Circuits Part 4: Capacitor Networks & Bioelectricity

Circuits Part 4: Capacitor Networks & Bioelectricity

Part 1 of 2 — Capacitor Networks & RC Circuits

Where this fits. The Circuits Part 3: EMF & Capacitance lesson introduced the single capacitor: $C = Q/V$, dielectrics and stored energy. This final Circuits lesson wires capacitors into networks and RC circuits, then reads the cell membrane, myelin and the defibrillator as circuits and closes with passage strategy. Next comes the Magnetism & Electromagnetic Induction lesson.

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. The inner plates of neighboring capacitors are joined to each other and to nothing else, so that isolated metal stays neutral: $+Q$ on one inner plate forces $-Q$ on its partner. Every series capacitor carries the same $Q$, the voltages add ($V = Q/C_1 + Q/C_2$), and the reciprocal rule follows. Stacked gaps act like one wider gap, so capacitance falls.

• Shortcut, two in series: $C_{\text{eq}} = \dfrac{C_1 C_2}{C_1 + C_2}$ ("product over sum", the resistor algebra applied to the opposite arrangement)
• Shortcut, $N$ identical capacitors: $C/N$ in series, $NC$ in parallel

Worked Example — Series Capacitors Share Charge, Not Voltage

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

• (a) What is the equivalent capacitance?
• (b) How much charge sits on each capacitor?
• (c) What is the voltage across each capacitor?

(a) 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.

(b) Charge, the same on each. With $C_{\text{eq}} = 2.0\;\mu\text{F}$ from (a): $Q = C_{\text{eq}}V = (2.0\;\mu\text{F})(12\;\text{V}) = 24\;\mu\text{C}$ on each capacitor, because series capacitors share one charge.

(c) Individual voltages. With $Q = 24\;\mu\text{C}$ from (b): $V_3 = 24/3.0 = 8.0\;\text{V}$ 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 2: Kirchhoff's Rules & Power 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

Problem. An ideal $12\;\text{V}$ battery, a switch, $R_1 = 2.0\;\Omega$ and $R_2 = 4.0\;\Omega$ form one series loop. A side branch across $R_2$ (so in PARALLEL with it) holds $R_3 = 4.0\;\Omega$ in series with an uncharged $5.0\;\mu\text{F}$ capacitor.

• (a) Just after the switch closes, what current leaves the battery?
• (b) At that instant, how does the current split between $R_2$ and the capacitor branch?
• (c) Long afterward (steady state), what current leaves the battery?
• (d) At steady state, what is the voltage across the capacitor?
• (e) What are the capacitor's final charge and stored energy?

(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 $R_3$):

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

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

(b) The split. With $I_0 = 3.0\;\text{A}$ from (a): because $R_2 = R_3$, the current splits equally. Half of it ($1.5\;\text{A}$) rushes into the capacitor branch, and half ($1.5\;\text{A}$) goes through $R_2$.

(c) Steady-state current. Open the capacitor branch. Only $R_1$ and $R_2$ in series remain: $I = 12/(2.0 + 4.0) = 2.0\;\text{A}$. The battery current has fallen from $3.0\;\text{A}$ to $2.0\;\text{A}$ as the capacitor filled.

(d) Capacitor voltage. With $I = 2.0\;\text{A}$ from (c), the voltage across $R_2$ is $V_2 = (2.0)(4.0) = 8.0\;\text{V}$. 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 $8.0\;\text{V}$. A common error is to subtract a drop across $R_3$; with zero current, there is none.

(e) Charge and energy. With $V_C = 8.0\;\text{V}$ from (d): $Q = CV_C = (5.0\;\mu\text{F})(8.0\;\text{V}) = 40\;\mu\text{C}$, and $U = \frac{1}{2}(5.0 \times 10^{-6})(8.0)^2 = 1.6 \times 10^{-4}\;\text{J}$. The series resistor $R_3$ shapes HOW FAST the capacitor charges, not the final voltage it reaches.

DC Capacitor Circuits & the Time Constant 🎯

Key Takeaways — Part 1

• Resistor rules reversed. Parallel capacitances add; series capacitances combine by reciprocals.
• Shared Q or shared V. Series capacitors share $Q$ (smaller $C$ takes more voltage); parallel ones share $V$.
• Wire, then open branch. An empty capacitor is a wire at $t = 0$; a full one is an open branch, so read $V_C$ across its nodes.
• The time constant. $\tau = RC$: about 63% charged (37% left when discharging) after $1\tau$, essentially done by $5\tau$.

Common traps

• Resistor formulas on capacitors. Capacitors combine the opposite way: series by reciprocals, parallel by adding.
• Larger C, larger voltage. In series the SMALLER capacitor takes the larger voltage, the reverse of resistors.
• Steady-state drop across R. A resistor in series with a full capacitor carries no current, so it drops no voltage.
• One time constant means full. After $1\tau$ the capacitor is only about 63% charged; its voltage also cannot jump.

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

Circuits Part 4: Capacitor Networks & Bioelectricity

Part 2 of 2 — 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 the Circuits Part 3: EMF & Capacitance 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 $\varepsilon/r$ (the Circuits Part 3: EMF & Capacitance 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

Problem 1. A defibrillator's $100\;\mu\text{F}$ capacitor is charged to $2000\;\text{V}$ over $5.0\;\text{s}$.

• (a) How much charge does the capacitor store?
• (b) How much energy does it store?
• (c) What is the average power while it charges?

(a) Charge.

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

(b) Energy.

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

(c) Average charging power. With $U = 200\;\text{J}$ from (b): $P_{\text{charge}} = 200\;\text{J} / 5.0\;\text{s} = 40\;\text{W}$, about as much as a household light bulb.

Problem 2. The charged capacitor ($100\;\mu\text{F}$, $2000\;\text{V}$, $200\;\text{J}$ stored, charged at $40\;\text{W}$) is discharged through a patient's chest, which acts as a $50\;\Omega$ resistor.

• (a) What is the discharge time constant?
• (b) What are the peak current and the peak power?
• (c) Why is nearly all of the energy delivered within about $10\;\text{ms}$?
• (d) What is the average delivery power?
• (e) How many times the charging power is that?

(a) 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.

(b) 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 $V^2/R = (4.0 \times 10^{6})/50 = 8.0 \times 10^{4}\;\text{W}$.

(c) Why about 10 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.

(d) Average delivery power. Treating the $200\;\text{J}$ as delivered over about $10\;\text{ms}$, from (c):

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

(e) 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 2

• Membrane as RC circuit. Bilayer = capacitor, channels = parallel resistors; more open channels shorten $\tau = R_mC_m$.
• Myelin in series. Stacked layers lower $C_m$ and raise $R_m$, so the signal jumps node to node (saltatory conduction).
• Defibrillator timing. Charge slowly, store $\frac{1}{2}CV^2$, discharge fast with $\tau = RC$: same energy, far more power.
• What is held constant? Decide whether $V$, $Q$, $C$ or $U$ is fixed, then scale with ratios and check units.

Common traps

• Myelin raises capacitance. Myelin LOWERS capacitance, like widening the gap between the plates.
• Channels in series. Added channels are resistors in PARALLEL, so the membrane resistance falls.
• Equal energy, equal power. Power is energy per time; the same energy delivered faster means far more power.
• Voltage linear in energy. $U \propto V^2$, so doubling the energy needs $\sqrt{2}$ times the voltage, not twice.

• 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 (from the Circuits Part 3: EMF & Capacitance lesson): EMF minus the internal drop, $V_{\text{term}} = \varepsilon - Ir$; ammeters go in series (0 Ω), voltmeters in parallel (∞ Ω).
• Capacitors: $C = Q/V = \kappa\varepsilon_0A/d$, $U = \frac{1}{2}CV^2$ (the Circuits Part 3: EMF & Capacitance lesson); series and parallel rules are the reverse of resistors; empty = wire, full = open branch, timescale $\tau = RC$ (Part 1 of this lesson).
• Bioelectricity (Part 2 of this lesson): the membrane is an RC circuit, myelin is capacitors in series, and a defibrillator charges slowly and discharges fast.
• Network rules: whenever a capacitor problem reduces to a resistor network (steady state, $t = 0$), the series/parallel rules come from the Circuits Part 1: Current & Resistance lesson and the Kirchhoff and power tools from the Circuits Part 2: Kirchhoff's Rules & Power lesson. If any network step felt shaky, review them there before the exit quiz.

Next up: the Magnetism & Electromagnetic Induction lesson.