Circuits Part 2: EMF & Capacitors

Circuits Part 2: EMF & Capacitors

Part 1 of 4 — EMF, Internal Resistance & Meters

Where this fits. In the Circuits Part 1: Current & Resistance lesson you learned the rules for steady current: $V = IR$, resistivity, series and parallel resistors, Kirchhoff's junction and loop rules, network reduction, and power. There, every battery was a perfect, fixed voltage. This lesson makes the battery real, then adds the element that stores charge instead of passing it: the capacitor.

What EMF Means

Inside a battery, chemical reactions push positive charge "uphill" from the negative terminal to the positive terminal. The EMF ($\varepsilon$, "electromotive force") is the work the battery does per coulomb moved through it:

$\varepsilon = \frac{W}{q}$

Despite its name, EMF is not a force. It is measured in volts ($1\;\text{V} = 1\;\text{J/C}$), and it is the battery's maximum possible voltage. Think of the battery as a pump: EMF is the pressure the pump can produce when nothing is flowing.

EMF vs. Terminal Voltage

A real battery behaves like an ideal EMF source in series with a small internal resistance $r$ (the resistance of its electrolyte and electrodes). The voltage you can actually measure between its two terminals is the terminal voltage:

$V_{\text{term}} = \varepsilon - Ir$

Because $r$ sits in series with the external load $R$, the same current flows through both:

$I = \frac{\varepsilon}{R + r}$

Read the cases:

• Open circuit ($I = 0$): nothing flows, nothing is lost inside, so $V_{\text{term}} = \varepsilon$. This is why an ideal voltmeter across an unloaded battery reads the EMF.
• Under load: the terminal voltage sags below $\varepsilon$ by $Ir$. More current means more sag. Headlights dim while a car's starter motor cranks because the starter draws hundreds of amps through the battery's internal resistance.
• Short circuit ($R = 0$): the current reaches its maximum, $I_{\max} = \varepsilon / r$, and $V_{\text{term}} = 0$. All of the battery's power is wasted as heat inside it, which is why a shorted battery gets hot.
• Being charged: a charger forces current backward into the positive terminal. Now the internal drop adds instead of subtracts: $V_{\text{term}} = \varepsilon + Ir$. A charger must supply MORE than the battery's EMF.

Energy bookkeeping: the chemistry supplies power $\varepsilon I$; the internal resistance wastes $I^2 r$; the load receives $I^2 R = V_{\text{term}} I$.

An aging battery often still shows its full EMF on a voltmeter, because the test draws almost no current. What has changed is its internal resistance, which rises with age, so the battery sags badly the moment a real load is attached.

Practice — EMF and Terminal Voltage 🎯

Batteries in Series and Parallel

Series (positive terminal of one cell to negative terminal of the next):

• EMFs add: four $1.5\;\text{V}$ cells make $6.0\;\text{V}$.
• Internal resistances add too, so a series stack sags more per amp.
• A cell inserted backward SUBTRACTS its EMF. Four cells with one reversed give $1.5 + 1.5 + 1.5 - 1.5 = 3.0\;\text{V}$, not $4.5\;\text{V}$.

Parallel (identical cells, all positive terminals joined):

• The EMF stays the same as one cell: two $1.5\;\text{V}$ cells in parallel still give $1.5\;\text{V}$.
• The internal resistances are in parallel, so the pack's internal resistance drops ($r/2$ for two cells). The pack sags less under load and, holding more total charge, lasts longer.

Rule of thumb: series for more voltage, parallel for more current capacity at the same voltage.

Capacity: a battery's rating in milliamp-hours (mAh) is a quantity of CHARGE, because $Q = It$. A $2000\;\text{mAh}$ cell can deliver $2.0\;\text{A}$ for one hour: $Q = (2.0\;\text{A})(3600\;\text{s}) = 7200\;\text{C}$. The energy it holds is roughly $QV$: at $3.7\;\text{V}$, about $2.7 \times 10^4\;\text{J}$.

Meters

• Ammeter measures current. It must carry the current it measures, so it goes in series with the element. An ideal ammeter has ~0 Ω resistance, so inserting it does not change the current.
• Voltmeter measures potential difference between two points, so it connects in parallel across the element. An ideal voltmeter has ~∞ Ω resistance, so it draws essentially no current and does not disturb the circuit.

What goes wrong when they are swapped:

• Ammeter wired in parallel across an element: its near-zero resistance is a short circuit. Almost all the current detours through the meter, the element goes dark (nearly $0\;\text{V}$ across it), and the large current can blow the meter's fuse.
• Voltmeter wired in series: its enormous resistance chokes the current to nearly zero. Every other element goes dark, and the voltmeter reads nearly the full EMF, because almost the entire voltage drop now occurs across the meter itself.

Real meters are not ideal. A real voltmeter's finite resistance, in parallel with the element, slightly lowers the combined resistance, so it reads a little low; a real ammeter adds a little series resistance, which also lowers the current it is trying to measure.

• An ohmmeter supplies its own small test current and reads $R = V/I$, so it must be used on a component that is unpowered and disconnected from the rest of the circuit.

Where EMF Comes From

A battery is one or more galvanic cells. The cell's EMF is the cell potential from electrochemistry ($E^\circ_{\text{cell}}$ under standard conditions), set by the redox reaction through $\Delta G = -nFE$ (where $n$ is the number of moles of electrons transferred and $F$ is Faraday's constant, the charge on one mole of electrons), and the battery's positive terminal is the cathode. The Electrochemistry topic covers that chemistry; here we treat the result as a circuit element.

Worked Example — Measuring a Battery's Internal Resistance

A voltmeter connected across a battery reads $9.0\;\text{V}$ with nothing else attached. A $4.0\;\Omega$ resistor is then connected across the terminals, and the voltmeter now reads $8.0\;\text{V}$. Find the EMF, the current, the internal resistance, and where the power goes.

Step 1 — EMF. With no load, $I \approx 0$ (an ideal voltmeter draws no current), so the reading IS the EMF: $\varepsilon = 9.0\;\text{V}$.

Step 2 — Current. With the resistor attached, the terminal voltage sits directly across the $4.0\;\Omega$ resistor:

$I = \frac{V_{\text{term}}}{R} = \frac{8.0}{4.0} = 2.0\;\text{A}$

Step 3 — Internal resistance. The missing $1.0\;\text{V}$ was dropped inside the battery:

$r = \frac{\varepsilon - V_{\text{term}}}{I} = \frac{9.0 - 8.0}{2.0} = 0.50\;\Omega$

Check with the loop: $I = \varepsilon / (R + r) = 9.0 / 4.5 = 2.0\;\text{A}$. ✓

Step 4 — Power accounting.

• Supplied by the chemistry: $\varepsilon I = (9.0)(2.0) = 18\;\text{W}$
• Delivered to the resistor: $I^2 R = (4.0)(4.0) = 16\;\text{W}$
• Wasted inside the battery: $I^2 r = (4.0)(0.50) = 2.0\;\text{W}$

$16 + 2 = 18\;\text{W}$: the books balance.

Step 5 — Two extremes. Shorting the terminals would draw $I_{\max} = \varepsilon / r = 9.0 / 0.50 = 18\;\text{A}$, all of it heating the battery. Recharging at $2.0\;\text{A}$ would need a charger voltage of $\varepsilon + Ir = 9.0 + 1.0 = 10\;\text{V}$.

Pattern: the open-circuit reading gives $\varepsilon$; any loaded reading plus the current gives $r$.

Batteries, Meters & Real Cells 🎯

Key Takeaways — Part 1

• EMF $\varepsilon$ = work per coulomb done by the battery (volts, not a force); it equals the terminal voltage only when no current flows
• Real battery = ideal EMF in series with internal resistance $r$: $I = \varepsilon/(R + r)$
• Discharging: $V_{\text{term}} = \varepsilon - Ir$ (sags more at higher current); charging: $V_{\text{term}} = \varepsilon + Ir$
• Short circuit: $I_{\max} = \varepsilon / r$, $V_{\text{term}} = 0$, all power heats the battery
• Cells in series: EMFs and internal resistances add (a reversed cell subtracts); identical cells in parallel: same EMF, lower internal resistance, longer life
• Ammeter: series, ~0 Ω. Voltmeter: parallel, ~∞ Ω
• Battery EMF comes from a galvanic cell's potential; the positive terminal is the cathode

Common traps

• Treating a battery's labeled voltage as fixed under load: the terminal voltage sags by $Ir$
• Forgetting the sign flip when a battery is being charged ($\varepsilon + Ir$, not $\varepsilon - Ir$)
• Thinking current is "used up": charge circulates; only energy is converted
• Ammeter in parallel = short circuit; voltmeter in series = current nearly stops and the meter reads about the full EMF
• An unloaded voltmeter test reads the EMF even on a worn-out cell; internal resistance only shows under load

• Low-yield extras: an ohmmeter supplies its own test current, so it is used only on an unpowered, isolated component

Circuits Part 2: EMF & Capacitors

Part 2 of 4 — Capacitance, Dielectrics & Stored Energy

Capacitance

A capacitor is two conductors (plates) separated by an insulator. Connect it to a battery and the battery moves charge from one plate to the other through the wires: one plate ends up with $+Q$, the other with $-Q$. No charge ever crosses the gap, and the capacitor's net charge stays zero. "The charge on a capacitor," $Q$, always means the charge on EACH plate.

Charging stops when the voltage across the plates equals the battery's voltage. The charge stored per volt is the capacitance:

$C = \frac{Q}{V}$

The unit is the farad ($1\;\text{F} = 1\;\text{C/V}$). A farad is huge; real capacitors are usually microfarads ($\mu\text{F} = 10^{-6}\;\text{F}$) to picofarads ($\text{pF} = 10^{-12}\;\text{F}$). A handy unit habit: $\mu\text{F} \times \text{V} = \mu\text{C}$.

Trap: $C = Q/V$ does NOT mean a bigger voltage gives a smaller capacitance. Capacitance is fixed by the capacitor's construction. Double the voltage and the charge doubles; $C$ does not change.

The Parallel-Plate Capacitor

For two flat plates of area $A$ separated by distance $d$:

$C = \frac{\kappa \varepsilon_0 A}{d}$

where $\varepsilon_0$ is the permittivity of free space and $\kappa$ is the dielectric constant of the material between the plates ($\kappa = 1$ for vacuum, essentially 1 for air).

• Bigger plates (larger $A$): more room to spread charge at the same voltage, so $C$ rises in proportion.
• Closer plates (smaller $d$): opposite charges attract each other more strongly across the gap, so more charge piles on per volt. $C \propto 1/d$.

The field between the plates is uniform: same strength and direction (from the $+$ plate toward the $-$ plate) everywhere between them, and essentially zero outside:

$E = \frac{V}{d}$

A uniform field means a charge between the plates feels a constant force $F = qE$, so it accelerates uniformly, just like a ball in gravity (the Electrostatics topic covers the motion).

• The permittivity of free space is $\varepsilon_0 \approx 8.85 \times 10^{-12}\;\text{F/m}$; a passage will give it if a calculation needs it.
• The field can also be written from the charge density on the plates: $E = \sigma/\varepsilon_0$, with $\sigma = Q/A$.

Practice — Capacitance and the Parallel Plate 🎯

Dielectrics

A dielectric is an insulating material placed between the plates. Its molecules polarize in the field: their electron clouds shift (and polar molecules rotate) so that a thin layer of negative charge lines up next to the $+$ plate and positive charge next to the $-$ plate. These induced charges partly cancel the plates' field inside the material.

For the same charge $Q$ on the plates, the field and the voltage between the plates drop by the factor $\kappa$, so the capacitance rises by the same factor:

$C = \kappa C_0 \qquad (\kappa \ge 1)$

Every real insulator has $\kappa > 1$. Water's is unusually large, for the same reason water dissolves salts: its polar molecules screen charges from each other.

Dielectric breakdown: every insulator has a maximum field it can withstand (its dielectric strength). Above it, the material ionizes and suddenly conducts, and a spark jumps the gap. Breakdown sets a capacitor's maximum voltage. A good dielectric therefore does two jobs: it raises $C$, and it lets the plates sit closer together (or reach a higher voltage) without breaking down.

• Typical dielectric constants: air about 1.0006, paper about 3, glass about 5 to 10, water about 80. Air breaks down near $3 \times 10^6\;\text{V/m}$.

Stored Energy

Charging a capacitor takes work, because each new bit of charge is pushed onto a plate that already repels it. That work is stored in the electric field between the plates:

$U = \frac{1}{2}QV = \frac{1}{2}CV^2 = \frac{Q^2}{2C}$

Why the $\frac{1}{2}$? The voltage rises steadily from $0$ to $V$ as the charge builds, so on average each coulomb is loaded at $V/2$. (On a graph of $V$ against $Q$, the energy is the triangle's area, $\frac{1}{2}QV$.)

Choose the form for what is held constant:

• Battery connected → $V$ fixed → use $U = \frac{1}{2}CV^2$, so $U \propto C$.
• Battery disconnected (isolated) → $Q$ fixed → use $U = Q^2/2C$, so $U \propto 1/C$.

The same logic works for moving the plates. Pull the plates of an ISOLATED capacitor apart and $C$ falls while $Q$ stays put, so $V = Q/C$ rises and $U = Q^2/2C$ rises: you did work pulling the attracting plates apart. Do the same with the battery connected and $V$ stays put, so $Q$ and $U$ both fall.

Where does the battery's energy go? A battery charging a capacitor through a resistor delivers $QV$ of energy but the capacitor stores only $\frac{1}{2}QV$. The other half is lost as heat in the resistance, whatever its size.

• Energy density of the field: $u = \frac{1}{2}\varepsilon_0 E^2$ (energy per volume), multiplied by $\kappa$ inside a dielectric.

Worked Example — Dielectric with the Battery Disconnected

A $2.0\;\mu\text{F}$ air-filled capacitor is charged to $100\;\text{V}$ and then disconnected from the battery. A slab with $\kappa = 4$ is slid in to fill the gap. Find the new charge, capacitance, voltage and stored energy.

Step 1 — What is fixed? Disconnected, so the charge has nowhere to go: $Q$ is fixed.

$Q = CV = (2.0\;\mu\text{F})(100\;\text{V}) = 200\;\mu\text{C}$

Step 2 — New capacitance. $C_{\text{new}} = \kappa C = 4(2.0\;\mu\text{F}) = 8.0\;\mu\text{F}$.

Step 3 — New voltage.

$V_{\text{new}} = \frac{Q}{C_{\text{new}}} = \frac{200\;\mu\text{C}}{8.0\;\mu\text{F}} = 25\;\text{V}$

The voltage fell by the factor $\kappa$, and so did the field between the plates.

Step 4 — Energy before and after.

$U_{\text{before}} = \frac{1}{2}CV^2 = \frac{1}{2}(2.0 \times 10^{-6})(100)^2 = 1.0 \times 10^{-2}\;\text{J}$

$U_{\text{after}} = \frac{1}{2}C_{\text{new}}V_{\text{new}}^2 = \frac{1}{2}(8.0 \times 10^{-6})(25)^2 = 2.5 \times 10^{-3}\;\text{J}$

Energy fell by the factor $\kappa = 4$. The ratio shortcut gets there in one line: $U = Q^2/2C$ with $Q$ fixed and $C \times 4$, so $U \div 4$. The missing energy went into the work the field did pulling the slab in.

Step 5 — Contrast: battery still connected. Now $V$ stays $100\;\text{V}$, so $Q = (8.0\;\mu\text{F})(100\;\text{V}) = 800\;\mu\text{C}$ and $U = \frac{1}{2}(8.0 \times 10^{-6})(100)^2 = 4.0 \times 10^{-2}\;\text{J}$, four times the original. Same slab, opposite energy change: everything hinges on whether $Q$ or $V$ is held constant.

Dielectrics & Stored Energy 🎯

Key Takeaways — Part 2

• $C = Q/V$ (farads); $Q$ is the charge on each plate; $C$ is set by construction, not by $Q$ or $V$
• Parallel plates: $C = \kappa \varepsilon_0 A / d$ (bigger plates, smaller gap, better dielectric all raise $C$)
• Uniform field between the plates: $E = V/d$, directed from $+$ to $-$
• Dielectrics polarize, partly cancel the field and multiply $C$ by $\kappa$; breakdown sets the maximum field and voltage
• $U = \frac{1}{2}QV = \frac{1}{2}CV^2 = Q^2/2C$; use $\frac{1}{2}CV^2$ when $V$ is fixed, $Q^2/2C$ when $Q$ is fixed
• Battery connected: $V$ fixed; inserting a dielectric raises $Q$ and $U$ by $\kappa$. Disconnected: $Q$ fixed; $V$, $E$ and $U$ fall by $\kappa$
• Charging through a resistor: the battery supplies $QV$, the capacitor stores half, and the other half becomes heat

Common traps

• Reading $C = Q/V$ as "more voltage means less capacitance"
• Forgetting to convert millimeters, microfarads or microcoulombs before computing
• Using $\frac{1}{2}CV^2$ for an isolated capacitor whose voltage is changing: pick the form whose variables stay fixed
• Ranking stored energy by charge or capacitance alone: energy scales with $V^2$
• Assuming a dielectric always increases stored energy (it lowers it when $Q$ is fixed)

• Low-yield extras: the value of $\varepsilon_0$ ($8.85 \times 10^{-12}\;\text{F/m}$); $E = \sigma/\varepsilon_0$; tabulated dielectric constants and air's breakdown field; field energy density $u = \frac{1}{2}\varepsilon_0 E^2$

Circuits Part 2: EMF & Capacitors

Part 3 of 4 — Capacitor Networks & RC Circuits

Capacitor Networks

Capacitors combine by rules that are the exact REVERSE of the resistor rules from the Circuits Part 1: Current & Resistance lesson.

Why parallel capacitors add. Side by side, every capacitor sees the same voltage, and the charges simply pile up: $Q = C_1V + C_2V$, so $C_{\text{eq}} = C_1 + C_2$. Two plates side by side act like one bigger plate.

Why series capacitors share charge. In a series chain, the inner plates of neighboring capacitors are joined to each other and to nothing else, so that isolated piece of metal stays neutral overall. If $+Q$ appears on one inner plate, $-Q$ must appear on its partner. Every capacitor in series carries the same $Q$, and the voltages add: $V = Q/C_1 + Q/C_2$, which gives the reciprocal rule. Stacking gaps in series acts like one wider gap, so capacitance falls.

Shortcuts (the same algebra as resistors, applied to the opposite arrangement):

• Two in series: $C_{\text{eq}} = \dfrac{C_1 C_2}{C_1 + C_2}$ ("product over sum")
• $N$ identical capacitors: $C/N$ in series, $NC$ in parallel

Worked Example — Series Capacitors Share Charge, Not Voltage

A $3.0\;\mu\text{F}$ and a $6.0\;\mu\text{F}$ capacitor are in series across a $12\;\text{V}$ battery.

Step 1 — Equivalent capacitance. $C_{\text{eq}} = \frac{(3.0)(6.0)}{3.0 + 6.0} = 2.0\;\mu\text{F}$, smaller than either one, as series capacitance must be.

Step 2 — Charge, the same on each. $Q = C_{\text{eq}}V = (2.0\;\mu\text{F})(12\;\text{V}) = 24\;\mu\text{C}$.

Step 3 — Individual voltages. $V_3 = 24/3.0 = 8.0\;\text{V}$ and $V_6 = 24/6.0 = 4.0\;\text{V}$. Check: $8.0 + 4.0 = 12\;\text{V}$. ✓

The SMALLER capacitor takes the LARGER share of the voltage, the mirror image of series resistors (where the larger resistor takes more).

Practice — Capacitor Networks 🎯

Capacitors in DC Circuits

A capacitor in a DC circuit has two simple snapshots, and MCAT questions almost always ask about one of them.

At $t = 0$ (switch just closed, capacitor uncharged): an empty capacitor has $V_C = Q/C = 0$, so it offers no opposition to current. It acts like a plain wire. The initial current is set by the resistors alone; for a single resistor in series with the capacitor, $I_0 = \varepsilon/R$.

At steady state (long after, fully charged): the capacitor's voltage has risen until it balances whatever drives its branch, so no current flows through it. It acts like an open branch (a cut wire).

The bridge between them: a capacitor's voltage cannot change instantly, because charge takes time to flow onto the plates. Current, by contrast, can jump the moment a switch moves.

Steady-state recipe:

1. Delete every capacitor branch (treat it as open).
2. Solve the remaining resistor network with the tools from the Circuits Part 1: Current & Resistance lesson.
3. Find $V_C$ as the potential difference between the two points the capacitor connects. A resistor in series with the capacitor carries no current, so it drops no voltage and can be ignored.
4. Then $Q = CV_C$ and $U = \frac{1}{2}CV_C^2$.

The Time Constant

How fast does a capacitor charge through a resistor? The scale is the time constant:

$\tau = RC$

(ohms times farads gives seconds).

Charging (from a battery through $R$): the charge and capacitor voltage rise quickly at first, then more and more slowly, leveling off at $Q_{\max} = C\varepsilon$. Meanwhile, the current starts at its maximum, $\varepsilon/R$, and dies away toward zero. The slowdown happens because the growing capacitor voltage opposes the battery, leaving less voltage to push current through $R$.

Discharging (the charged capacitor connected across a resistor): charge, voltage and current all start at their maximum and decay toward zero, fast at first and then ever more slowly. The discharge current flows in the OPPOSITE direction to the charging current, because charge now leaves the plates instead of arriving.

Intuition for $\tau = RC$: a bigger $R$ lets less current flow to deliver the charge; a bigger $C$ needs more charge to reach the same voltage. Either one slows things down. The battery's voltage does NOT change $\tau$; it only changes the final charge.

The exact curves (a passage will give these; recognize their shape): charging $Q(t) = C\varepsilon(1 - e^{-t/RC})$, which rises and levels off; discharging $Q(t) = Q_0 e^{-t/RC}$, which decays toward zero. Setting $t = RC$ gives $1 - e^{-1} \approx 0.63$ and $e^{-1} \approx 0.37$, the numbers in the table.

• The time to reach half of the final charge (or to lose half while discharging) is about $0.69\tau$.

Worked Example — Steady-State Voltage Across a Capacitor

Build the circuit piece by piece:

1. An ideal $12\;\text{V}$ battery and a switch.
2. In series with the battery: $R_1 = 2.0\;\Omega$, then $R_2 = 4.0\;\Omega$. This makes one main loop.
3. A side branch connected across $R_2$ (its two ends attach to the same two points as $R_2$'s ends), so the branch is in PARALLEL with $R_2$.
4. Inside that branch, in series with each other: $R_3 = 4.0\;\Omega$ and an uncharged $5.0\;\mu\text{F}$ capacitor.

Find the battery current just after the switch closes, and the capacitor's charge and energy long afterward.

Part A — The instant the switch closes. The empty capacitor acts like a wire, so the branch is just $R_3$, in parallel with $R_2$. Call that combination $R_{2\|3}$ ($R_2$ in parallel with $R_3$):

$R_{2\|3} = \frac{(4.0)(4.0)}{4.0 + 4.0} = 2.0\;\Omega, \qquad R_{\text{total}} = 2.0 + 2.0 = 4.0\;\Omega$

$I_0 = \frac{12}{4.0} = 3.0\;\text{A}$

Because $R_2 = R_3$, the current splits equally: half of it ($1.5\;\text{A}$) rushes into the capacitor branch, and half goes through $R_2$.

Part B — Steady state.

1. Open the capacitor branch. Only $R_1$ and $R_2$ in series remain: $I = 12/(2.0 + 4.0) = 2.0\;\text{A}$.
2. Voltage across $R_2$: $V_2 = (2.0)(4.0) = 8.0\;\text{V}$.
3. The capacitor branch connects the same two points as $R_2$. No current flows in that branch, so $R_3$ drops nothing, and the capacitor sits at the full $8.0\;\text{V}$.
4. $Q = CV_C = (5.0\;\mu\text{F})(8.0\;\text{V}) = 40\;\mu\text{C}$, and $U = \frac{1}{2}(5.0 \times 10^{-6})(8.0)^2 = 1.6 \times 10^{-4}\;\text{J}$.

Takeaways: the battery current falls from $3.0\;\text{A}$ to $2.0\;\text{A}$ as the capacitor fills. The series resistor $R_3$ shapes HOW FAST the capacitor charges, not the final voltage it reaches. A common error is to subtract a drop across $R_3$ at steady state; with zero current, there is none.

DC Capacitor Circuits & the Time Constant 🎯

Key Takeaways — Part 3

• Capacitor rules are resistor rules reversed: parallel capacitances add; series combine by reciprocals
• Series capacitors share the same $Q$ (smaller $C$ takes the larger voltage); parallel capacitors share the same $V$ (larger $C$ holds more charge)
• Uncharged capacitor at $t = 0$ = wire; fully charged capacitor = open branch
• Steady state: delete capacitor branches, solve the resistor network, read $V_C$ between the capacitor's two nodes (series resistors in its branch drop nothing), then $Q = CV_C$
• $\tau = RC$: about 63% charged (or 37% left when discharging) after $1\tau$; essentially complete after about $5\tau$
• Charging: charge and capacitor voltage rise and level off; current starts at $\varepsilon/R$ and decays. Discharging: everything decays, and the current reverses direction
• Exact forms (a passage supplies them): $Q = C\varepsilon(1 - e^{-t/RC})$ charging, $Q = Q_0e^{-t/RC}$ discharging; the battery's voltage changes the final charge, not $\tau$

Common traps

• Applying resistor formulas to capacitors (or vice versa)
• Giving the larger series capacitor the larger voltage
• Subtracting a voltage drop across a resistor that is in series with a fully charged capacitor (no current, no drop)
• Treating "one time constant" as fully charged: it is only about 63%
• Thinking a capacitor's voltage can jump when a switch closes (its current can; its voltage cannot)

• Low-yield extras: the half-time of about $0.69\tau$

Circuits Part 2: EMF & Capacitors

Part 4 of 4 — Bioelectricity & Passage Strategy

The MCAT loves to hide capacitors inside biology. Every idea in this part is physics you already have (batteries, resistors, capacitors, $\tau = RC$) wearing a neuron or a heart.

The Cell Membrane Is an RC Circuit

Map each piece of a neuron onto a circuit element:

The field inside the membrane is enormous: about $70\;\text{mV}$ across a bilayer about $7\;\text{nm}$ thick gives $E = V/d = (7 \times 10^{-2}\;\text{V})/(7 \times 10^{-9}\;\text{m}) = 1 \times 10^7\;\text{V/m}$, the parallel-plate formula from Part 2 of this lesson.

Consequences the MCAT tests:

• The resting potential is stored charge. About $-70\;\text{mV}$ across the membrane means a thin layer of excess negative charge on the inner face and positive charge on the outer face. Because $C_m$ is small, $Q = C_mV$ is a tiny amount of charge: far fewer ions than the cell contains, so firing does not measurably change bulk ion concentrations.
• The membrane voltage cannot jump. Like any capacitor, the membrane must charge or discharge through its resistance, with time constant $\tau = R_mC_m$.
• More open channels = more parallel resistors = lower $R_m$ = shorter $\tau$. The membrane responds faster, but each input fades faster. Closing channels raises $R_m$ and lengthens $\tau$.
• A longer $\tau$ favors temporal summation: each postsynaptic potential decays more slowly, so inputs arriving close together can add up to reach threshold.

• Biological membranes have a nearly universal specific capacitance of about $1\;\mu\text{F/cm}^2$, so a cell's total $C_m$ scales with its surface area.
• How far a passive voltage change spreads along an axon is described by the length constant $\lambda$, which grows with membrane resistance and shrinks with internal (axial) resistance.

Myelin and Conduction Speed

Myelin is many tight wraps of glial cell membrane (Schwann cells in the peripheral nervous system, oligodendrocytes in the central nervous system) around the axon, interrupted at the nodes of Ranvier. Electrically, those wraps are many membranes stacked in SERIES:

• Capacitors in series → lower capacitance. Stacking layers acts like widening the gap $d$, so $C_m$ falls. Less charge is needed to change the voltage of each stretch of axon, so the depolarization travels farther and faster.
• Resistors in series → higher resistance. $R_m$ rises, so less current leaks out across the membrane and more flows down the axon to the next node.

Result: the action potential is regenerated only at the nodes, where voltage-gated $\text{Na}^+$ channels are concentrated, and seems to jump from node to node: saltatory conduction. It is fast and energy-efficient, because ions cross the membrane only at the nodes, leaving less for the $\text{Na}^+/\text{K}^+$ pump to restore.

The other speed lever is axon diameter: a wider axon has lower internal resistance, so current flows down it more easily. In demyelinating diseases such as multiple sclerosis, internodes lose myelin, $C_m$ rises and $R_m$ falls, current leaks away, and conduction slows or fails. The Nervous System lesson covers the channel sequence of the action potential itself.

Practice — The Membrane as a Circuit 🎯

Defibrillators and Capacitor Discharge

In ventricular fibrillation, heart muscle cells fire chaotically and the heart stops pumping. A defibrillator delivers a brief, strong current that depolarizes most of the myocardium at once, which halts the chaos and lets the SA node take back control.

Why use a capacitor? A small battery has only a few volts of EMF, and its internal resistance caps its current at $\varepsilon/r$ (Part 1 of this lesson), so it cannot drive tens of amps through the chest directly; it CAN slowly fill a capacitor to a high voltage:

1. Charge slowly: the device's battery and circuitry charge a capacitor to a high voltage over several seconds, at modest power.
2. Store: the capacitor holds $U = \frac{1}{2}CV^2$. The energy setting the operator chooses is reached by charging to the matching voltage, since $C$ is fixed.
3. Discharge fast: paddles connect the capacitor across the chest, which acts as a resistor $R$. The discharge time scale is $\tau = RC$, a few milliseconds.

Same energy in, same energy out, but delivered hundreds of times faster (seconds to charge, milliseconds to discharge), so the delivery power is hundreds of times higher. Power is energy per time. The capacitor is not an energy source; it is a time compressor. A camera flash works the same way.

• Typical external defibrillator shocks deliver roughly 120 to 360 J, and the chest's resistance is roughly 50 to 100 Ω; a passage will supply any numbers it needs.

Passage Strategy — What Is Held Constant?

Capacitor passages reduce to one question: what stays fixed?

Ratio reasoning beats plugging in. Write the formula, mark what is constant, and scale. Doubling $V$ on a fixed capacitor: $U \propto V^2$, so $\times 4$. Inserting a dielectric of $\kappa = 3$ into an isolated capacitor: $U = Q^2/2C$, so $\div 3$.

Unit checks catch most slips: $\mu\text{F} \times \text{V} = \mu\text{C}$; $\Omega \times \text{F} = \text{s}$; $\text{F} \times \text{V}^2 = \text{J}$; $\text{J}/\text{s} = \text{W}$.

Translate the biology: membrane = capacitor, channels = parallel resistors, myelin = series layers (lower $C$, higher $R$), firing threshold or pacing interval = RC timing, defibrillator = slow charge and fast discharge.

Worked Example — Defibrillator Energy and Power

A defibrillator's $100\;\mu\text{F}$ capacitor is charged to $2000\;\text{V}$ over $5.0\;\text{s}$, then discharged through a patient's chest, which acts as a $50\;\Omega$ resistor. Find the stored charge and energy, the charging power, the discharge time scale, and the delivery power.

Step 1 — Charge and energy.

$Q = CV = (1.0 \times 10^{-4}\;\text{F})(2.0 \times 10^{3}\;\text{V}) = 0.20\;\text{C}$

$U = \frac{1}{2}CV^2 = \frac{1}{2}(1.0 \times 10^{-4})(4.0 \times 10^{6}) = 200\;\text{J}$

Step 2 — Average charging power. $P_{\text{charge}} = 200\;\text{J} / 5.0\;\text{s} = 40\;\text{W}$, about as much as a household light bulb.

Step 3 — Discharge time scale.

$\tau = RC = (50)(1.0 \times 10^{-4}) = 5.0 \times 10^{-3}\;\text{s} = 5.0\;\text{ms}$

So the capacitor is essentially empty after about $5\tau = 25\;\text{ms}$, and most of the energy leaves in the first several milliseconds.

Step 4 — Peak current and power. At the first instant the full $2000\;\text{V}$ is across the chest: $I_0 = V/R = 2000/50 = 40\;\text{A}$, and the peak power is $V^2/R = (4.0 \times 10^{6})/50 = 8.0 \times 10^{4}\;\text{W}$.

Step 5 — Average delivery power. Why "about $10\;\text{ms}$"? Stored energy goes as $Q^2$, so it drains twice as fast as the charge: after $2\tau = 10\;\text{ms}$ the charge is down to $e^{-2} \approx 0.14$ of its start, and the energy to $e^{-4} \approx 0.02$, so about 98% has been delivered. Treating the $200\;\text{J}$ as delivered over about $10\;\text{ms}$:

$P_{\text{deliver}} \approx \frac{200\;\text{J}}{1.0 \times 10^{-2}\;\text{s}} = 2.0 \times 10^{4}\;\text{W}$

Step 6 — The ratio. $P_{\text{deliver}}/P_{\text{charge}} = (2.0 \times 10^{4})/40 = 500$, exactly the ratio of the times ($5.0\;\text{s}$ versus $10\;\text{ms}$). Same energy, five hundred times the power. To double the energy setting, the device would raise the voltage by $\sqrt{2}$, not by 2.

Defibrillators & Passage Reasoning 🎯

Key Takeaways — Part 4

• Membrane = RC circuit: bilayer = capacitor, channels = parallel resistors, concentration gradients = batteries; $\tau = R_mC_m$
• The resting potential is a tiny separated charge ($Q = C_mV$); the membrane voltage cannot change instantly
• Opening channels lowers $R_m$ and shortens $\tau$; a longer $\tau$ favors temporal summation
• Myelin = membranes in series: lower $C_m$ and higher $R_m$, so current travels farther and faster; action potentials regenerate only at nodes of Ranvier (saltatory conduction). Wider axons also conduct faster; demyelination (multiple sclerosis) slows or blocks conduction
• Defibrillator: slow charge, store $\frac{1}{2}CV^2$, fast discharge with $\tau = RC$; same energy, far more power
• Passage strategy: decide what is held constant ($V$, $Q$, $C$ or $U$), then scale with ratios and check units

Common traps

• Saying myelin RAISES capacitance (it lowers it, like widening the gap)
• Forgetting that adding channels adds resistors in PARALLEL, which lowers resistance
• Equating equal energy with equal power
• Scaling voltage linearly with energy ($U \propto V^2$, so double the energy needs $\sqrt{2}$ times the voltage)

• Low-yield extras: specific membrane capacitance of about $1\;\mu\text{F/cm}^2$; the length constant $\lambda$; typical defibrillator energies (120 to 360 J) and chest resistance (50 to 100 Ω)

Lesson Recap

• Real batteries: EMF minus the internal drop, $V_{\text{term}} = \varepsilon - Ir$; ammeters go in series (0 Ω), voltmeters in parallel (∞ Ω).
• Capacitors: $C = Q/V = \kappa\varepsilon_0A/d$, $U = \frac{1}{2}CV^2$; series and parallel rules are the reverse of resistors; empty = wire, full = open branch, timescale $\tau = RC$.
• Network rules: whenever a capacitor problem reduces to a resistor network (steady state, $t = 0$), the series/parallel, Kirchhoff and power tools come from the Circuits Part 1: Current & Resistance lesson. Review them there.

Next up: if any network step in this lesson felt shaky, revisit the Circuits Part 1: Current & Resistance lesson before the exit quiz.