High School Physics · Grades 10–12

30RC and RL Circuits

Press a camera’s shutter halfway: a thin 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 and coil — 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 insulator. Wired to a source, the plates take opposite charges $+q$ and $-q$ — an electric field fills the gap (Chapter 14) — and the 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 is enormous — one coulomb parked per volt. Ceramics of radio circuits: $\mathrm{pF}$ to $\mathrm{nF}$ ; electrolytics of power supplies and flashes: $\text{µ}\mathrm{F}$ to $\mathrm{mF}$ ; supercapacitors: thousands of farads.

Proposition 30.3 (Current into a capacitor)

The current arriving on a 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).

Proof. Current is charge delivered per second (Definition 12.1): 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 passes no steady current. And $u$ can never jump — a step would demand infinite current: the capacitor’s voltage is the circuit’s memory.

30.2 Charging a capacitor: the RC circuit

One loop — source $\mathcal{E}$ (Definition 12.18), switch, resistor, empty capacitor: close $K$ at $t = 0$ .

The RC charging circuit: closing K lets the source push charge through R onto the plates of C. — 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 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 along the loop add (Proposition 12.4): $\mathcal{E} = u_R + u = Ri + u$ at every instant; substitute $i = C\,du/dt$ (Proposition 30.3). 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 is $\tau = RC$ . Units: $\Omega \times \mathrm{F} = (\mathrm{V}/\mathrm{A})(\mathrm{C}/\mathrm{V}) = \mathrm{C}/\mathrm{A} = \mathrm{s}$ — the circuit’s own tempo.

Method 30.7 (Reading $\tau$ on the curve)

Three equivalent readings, straight off a graph:

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

Proposition 30.8 (Discharge)

A 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. — 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 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 of wire: its current threads its own turns with a magnetic field (Chapter 15); whenever it changes, a voltage appears across the coil:

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 — induction — is derived in the university volumes; here we take the law and its character: the coil opposes changes of its current. Mirror twins: a capacitor’s $u$ cannot jump, a coil’s $i$ cannot jump.

The RL circuit: after K closes, the coil lets the current climb only at its own pace = L/R. — The RL circuit: after $K$ closes, the coil 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 $L$ 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 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 transfers. ∎

Example 30.13 (An electromagnet wakes up)

A relay coil, $L = 0.30\,\mathrm{H}$ 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 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 — a spark jumps the contact. A car’s ignition coil 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 $C$ charged to voltage $u$ and a coil $L$ 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 $u$ costs $u\,dq$ (Definition 12.3); 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 (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. — Each slice $dq$ costs $u\,dq$ ; the whole charge costs the triangle’s area $E_C = \tfrac12 q_0 u_0$ — the early coulombs board at low voltage.

Example 30.16 (Energy orders of magnitude)

The flash capacitor of Example 30.9: $\tfrac12 \times 8.0\times10^{-4} \times 300^2 = 36\,\mathrm{J}$ . 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 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 the camera flash and the defibrillator. The delay $\tau = RC$ runs timing circuits: intermittent wipers, blinking beacons, stairwell lights. A capacitor across a supply smooths voltage ( $u$ cannot jump); a series coil smooths current ( $i$ cannot jump). And coil plus capacitor together make an oscillator (Chapter 31).

30.5 Exercises

Exercise 30.1 ★

A $470\,\text{µ}\mathrm{F}$ capacitor is charged to $12\,\mathrm{V}$ : what charge does it hold? What voltage would put $2.0\,\mathrm{mC}$ on $100\,\text{µ}\mathrm{F}$ ?

Solution

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 at $2.7\,\mathrm{V}$ .

Solution

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 across a $220\,\text{µ}\mathrm{F}$ capacitor rises steadily from $0$ to $5.0\,\mathrm{V}$ in $10\,\mathrm{ms}$ : what current flows in? What current once the voltage holds still, and why?

Solution

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 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 by unit, that both combinations are seconds.

Solution

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 charges toward $6.0\,\mathrm{V}$ , passing $3.8\,\mathrm{V}$ at $t = 2.0\,\mathrm{s}$ . Read off the time constant; when is the charge $95\%$ done?

Solution

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: $\mathcal{E} = 9.0\,\mathrm{V}$ , $R = 4.7\,\mathrm{k}\Omega$ , $C = 220\,\text{µ}\mathrm{F}$ . Compute $\tau$ , $u(\tau)$ , $u(3\tau)$ , and the current at the instant the switch closes.

Solution

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: 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

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 starts empty; $u \to \mathcal{E}$ : fully charged, the current has stopped.

Exercise 30.8 ★★

A $100\,\text{µ}\mathrm{F}$ 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

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 climbs steadily from $0$ to $2.0\,\mathrm{A}$ in $40\,\mathrm{ms}$ : what voltage appears across it? With $u = L\,di/dt$ , why does opening a switch on a coil spark, not closing?

Solution

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 $i$ to zero almost instantly, $di/dt$ is huge and $u$ reaches sparking values.

Exercise 30.10 ★★

The relay of Example 30.13: 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

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

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, computing each: a $3000\,\mathrm{F}$ supercapacitor at $2.7\,\mathrm{V}$ ; a flash capacitor, $800\,\text{µ}\mathrm{F}$ at $300\,\mathrm{V}$ ; a $10\,\mathrm{mH}$ coil 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

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 $\tfrac12 \times 0.010 \times 10^2 = 0.50\,\mathrm{J}$ . Battery ( $1.3 \times 10^{4}\,\mathrm{J}$ ) $>$ supercapacitor $\gg$ flash $\gg$ coil — but all three can dump their energy in milliseconds: power, not energy.

Exercise 30.13 ★★★

A stairwell timer lights while $u < 0.63\,\mathcal{E}$ across its charging 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 is $\tau \ln 2$ and compute it.

Solution

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 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 droop, then the capacitance keeping it under $0.50\,\mathrm{V}$ .

Solution

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, $L = 0.80\,\mathrm{H}$ , carries $2.0\,\mathrm{A}$ when the breaker cuts it off in about $1.0\,\mathrm{ms}$ . Estimate the coil voltage during the cut and the spark’s energy; why does a slow interruption give no spark?

Solution

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: 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 stores energy in a capacitor $C = 800\,\text{µ}\mathrm{F}$ , charged through $R = 1.0\,\mathrm{k}\Omega$ from a converter modeled as an ideal source $\mathcal{E} = 300\,\mathrm{V}$ ; the “ready” light trips at $95\%$ of full voltage. Fired, the 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.

Compute the charge on the capacitor at $300\,\mathrm{V}$ .
How many elementary charges is that?
Compute the stored energy $E_C$ .
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’s energy is that?
So why bother with the capacitor? One sentence, using “power”.

Part II — The wait for green.

Write the loop law and turn it into the differential equation for $u(t)$ .
Verify by substitution that $u(t) = \mathcal{E}(1 - e^{-t/RC})$ solves it, starting from $u(0) = 0$ .
Compute the time constant $\tau$ .
After what time does the ready light turn green, and at what voltage? Recognize the two-second whine.
Compute the initial charging current; why does it then die away?

Part III — The millisecond of glory.

Write the discharge law $u(t)$ through the tube and compute the new time constant $\tau'$ .
Compute the peak current through the tube.
Compute the peak power; compare it with a $2.2\,\mathrm{kW}$ kettle.
Taking the flash as over at $3\tau'$ , give its duration and average power.
Form the ratio of charging time to flash time: same capacitor, same charge — what changed, and what does that make a capacitor: an energy tank, or a 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 $50\,\Omega$ .

Compute the required capacitance.
Compute the stored charge.
Compute the shock’s time constant and duration ( $3\tau$ ); compare with the flash.
Compute the peak current and peak power in the patient.
The recharge circuit must make the device ready ( $95\%$ ) in $6.0\,\mathrm{s}$ : find the required charging resistance and the initial charging current — the numbers a designer would order.

Solution

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 turns modest energy into enormous 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 $\approx 36/1.2\times10^{-3} = 30\,\mathrm{kW}$ .

15. $2.4/1.2\times10^{-3} = 2000$ . Only the resistance changed ( $1.0\,\mathrm{k}\Omega$ in, $0.50\,\Omega$ out): a 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.