com/books/physics/2/en/chapter/1-orders-of-magnitude-measuring-the-universe#def-g10-orders-of-magnitude-unit): $\Omega \times \mathrm{F} =
(\mathrm{V}/\mathrm{A})(\mathrm{C}/\mathrm{V}) = \mathrm{C}/\mathrm{A} = \mathrm{s}$ — the circuit’s own tempo.

**Method 30.7 (Reading τ\tauτ on the curve).**

Three equivalent readings, straight off a graph:

1. *tangent at the origin:* $u'(0) = \mathcal{E}/\tau$ , so the initial tangent reaches the asymptote at $t = \tau$ ;
2. *63%:* $u(\tau) = \mathcal{E}(1 - e^{-1}) \approx  0.63\,\mathcal{E}$ ;
3. *95%:* $u(3\tau) \approx 0.95\,\mathcal{E}$ — charged, for practical purposes ( $5\tau$ : $99\%$ ).

**Proposition 30.8 (Discharge).**

A [capacitor](#def-g12-rc-rl-circuits-capacitor) charged to $U_0$ and closed on a resistor $R$ alone obeys $RC\,du/dt + u = 0$ and follows

$$
u(t) = U_0\,e^{-t/\tau}, \qquad \tau = RC
$$

— $37\%$ left at $\tau$, $5\%$ at $3\tau$.

**Proof.** Same loop law, no source; substitution again: $RC(-U_0/\tau)e^{-t/\tau} + U_0 e^{-t/\tau} = 0$, with $u(0) = U_0$. ∎

![Charge (blue, from u = 0 toward E) and discharge (red, from U_0 = E). Both initial tangents meet their asymptote at t =; 63\% done at , 95\% at 3.](https://one-course.com/images/onecourse/chapters/physics-2/g12-rc-rl-circuits/fig-de2eed6e47d0.svg)

*Charge (blue, from $u = 0$ toward $\mathcal{E}$) and discharge (red, from $U_0 = \mathcal{E}$). Both initial tangents meet their asymptote at $t = \tau$; $63\%$ done at $\tau$, $95\%$ at $3\tau$.*

**Example 30.9 (A flash unit’s numbers).**

A photo flash charges $C = 800\,\text{µ}\mathrm{F}$ through $R = 1.0\,\mathrm{k}\Omega$: $\tau = 10^3 \times 8.0\times10^{-4} =
0.80\,\mathrm{s}$, so the “ready” light — wired to trip at $95\%$ — waits $3\tau \approx 2.4\,\mathrm{s}$: the whine of the hook. Fired through the flash tube’s mere $0.50\,\Omega$, the *same* [capacitor](#def-g12-rc-rl-circuits-capacitor) empties with $\tau' = 0.40\,\mathrm{ms}$: two thousand times faster out than in.

## 30.3 The coil: inductance and the RL circuit

**Definition 30.10 (Inductor and inductance).**

An *inductor* is a [coil](https://one-course.com/books/physics/2/en/chapter/15-magnetism-and-magnetic-fields#def-g11-magnetic-fields-solenoid) of wire: its current threads its own turns with a [magnetic field](https://one-course.com/books/physics/2/en/chapter/15-magnetism-and-magnetic-fields#def-g11-magnetic-fields-bfield) ([Chapter 15](https://one-course.com/books/physics/2/en/chapter/15-magnetism-and-magnetic-fields#ch-g11-magnetic-fields)); whenever it *changes*, a [voltage](https://one-course.com/books/physics/2/en/chapter/12-electric-circuits-and-power#def-g11-circuits-and-power-voltage) appears across the [coil](https://one-course.com/books/physics/2/en/chapter/15-magnetism-and-magnetic-fields#def-g11-magnetic-fields-solenoid):

$$
u = L\,\frac{di}{dt} .
$$

The constant $L$ is the *inductance*, in *henries* ($\mathrm{H}$): $1\,\mathrm{H} = 1\,\mathrm{V}\,\mathrm{s}/\mathrm{A}$.

**Proof.** *Admitted at this level.* ∎

**Remark 30.11 (The coil resists change).**

Why a changing current makes a [voltage](https://one-course.com/books/physics/2/en/chapter/12-electric-circuits-and-power#def-g11-circuits-and-power-voltage) — induction — is derived in the university volumes; here we take the law and its character: the [coil](https://one-course.com/books/physics/2/en/chapter/15-magnetism-and-magnetic-fields#def-g11-magnetic-fields-solenoid) *opposes changes* of its current. Mirror twins: a [capacitor](#def-g12-rc-rl-circuits-capacitor)’s $u$ cannot jump, a [coil](https://one-course.com/books/physics/2/en/chapter/15-magnetism-and-magnetic-fields#def-g11-magnetic-fields-solenoid)’s $i$ cannot jump.

![The RL circuit: after K closes, the coil lets the current climb only at its own pace = L/R.](https://one-course.com/images/onecourse/chapters/physics-2/g12-rc-rl-circuits/fig-9934f9796a6b.svg)

*The [RL circuit](#prop-g12-rc-rl-circuits-rl): after $K$ closes, the [coil](https://one-course.com/books/physics/2/en/chapter/15-magnetism-and-magnetic-fields#def-g11-magnetic-fields-solenoid) lets the current climb only at its own pace $\tau = L/R$.*

**Proposition 30.12 (Current rise in an RL circuit).**

Closing the switch on a source $\mathcal{E}$, a resistor $R$ and a [coil](https://one-course.com/books/physics/2/en/chapter/15-magnetism-and-magnetic-fields#def-g11-magnetic-fields-solenoid) $L$ in series gives $\mathcal{E} = Ri + L\,di/dt$, i.e.

$$
\frac{L}{R}\,\frac{di}{dt} + i = \frac{\mathcal{E}}{R},
\qquad\text{solved by}\quad
i(t) = \frac{\mathcal{E}}{R}\left(1 - e^{-t/\tau}\right),
\quad \tau = \frac{L}{R} .
$$

**Proof.** The loop law again; and it is the *same equation* as in [Theorem 30.5](#thm-g12-rc-rl-circuits-charging) with $u \to i$, $\mathcal{E} \to \mathcal{E}/R$, $RC \to L/R$ — the same substitution verifies the solution, and every reading of [Method 30.7](#met-g12-rc-rl-circuits-reading) transfers. ∎

**Example 30.13 (An electromagnet wakes up).**

A relay [coil](https://one-course.com/books/physics/2/en/chapter/15-magnetism-and-magnetic-fields#def-g11-magnetic-fields-solenoid), $L = 0.30\,\mathrm{H}$ and $R = 6.0\,\Omega$, on a $12\,\mathrm{V}$ supply: final current $\mathcal{E}/R = 2.0\,\mathrm{A}$, at the pace $\tau = 0.30/6.0 = 50\,\mathrm{ms}$ — awake in $3\tau \approx
0.15\,\mathrm{s}$.

**Remark 30.14 (The spark at opening).**

Open a switch on a live [coil](https://one-course.com/books/physics/2/en/chapter/15-magnetism-and-magnetic-fields#def-g11-magnetic-fields-solenoid) and $i$ must crash to zero in a fraction of a millisecond: $di/dt$ is huge and $u = L\,di/dt$ reaches hundreds or thousands of [volts](https://one-course.com/books/physics/2/en/chapter/12-electric-circuits-and-power#def-g11-circuits-and-power-voltage) — a spark jumps the contact. A car’s ignition [coil](https://one-course.com/books/physics/2/en/chapter/15-magnetism-and-magnetic-fields#def-g11-magnetic-fields-solenoid) does it on purpose, thousands of times a minute.

## 30.4 Stored energy

**Proposition 30.15 (Energy in a capacitor, energy in a coil).**

A [capacitor](#def-g12-rc-rl-circuits-capacitor) $C$ charged to [voltage](https://one-course.com/books/physics/2/en/chapter/12-electric-circuits-and-power#def-g11-circuits-and-power-voltage) $u$ and a [coil](https://one-course.com/books/physics/2/en/chapter/15-magnetism-and-magnetic-fields#def-g11-magnetic-fields-solenoid) $L$ carrying a current $i$ store

$$
E_C = \tfrac12\,C u^2 = \frac{q^2}{2C},
\qquad
E_L = \tfrac12\,L i^2 .
$$

**Proof.** Landing a small charge $dq$ on plates already at [voltage](https://one-course.com/books/physics/2/en/chapter/12-electric-circuits-and-power#def-g11-circuits-and-power-voltage) $u$ costs $u\,dq$ ([Definition 12.3](https://one-course.com/books/physics/2/en/chapter/12-electric-circuits-and-power#def-g11-circuits-and-power-voltage)); stacking the costs, the total is the area under the line $u = q/C$ — the triangle below — so $E_C = \tfrac12 qu = \tfrac12 Cu^2$. The same triangle for the [coil](https://one-course.com/books/physics/2/en/chapter/15-magnetism-and-magnetic-fields#def-g11-magnetic-fields-solenoid) (raising $i$ by $di$ costs $Li\,di$) gives $E_L = \tfrac12 Li^2$. ∎

![Each slice dq costs u\,dq; the whole charge costs the triangle’s area E_C = 1/2 q_0 u_0 — the early coulombs board at low voltage.](https://one-course.com/images/onecourse/chapters/physics-2/g12-rc-rl-circuits/fig-fe806a23267a.svg)

*Each slice $dq$ costs $u\,dq$; the whole charge costs the triangle’s area $E_C = \tfrac12 q_0 u_0$ — the early [coulombs](https://one-course.com/books/physics/2/en/chapter/13-the-fundamental-interactions#def-g11-fundamental-interactions-charge) board at low [voltage](https://one-course.com/books/physics/2/en/chapter/12-electric-circuits-and-power#def-g11-circuits-and-power-voltage).*

**Example 30.16 (Energy orders of magnitude).**

The flash [capacitor](#def-g12-rc-rl-circuits-capacitor) of [Example 30.9](#ex-g12-rc-rl-circuits-flashrc): $\tfrac12 \times 8.0\times10^{-4} \times 300^2 = 36\,\mathrm{J}$. A $3000\,\mathrm{F}$ supercapacitor at $2.7\,\mathrm{V}$: $1.1 \times 10^{4}\,\mathrm{J}$ — an AA battery’s worth. A $10\,\mathrm{mH}$ [coil](https://one-course.com/books/physics/2/en/chapter/15-magnetism-and-magnetic-fields#def-g11-magnetic-fields-solenoid) at $10\,\mathrm{A}$: $0.50\,\mathrm{J}$. Less than batteries hold — but releasable in a millisecond.

**Remark 30.17 (What the two time-keepers are for).**

Slow-in, fast-out [powers](https://one-course.com/books/physics/2/en/chapter/17-work-of-a-force#def-g11-work-of-force-power) the camera flash and the defibrillator. The [delay](https://one-course.com/books/physics/2/en/chapter/20-mechanical-waves#def-g12-mechanical-waves-delay) $\tau = RC$ runs [timing circuits](#rem-g12-rc-rl-circuits-applications): intermittent wipers, blinking beacons, stairwell lights. A [capacitor](#def-g12-rc-rl-circuits-capacitor) across a supply *smooths* [voltage](https://one-course.com/books/physics/2/en/chapter/12-electric-circuits-and-power#def-g11-circuits-and-power-voltage) ($u$ cannot jump); a series [coil](https://one-course.com/books/physics/2/en/chapter/15-magnetism-and-magnetic-fields#def-g11-magnetic-fields-solenoid) smooths current ($i$ cannot jump). And [coil](https://one-course.com/books/physics/2/en/chapter/15-magnetism-and-magnetic-fields#def-g11-magnetic-fields-solenoid) plus [capacitor](#def-g12-rc-rl-circuits-capacitor) together make an [oscillator](https://one-course.com/books/physics/2/en/chapter/28-mechanical-oscillators-and-the-measurement-of-time#def-g12-oscillators-and-time-oscillator) ([Chapter 31](https://one-course.com/books/physics/2/en/chapter/31-free-electrical-oscillations-rlc#ch-g12-rlc-oscillations)).

## 30.5 Exercises

**Exercise 30.1 ★.**

A $470\,\text{µ}\mathrm{F}$ [capacitor](#def-g12-rc-rl-circuits-capacitor) is charged to $12\,\mathrm{V}$: what charge does it hold? What [voltage](https://one-course.com/books/physics/2/en/chapter/12-electric-circuits-and-power#def-g11-circuits-and-power-voltage) would put $2.0\,\mathrm{mC}$ on $100\,\text{µ}\mathrm{F}$?

**Solution of Exercise 30.1.**

$q = Cu = 4.7\times10^{-4} \times 12 \approx 5.6\,\mathrm{mC}$; $u = q/C = 2.0\times10^{-3}/10^{-4} = 20\,\mathrm{V}$.

**Exercise 30.2 ★.**

Attribute $47\,\mathrm{pF}$, $2200\,\text{µ}\mathrm{F}$ and $10\,\mathrm{F}$ among a power-supply smoother, a radio tuner and a supercapacitor; compute the last one’s [energy](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-energy) at $2.7\,\mathrm{V}$.

**Solution of Exercise 30.2.**

$47\,\mathrm{pF}$: radio tuner; $2200\,\text{µ}\mathrm{F}$: smoother; $10\,\mathrm{F}$: supercapacitor. $E = \tfrac12 \times 10 \times 2.7^2 \approx 36\,\mathrm{J}$.

**Exercise 30.3 ★.**

The [voltage](https://one-course.com/books/physics/2/en/chapter/12-electric-circuits-and-power#def-g11-circuits-and-power-voltage) across a $220\,\text{µ}\mathrm{F}$ [capacitor](#def-g12-rc-rl-circuits-capacitor) rises steadily from $0$ to $5.0\,\mathrm{V}$ in $10\,\mathrm{ms}$: what current flows in? What current once the [voltage](https://one-course.com/books/physics/2/en/chapter/12-electric-circuits-and-power#def-g11-circuits-and-power-voltage) holds still, and why?

**Solution of Exercise 30.3.**

$i = C\,\Delta u/\Delta t = 2.2\times10^{-4} \times 5.0/0.010 =
0.11\,\mathrm{A}$. Then $i = C\,du/dt = 0$: a charged [capacitor](#def-g12-rc-rl-circuits-capacitor) is an open circuit — no steady current.

**Exercise 30.4 ★.**

Compute $\tau$ for $R = 10\,\mathrm{k}\Omega$, $C = 100\,\text{µ}\mathrm{F}$, then for $L = 0.10\,\mathrm{H}$, $R = 50\,\Omega$. Check, [unit](https://one-course.com/books/physics/2/en/chapter/1-orders-of-magnitude-measuring-the-universe#def-g10-orders-of-magnitude-unit) by [unit](https://one-course.com/books/physics/2/en/chapter/1-orders-of-magnitude-measuring-the-universe#def-g10-orders-of-magnitude-unit), that both combinations are seconds.

**Solution of Exercise 30.4.**

$RC = 10^4 \times 10^{-4} = 1.0\,\mathrm{s}$; $L/R = 0.10/50 = 2.0\,\mathrm{ms}$. $\Omega\,\mathrm{F} =
(\mathrm{V}/\mathrm{A})(\mathrm{C}/\mathrm{V}) = \mathrm{C}/\mathrm{A} = \mathrm{s}$; $\mathrm{H}/\Omega = (\mathrm{V}\,\mathrm{s}/\mathrm{A})/(\mathrm{V}/\mathrm{A}) = \mathrm{s}$.

**Exercise 30.5 ★.**

A [capacitor](#def-g12-rc-rl-circuits-capacitor) charges toward $6.0\,\mathrm{V}$, passing $3.8\,\mathrm{V}$ at $t = 2.0\,\mathrm{s}$. Read off the [time constant](#def-g12-rc-rl-circuits-tau); when is the charge $95\%$ done?

**Solution of Exercise 30.5.**

$3.8/6.0 = 0.63$, so $\tau = 2.0\,\mathrm{s}$; $95\%$ at $3\tau =
6.0\,\mathrm{s}$.

**Exercise 30.6 ★★.**

An [RC circuit](#thm-g12-rc-rl-circuits-charging): $\mathcal{E} = 9.0\,\mathrm{V}$, $R = 4.7\,\mathrm{k}\Omega$, $C = 220\,\text{µ}\mathrm{F}$. Compute $\tau$, $u(\tau)$, $u(3\tau)$, and the current at the instant the switch closes.

**Solution of Exercise 30.6.**

$\tau = 4700 \times 2.2\times10^{-4} \approx 1.0\,\mathrm{s}$; $u(\tau) = 0.63 \times 9.0 = 5.7\,\mathrm{V}$; $u(3\tau) = 0.95 \times 9.0 = 8.6\,\mathrm{V}$; $I_0 = \mathcal{E}/R = 9.0/4700 \approx 1.9\,\mathrm{mA}$ (empty [capacitor](#def-g12-rc-rl-circuits-capacitor): all of $\mathcal{E}$ sits across $R$).

**Exercise 30.7 ★★.**

Verify by substitution that $u(t) = \mathcal{E}(1 - e^{-t/RC})$ satisfies $RC\,du/dt + u = \mathcal{E}$; what do its value at $t = 0$ and its large-$t$ limit encode physically?

**Solution of Exercise 30.7.**

$du/dt = (\mathcal{E}/RC)e^{-t/RC}$, so $RC\,du/dt + u =
\mathcal{E}e^{-t/RC} + \mathcal{E}(1 - e^{-t/RC}) = \mathcal{E}$. $u(0) = 0$: the [capacitor](#def-g12-rc-rl-circuits-capacitor) starts empty; $u \to \mathcal{E}$: fully charged, the current has stopped.

**Exercise 30.8 ★★.**

A $100\,\text{µ}\mathrm{F}$ [capacitor](#def-g12-rc-rl-circuits-capacitor) charged to $12\,\mathrm{V}$ discharges through $47\,\mathrm{k}\Omega$: compute $\tau$, $u(\tau)$, and the time for $u$ to drop below $0.60\,\mathrm{V}$.

**Solution of Exercise 30.8.**

$\tau = 4.7\times10^{4} \times 10^{-4} = 4.7\,\mathrm{s}$; $u(\tau) = 12\,e^{-1} \approx 4.4\,\mathrm{V}$; $0.60\,\mathrm{V}$ is $5\%$, reached at $3\tau \approx 14\,\mathrm{s}$.

**Exercise 30.9 ★★.**

The current in a $0.50\,\mathrm{H}$ [coil](https://one-course.com/books/physics/2/en/chapter/15-magnetism-and-magnetic-fields#def-g11-magnetic-fields-solenoid) climbs steadily from $0$ to $2.0\,\mathrm{A}$ in $40\,\mathrm{ms}$: what [voltage](https://one-course.com/books/physics/2/en/chapter/12-electric-circuits-and-power#def-g11-circuits-and-power-voltage) appears across it? With $u = L\,di/dt$, why does *opening* a switch on a [coil](https://one-course.com/books/physics/2/en/chapter/15-magnetism-and-magnetic-fields#def-g11-magnetic-fields-solenoid) spark, not closing?

**Solution of Exercise 30.9.**

$u = L\,\Delta i/\Delta t = 0.50 \times 2.0/0.040 = 25\,\mathrm{V}$. Closing, $i$ climbs at the circuit’s own pace, $di/dt$ modest; opening *[forces](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force)* $i$ to zero almost instantly, $di/dt$ is huge and $u$ reaches sparking values.

**Exercise 30.10 ★★.**

The relay of [Example 30.13](#ex-g12-rc-rl-circuits-rl): verify by substitution that $i(t) = (\mathcal{E}/R)(1 - e^{-Rt/L})$ satisfies $L\,di/dt + Ri = \mathcal{E}$, then compute $i(\tau)$ and $i(3\tau)$.

**Solution of Exercise 30.10.**

$di/dt = (\mathcal{E}/L)e^{-Rt/L}$, so $L\,di/dt + Ri =
\mathcal{E}e^{-Rt/L} + \mathcal{E}(1 - e^{-Rt/L}) = \mathcal{E}$. $i(\tau) = 0.63 \times 2.0 = 1.3\,\mathrm{A}$; $i(3\tau) = 0.95 \times 2.0 = 1.9\,\mathrm{A}$.

**Exercise 30.11 ★★.**

Show that $u'(0) = \mathcal{E}/\tau$, and deduce that the tangent at the origin meets the asymptote at $t = \tau$. On a recorded curve that gives $\tau = 0.50\,\mathrm{s}$, with $R = 2.0\,\mathrm{k}\Omega$: find $C$.

**Solution of Exercise 30.11.**

$u'(t) = (\mathcal{E}/\tau)e^{-t/\tau}$, so $u'(0) = \mathcal{E}/\tau$: the tangent $u = (\mathcal{E}/\tau)\,t$ reaches $\mathcal{E}$ at $t = \tau$. $C = \tau/R = 0.50/2000 = 250\,\text{µ}\mathrm{F}$.

**Exercise 30.12 ★★★.**

Rank by stored [energy](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-energy), computing each: a $3000\,\mathrm{F}$ supercapacitor at $2.7\,\mathrm{V}$; a flash [capacitor](#def-g12-rc-rl-circuits-capacitor), $800\,\text{µ}\mathrm{F}$ at $300\,\mathrm{V}$; a $10\,\mathrm{mH}$ [coil](https://one-course.com/books/physics/2/en/chapter/15-magnetism-and-magnetic-fields#def-g11-magnetic-fields-solenoid) at $10\,\mathrm{A}$; an AA battery ($E \approx 1.3 \times 10^{4}\,\mathrm{J}$). What do the first three offer that the battery cannot?

**Solution of Exercise 30.12.**

Supercapacitor $\tfrac12 \times 3000 \times 2.7^2 \approx 1.1 \times 10^{4}\,\mathrm{J}$; flash $\tfrac12 \times 8.0\times10^{-4} \times 300^2 = 36\,\mathrm{J}$; [coil](https://one-course.com/books/physics/2/en/chapter/15-magnetism-and-magnetic-fields#def-g11-magnetic-fields-solenoid) $\tfrac12 \times 0.010 \times 10^2 = 0.50\,\mathrm{J}$. Battery ($1.3 \times 10^{4}\,\mathrm{J}$) $>$ supercapacitor $\gg$ flash $\gg$ [coil](https://one-course.com/books/physics/2/en/chapter/15-magnetism-and-magnetic-fields#def-g11-magnetic-fields-solenoid) — but all three can dump their [energy](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-energy) in milliseconds: [power](https://one-course.com/books/physics/2/en/chapter/17-work-of-a-force#def-g11-work-of-force-power), not [energy](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-energy).

**Exercise 30.13 ★★★.**

A stairwell timer lights while $u < 0.63\,\mathcal{E}$ across its charging [capacitor](#def-g12-rc-rl-circuits-capacitor), going dark at $t = \tau$. Design the light to last $5.0\,\mathrm{s}$ using $C = 100\,\text{µ}\mathrm{F}$: what $R$? For a threshold $0.50\,\mathcal{E}$, show the [delay](https://one-course.com/books/physics/2/en/chapter/20-mechanical-waves#def-g12-mechanical-waves-delay) is $\tau \ln 2$ and compute it.

**Solution of Exercise 30.13.**

$\tau = 5.0\,\mathrm{s}$, so $R = \tau/C = 5.0/10^{-4} = 50\,\mathrm{k}\Omega$. $1 - e^{-t/\tau} = 0.50$ gives $e^{-t/\tau} = 0.50$, $t = \tau\ln 2
\approx 0.69\,\tau = 3.5\,\mathrm{s}$.

**Exercise 30.14 ★★★.**

A $2200\,\text{µ}\mathrm{F}$ [capacitor](#def-g12-rc-rl-circuits-capacitor) smooths a supply: between recharges, it alone feeds a load drawing $0.50\,\mathrm{A}$ for $10\,\mathrm{ms}$. From $i = C\,du/dt$, compute the [voltage](https://one-course.com/books/physics/2/en/chapter/12-electric-circuits-and-power#def-g11-circuits-and-power-voltage) droop, then the [capacitance](#def-g12-rc-rl-circuits-capacitor) keeping it under $0.50\,\mathrm{V}$.

**Solution of Exercise 30.14.**

$\Delta u = i\,\Delta t/C = 0.50 \times 0.010/2.2\times10^{-3} \approx
2.3\,\mathrm{V}$. Need $C \geq i\,\Delta t/\Delta u =
5.0\times10^{-3}/0.50 = 10\,\mathrm{mF}$.

**Exercise 30.15 ★★★.**

An ignition [coil](https://one-course.com/books/physics/2/en/chapter/15-magnetism-and-magnetic-fields#def-g11-magnetic-fields-solenoid), $L = 0.80\,\mathrm{H}$, carries $2.0\,\mathrm{A}$ when the breaker cuts it off in about $1.0\,\mathrm{ms}$. Estimate the [coil](https://one-course.com/books/physics/2/en/chapter/15-magnetism-and-magnetic-fields#def-g11-magnetic-fields-solenoid) [voltage](https://one-course.com/books/physics/2/en/chapter/12-electric-circuits-and-power#def-g11-circuits-and-power-voltage) during the cut and the spark’s [energy](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-energy); why does a slow interruption give no spark?

**Solution of Exercise 30.15.**

$u \approx L\,\Delta i/\Delta t = 0.80 \times 2.0/10^{-3} =
1.6\,\mathrm{kV}$; $E_L = \tfrac12 \times 0.80 \times 2.0^2 =
1.6\,\mathrm{J}$. Cut slowly, $di/dt$ stays small, $u$ stays a few [volts](https://one-course.com/books/physics/2/en/chapter/12-electric-circuits-and-power#def-g11-circuits-and-power-voltage): no spark.

## 30.6 Problem: The camera flash

**Problem 30.1.**

Weekend problem — the camera flash: a two-second whine, a millisecond of glory, and the wall-mounted cousin that restarts hearts — one exponential runs them all

A compact flash [unit](https://one-course.com/books/physics/2/en/chapter/1-orders-of-magnitude-measuring-the-universe#def-g10-orders-of-magnitude-unit) stores [energy](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-energy) in a [capacitor](#def-g12-rc-rl-circuits-capacitor) $C = 800\,\text{µ}\mathrm{F}$, charged through $R = 1.0\,\mathrm{k}\Omega$ from a [converter](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-transfer) modeled as an ideal source $\mathcal{E} = 300\,\mathrm{V}$; the “ready” light trips at $95\%$ of full [voltage](https://one-course.com/books/physics/2/en/chapter/12-electric-circuits-and-power#def-g11-circuits-and-power-voltage). Fired, the [capacitor](#def-g12-rc-rl-circuits-capacitor) sees only the flash tube, $R_f = 0.50\,\Omega$. Data: an AA cell delivers $1.5\,\mathrm{V}$ and $2.4\,\mathrm{A}\,\mathrm{h}$; $e = 1.60 \times 10^{-19}\,\mathrm{C}$.

**Part I — The reservoir.**

1. Compute the charge on the [capacitor](#def-g12-rc-rl-circuits-capacitor) at $300\,\mathrm{V}$ .
2. How many [elementary charges](https://one-course.com/books/physics/2/en/chapter/13-the-fundamental-interactions#def-g11-fundamental-interactions-elementary) is that?
3. Compute the stored [energy](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-energy) $E_C$ .
4. The AA cell holds $q = It \approx 8.6 \times 10^{3}\,\mathrm{C}$ , hence about $1.3 \times 10^{4}\,\mathrm{J}$ . How many times the [capacitor](#def-g12-rc-rl-circuits-capacitor) ’s [energy](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-energy) is that?
5. So why bother with the [capacitor](#def-g12-rc-rl-circuits-capacitor) ? One sentence, using “ [power](https://one-course.com/books/physics/2/en/chapter/17-work-of-a-force#def-g11-work-of-force-power) ”.

**Part II — The wait for green.**

6. Write the loop law and turn it into the differential equation for $u(t)$ .
7. Verify by substitution that $u(t) = \mathcal{E}(1 - e^{-t/RC})$ solves it, starting from $u(0) = 0$ .
8. Compute the [time constant](#def-g12-rc-rl-circuits-tau) $\tau$ .
9. After what time does the ready light turn green, and at what [voltage](https://one-course.com/books/physics/2/en/chapter/12-electric-circuits-and-power#def-g11-circuits-and-power-voltage) ? Recognize the two-second whine.
10. Compute the initial charging current; why does it then die away?

**Part III — The millisecond of glory.**

11. Write the discharge law $u(t)$ through the tube and compute the new [time constant](#def-g12-rc-rl-circuits-tau) $\tau'$ .
12. Compute the peak current through the tube.
13. Compute the peak [power](https://one-course.com/books/physics/2/en/chapter/17-work-of-a-force#def-g11-work-of-force-power) ; compare it with a $2.2\,\mathrm{kW}$ kettle.
14. Taking the flash as over at $3\tau'$ , give its duration and average [power](https://one-course.com/books/physics/2/en/chapter/17-work-of-a-force#def-g11-work-of-force-power) .
15. Form the ratio of charging time to flash time: same [capacitor](#def-g12-rc-rl-circuits-capacitor) , same charge — what changed, and what does that make a [capacitor](#def-g12-rc-rl-circuits-capacitor) : an [energy](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-energy) tank, or a [power](https://one-course.com/books/physics/2/en/chapter/17-work-of-a-force#def-g11-work-of-force-power) lever?

**Part IV — The cousin on the wall.** A defibrillator must deliver $E_C = 200\,\mathrm{J}$ at $U_0 = 1500\,\mathrm{V}$; chest and paddles present about $50\,\Omega$.

16. Compute the required [capacitance](#def-g12-rc-rl-circuits-capacitor) .
17. Compute the stored charge.
18. Compute the shock’s [time constant](#def-g12-rc-rl-circuits-tau) and duration ( $3\tau$ ); compare with the flash.
19. Compute the peak current and peak [power](https://one-course.com/books/physics/2/en/chapter/17-work-of-a-force#def-g11-work-of-force-power) in the patient.
20. The recharge circuit must make the device ready ( $95\%$ ) in $6.0\,\mathrm{s}$ : find the required charging [resistance](https://one-course.com/books/physics/2/en/chapter/12-electric-circuits-and-power#def-g11-circuits-and-power-resistance) and the initial charging current — the numbers a designer would order.

**Solution of Problem 30.1.**

**1.** $q = CU_0 = 8.0\times10^{-4} \times 300 = 0.24\,\mathrm{C}$.

**2.** $N = 0.24/1.60\times10^{-19} = 1.5 \times 10^{18}$.

**3.** $E_C = \tfrac12 \times 8.0\times10^{-4} \times 300^2 =
36\,\mathrm{J}$.

**4.** $1.3 \times 10^{4}/36 \approx 360$ times.

**5.** The cell stores plenty but delivers slowly; the [capacitor](#def-g12-rc-rl-circuits-capacitor) turns modest [energy](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-energy) into enormous *[power](https://one-course.com/books/physics/2/en/chapter/17-work-of-a-force#def-g11-work-of-force-power)*.

**6.** $\mathcal{E} = Ri + u$ with $i = C\,du/dt$: $RC\,du/dt + u = \mathcal{E}$.

**7.** $du/dt = (\mathcal{E}/RC)e^{-t/RC}$: $\mathcal{E}e^{-t/RC} + \mathcal{E}(1 - e^{-t/RC}) = \mathcal{E}$ for all $t$, and $u(0) = 0$.

**8.** $\tau = RC = 10^3 \times 8.0\times10^{-4} = 0.80\,\mathrm{s}$.

**9.** At $3\tau = 2.4\,\mathrm{s}$, $u = 0.95 \times 300 =
285\,\mathrm{V}$ — the two-second whine.

**10.** $I_0 = \mathcal{E}/R = 0.30\,\mathrm{A}$; as $u$ climbs, the remainder $\mathcal{E} - u$ across $R$ shrinks, so $i = (\mathcal{E}-u)/R$ dies away.

**11.** $u(t) = U_0\,e^{-t/\tau'}$ with $\tau' = R_f C =
0.50 \times 8.0\times10^{-4} = 0.40\,\mathrm{ms}$.

**12.** $I_0 = U_0/R_f = 300/0.50 = 600\,\mathrm{A}$.

**13.** $P_0 = U_0 I_0 = 300 \times 600 = 180\,\mathrm{kW}$ — about $80$ kettles at once.

**14.** Duration $3\tau' = 1.2\,\mathrm{ms}$; average [power](https://one-course.com/books/physics/2/en/chapter/17-work-of-a-force#def-g11-work-of-force-power) $\approx 36/1.2\times10^{-3} = 30\,\mathrm{kW}$.

**15.** $2.4/1.2\times10^{-3} = 2000$. Only the [resistance](https://one-course.com/books/physics/2/en/chapter/12-electric-circuits-and-power#def-g11-circuits-and-power-resistance) changed ($1.0\,\mathrm{k}\Omega$ in, $0.50\,\Omega$ out): a [power](https://one-course.com/books/physics/2/en/chapter/17-work-of-a-force#def-g11-work-of-force-power) lever.

**16.** $C = 2E_C/U_0^2 = 400/1500^2 \approx 1.8 \times 10^{-4}\,\mathrm{F} =
180\,\text{µ}\mathrm{F}$.

**17.** $q = CU_0 = 1.78\times10^{-4} \times 1500 \approx
0.27\,\mathrm{C}$.

**18.** $\tau = 50 \times 1.78\times10^{-4} \approx 8.9\,\mathrm{ms}$, duration $3\tau \approx 27\,\mathrm{ms}$ — some twenty times the flash.

**19.** $I_0 = 1500/50 = 30\,\mathrm{A}$; $P_0 = 1500 \times 30 = 45\,\mathrm{kW}$.

**20.** $3\tau_c = 6.0\,\mathrm{s}$ gives $\tau_c = 2.0\,\mathrm{s}$, so $R_c = \tau_c/C = 2.0/1.78\times10^{-4} \approx 11\,\mathrm{k}\Omega$ and $I_0 = 1500/1.12\times10^{4} \approx 0.13\,\mathrm{A}$ — ready in six seconds, on a tenth of an [ampere](https://one-course.com/books/physics/2/en/chapter/12-electric-circuits-and-power#def-g11-circuits-and-power-current).