Circuits

Circuits

Part 1 of 4 — Current, Voltage & Resistance

Current

$I = \frac{Q}{t}$ — charge passing a point per second, in amperes ($1\;\text{A} = 1\;\text{C/s}$).

• Conventional current points in the direction POSITIVE charge flows — opposite to the actual electron drift in a wire.
• In electrolyte solutions (and in the body), current is carried by IONS moving both ways: cations along the current direction, anions against it.

Voltage (Potential Difference)

Voltage is the electrical "pressure difference" that drives current. A battery is a chemical pump that maintains a fixed potential difference — its EMF — between its terminals.

Resistance and Ohm's Law

$V = IR$

Resistance measures how hard it is to push current through an element, in ohms ($\Omega$). For a wire:

$R = \rho\frac{L}{A}$

• $\rho$ (resistivity) is a material property — low for copper, high for glass, in between for physiological saline.
• Longer wire ($L$ up) → more resistance. Fatter wire ($A$ up) → less resistance.
• For most metals, resistivity INCREASES with temperature.

Series vs. Parallel Resistors

Two useful facts about parallel:

• Two resistors: $R_{eq} = \dfrac{R_1 R_2}{R_1 + R_2}$ ("product over sum")
• $R_{eq}$ is always SMALLER than the smallest branch — adding a path always lets more current flow.

Kirchhoff's Rules (Conservation Laws in Disguise)

• Junction rule: current in = current out at every node (conservation of charge).
• Loop rule: voltage gains equal voltage drops around any closed loop (conservation of energy).

Worked Example — A Simple Series Circuit

A $24\;\text{V}$ battery drives current through three resistors in series: $2\;\Omega$, $4\;\Omega$, and $6\;\Omega$.

Step 1 — Equivalent resistance.

$R_{eq} = 2 + 4 + 6 = 12\;\Omega$

Step 2 — Current (same everywhere in series).

$I = \frac{V}{R_{eq}} = \frac{24}{12} = 2.0\;\text{A}$

Step 3 — Voltage drops.

$V_2 = (2.0)(2) = 4\;\text{V}, \quad V_4 = (2.0)(4) = 8\;\text{V}, \quad V_6 = (2.0)(6) = 12\;\text{V}$

Check with the loop rule: $4 + 8 + 12 = 24\;\text{V}$ — every volt the battery supplies is dropped. The largest resistor takes the largest share of the voltage.

Worked Example — Product Over Sum

A $3\;\Omega$ and a $6\;\Omega$ resistor sit in parallel across a $12\;\text{V}$ battery.

Step 1 — Equivalent resistance.

$R_{eq} = \frac{(3)(6)}{3 + 6} = \frac{18}{9} = 2\;\Omega$

(Smaller than $3\;\Omega$, as parallel must be.)

Step 2 — Branch currents (each branch sees the full $12\;\text{V}$).

$I_3 = \frac{12}{3} = 4\;\text{A}, \quad I_6 = \frac{12}{6} = 2\;\text{A}$

Step 3 — Junction check. Total $I = 4 + 2 = 6\;\text{A}$, and indeed $V/R_{eq} = 12/2 = 6\;\text{A}$. More current takes the easier (lower-resistance) path, in exact inverse proportion.

Current, Voltage & Resistance 🎯

Key Takeaways — Part 1

• $I = Q/t$; conventional current follows positive charge; in solution, ions carry it both ways
• Ohm: $V = IR$; wires: $R = \rho L/A$ (long and thin = high resistance)
• Series: same $I$, voltages add, $R_{eq}$ grows. Parallel: same $V$, currents add, $R_{eq}$ shrinks below the smallest branch
• Two in parallel: product over sum
• Kirchhoff = conservation: charge at junctions, energy around loops

Circuits

Part 2 of 4 — Power, EMF & Real Batteries

Electrical Power

Power is the rate at which electrical energy is converted to heat, light, or work:

$P = IV = I^2R = \frac{V^2}{R}$

All three forms are the same law combined with $V = IR$ — pick the one whose two variables you actually know.

• $P = I^2R$ is the natural form for SERIES elements (same $I$): more resistance means more power.
• $P = V^2/R$ is the natural form for PARALLEL elements (same $V$): LESS resistance means more power.

The same physical resistor obeys both — the difference is which quantity the circuit holds fixed for it.

Energy and the Kilowatt-Hour

$E = Pt$. Utilities bill in kilowatt-hours: $1\;\text{kWh} = (1000\;\text{W})(3600\;\text{s}) = 3.6 \times 10^6\;\text{J}$.

EMF vs. Terminal Voltage

A real battery is an ideal EMF source $\varepsilon$ in series with a small internal resistance $r$. When the battery drives current $I$:

$V_{terminal} = \varepsilon - Ir$

• No current drawn (open circuit, or an ideal voltmeter): $V_{terminal} = \varepsilon$.
• Discharging: terminal voltage sags BELOW the EMF, and it sags more at higher current — why headlights dim while the starter motor cranks.
• Being recharged (current forced backward through it): $V_{terminal} = \varepsilon + Ir$, ABOVE the EMF.

Maximum possible current (a short circuit, $R_{ext} = 0$): $I_{max} = \varepsilon/r$.

Batteries in Series and Parallel

• Series (plus to minus): EMFs add — two $1.5\;\text{V}$ cells make $3.0\;\text{V}$. Internal resistances add too.
• Parallel (identical cells): EMF stays the same, but the pack can supply more current and lasts longer; internal resistance drops.

Meters

• Ammeter: measures current, wired IN SERIES, built with very LOW resistance so it does not impede the current it measures.
• Voltmeter: measures potential difference, wired IN PARALLEL across an element, built with very HIGH resistance so it steals negligible current.

Swapping them is catastrophic: an ammeter in parallel is a short circuit.

Worked Example — A Real Battery Under Load

A battery with EMF $\varepsilon = 12\;\text{V}$ and internal resistance $r = 0.5\;\Omega$ is connected to a $5.5\;\Omega$ resistor.

Step 1 — Current. Internal and external resistance are in series:

$I = \frac{\varepsilon}{R + r} = \frac{12}{5.5 + 0.5} = 2.0\;\text{A}$

Step 2 — Terminal voltage.

$V_{terminal} = \varepsilon - Ir = 12 - (2.0)(0.5) = 11\;\text{V}$

Step 3 — Power accounting.

• Delivered to the load: $P_R = I^2R = (4)(5.5) = 22\;\text{W}$
• Wasted inside the battery: $P_r = I^2r = (4)(0.5) = 2\;\text{W}$
• Total supplied: $P = \varepsilon I = (12)(2.0) = 24\;\text{W}$ — the books balance.

Worked Example — Which Bulb Is Brighter?

A $60\;\text{W}$ bulb and a $100\;\text{W}$ bulb (ratings at $120\;\text{V}$) are wired IN SERIES across an outlet. Which glows brighter?

Step 1 — Compare resistances from the ratings. At rated voltage $P = V^2/R$, so $R = V^2/P$. The $60\;\text{W}$ bulb has the LARGER resistance (smaller rated power, same rated voltage).

Step 2 — Series means same current. In series, $P = I^2R$: the larger resistance dissipates more power.

Conclusion: the $60\;\text{W}$ bulb is brighter in series — the reverse of the parallel (normal household) case. Ratings assume parallel wiring; the circuit decides what is held constant.

Power, EMF & Real Batteries 🎯

Key Takeaways — Part 2

• $P = IV = I^2R = V^2/R$; use $I^2R$ for series comparisons, $V^2/R$ for parallel
• $1\;\text{kWh} = 3.6 \times 10^6\;\text{J}$ of energy, not power
• Real battery: $V_{terminal} = \varepsilon - Ir$ discharging, $\varepsilon + Ir$ charging, $= \varepsilon$ at zero current
• Series cells: EMFs add; parallel identical cells: same EMF, more available current
• Ammeter: low $R$, in series. Voltmeter: high $R$, in parallel

Circuits

Part 3 of 4 — Capacitors & Dielectrics

Capacitance

A capacitor stores separated charge. Its capacitance is the charge stored per volt:

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

in farads ($1\;\text{F} = 1\;\text{C/V}$). Real capacitors are microfarads ($10^{-6}$) to picofarads ($10^{-12}$); $Q$ means the charge on EACH plate ($+Q$ on one, $-Q$ on the other).

The Parallel-Plate Capacitor

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

Capacitance is pure geometry: bigger plates (larger $A$) or closer plates (smaller $d$) both raise $C$. Between the plates the field is uniform:

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

Dielectrics

Sliding an insulating material (dielectric constant $\kappa > 1$) between the plates multiplies the capacitance:

$C = \kappa C_0$

The polarized dielectric partially cancels the internal field. Track what is held constant:

• Battery connected ($V$ fixed): $Q = CV$ increases — the battery pumps on more charge.
• Battery disconnected ($Q$ fixed): $V = Q/C$ DROPS, and stored energy drops too (the field pulls the dielectric in).

Stored Energy

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

The factor $\frac{1}{2}$ appears because the voltage climbs from $0$ to $V$ while charging — each coulomb is loaded at the average voltage $V/2$.

Capacitor Networks — the Rules Flip

Exactly OPPOSITE to resistors: parallel capacitors add because side-by-side plates act like one bigger plate.

Capacitors in DC Circuits

A charging capacitor fills quickly at first, then ever more slowly, with time constant $\tau = RC$. Once fully charged, NO current flows through its branch — a full capacitor behaves like an open switch, and the rest of the circuit is analyzed without that branch.

Worked Example — Series Capacitors Share Charge, Not Voltage

A $3\;\mu\text{F}$ and a $6\;\mu\text{F}$ capacitor are in series across a $12\;\text{V}$ battery.

Step 1 — Equivalent capacitance (product over sum works here too).

$C_{eq} = \frac{(3)(6)}{3 + 6} = 2\;\mu\text{F}$

Step 2 — Charge (same on every series capacitor).

$Q = C_{eq}V = (2\;\mu\text{F})(12\;\text{V}) = 24\;\mu\text{C}$

Step 3 — Individual voltages.

$V_3 = \frac{24}{3} = 8\;\text{V}, \quad V_6 = \frac{24}{6} = 4\;\text{V}$

Check: $8 + 4 = 12\;\text{V}$. The SMALLER capacitor takes the LARGER voltage share — the mirror image of series resistors, where the larger resistance takes more voltage.

Worked Example — Dielectric with the Battery Disconnected

A $2\;\mu\text{F}$ parallel-plate capacitor is charged to $100\;\text{V}$, then disconnected. A dielectric with $\kappa = 4$ is slid in.

Step 1 — What is fixed? Disconnected, so $Q$ is trapped: $Q = CV = (2\;\mu\text{F})(100) = 200\;\mu\text{C}$.

Step 2 — New capacitance and voltage.

$C_{new} = \kappa C = 8\;\mu\text{F}, \quad V_{new} = \frac{Q}{C_{new}} = \frac{200\;\mu\text{C}}{8\;\mu\text{F}} = 25\;\text{V}$

Step 3 — Energy. Before: $U = \frac{1}{2}CV^2 = \frac{1}{2}(2 \times 10^{-6})(100)^2 = 0.010\;\text{J}$. After: $U = \frac{Q^2}{2C} = 0.0025\;\text{J}$ — energy fell by the factor $\kappa$. The missing energy became the work the field did pulling the slab in.

Capacitors & Dielectrics 🎯

Key Takeaways — Part 3

• $C = Q/V$; parallel plate: $C = \varepsilon_0 A/d$, uniform field $E = V/d$
• Dielectric multiplies $C$ by $\kappa$; battery connected keeps $V$ (charge rises), disconnected keeps $Q$ (voltage falls)
• Energy: $U = \frac{1}{2}QV = \frac{1}{2}CV^2 = Q^2/2C$
• Capacitor rules are resistor rules flipped: parallel adds, series is reciprocal; series shares $Q$, parallel shares $V$
• Charging follows $\tau = RC$; a full capacitor is an open switch in DC

Circuits

Part 4 of 4 — MCAT Integration: Complex Circuits & Bioelectricity

MCAT circuit questions hide simple rules inside layered networks and biological wrappers. The playbook:

Strategy 1 — Collapse the Network from the Inside Out

Find the innermost pure-series or pure-parallel cluster, replace it with one equivalent resistor, redraw, repeat. Then walk BACK out: total current first, then split voltages and currents step by step. Never try to write every current at once.

Strategy 2 — Know What Series and Parallel Really Mean

• Series: elements share the SAME single current path — no junction between them.
• Parallel: elements connect the SAME two nodes — identical voltage, no matter how the wires are drawn.

Exam figures deliberately stretch wires to disguise parallel pairs. Trace nodes, not drawings.

Strategy 3 — Rank Power Without Computing It

• Same current (series): $P = I^2R$ — bigger $R$ wins.
• Same voltage (parallel): $P = V^2/R$ — smaller $R$ wins.

Most "which resistor dissipates the most power" questions fall to this in ten seconds.

Strategy 4 — The Cell Membrane Is an RC Circuit

The lipid bilayer is a capacitor (an insulator separating two conducting solutions); ion channels are resistors in parallel with it. Consequences the MCAT tests:

• Membrane potential cannot change instantly — it charges with time constant $\tau = R_mC_m$.
• Blocking channels raises $R_m$, so $\tau$ LENGTHENS and the neuron integrates inputs more slowly.
• Myelin thickens the "dielectric," LOWERING membrane capacitance — less charge is needed per millivolt, so signals travel faster.

Strategy 5 — Defibrillators and Capacitor Discharge

A defibrillator charges a capacitor slowly (small current, several seconds), then dumps $U = \frac{1}{2}CV^2$ through the chest in milliseconds. Same energy, wildly different power — power is energy per TIME. That asymmetry (slow charge, fast discharge) is the whole point of using a capacitor.

Worked Example — A Two-Step Network (Passage Style)

A $12\;\text{V}$ battery feeds a $2\;\Omega$ resistor in series with a parallel pair: $6\;\Omega$ and $3\;\Omega$.

Step 1 — Collapse the parallel pair.

$R_{pair} = \frac{(6)(3)}{6 + 3} = 2\;\Omega$

Step 2 — Total resistance and current.

$R_{total} = 2 + 2 = 4\;\Omega, \quad I = \frac{12}{4} = 3\;\text{A}$

Step 3 — Walk back out. The series resistor drops $V = (3)(2) = 6\;\text{V}$, leaving $6\;\text{V}$ across the pair:

$I_6 = \frac{6}{6} = 1\;\text{A}, \quad I_3 = \frac{6}{3} = 2\;\text{A}$

Junction check: $1 + 2 = 3\;\text{A}$. The $3\;\Omega$ branch carries twice the current of the $6\;\Omega$ branch — inverse to resistance, always.

Worked Example — Defibrillator Energy and Power

A defibrillator capacitor ($C = 100\;\mu\text{F}$) is charged to $2000\;\text{V}$ and discharges $80\%$ of its energy into the chest in $10\;\text{ms}$.

Step 1 — Stored energy.

$U = \frac{1}{2}CV^2 = \frac{1}{2}(10^{-4})(2 \times 10^3)^2 = \frac{1}{2}(10^{-4})(4 \times 10^6) = 200\;\text{J}$

Step 2 — Delivered energy and average power.

$E = 0.8 \times 200 = 160\;\text{J}, \quad P = \frac{160\;\text{J}}{10^{-2}\;\text{s}} = 16{,}000\;\text{W}$

Sixteen kilowatts — from a device that draws only a few hundred watts from the wall while charging. The capacitor is a power amplifier by time compression, not an energy source.

MCAT Integration: Circuits 🎯

Key Takeaways — Part 4

• Collapse networks inside-out, then walk currents and voltages back out; junction-check at the end
• Series vs parallel is about shared nodes, not how the figure is drawn
• Power ranking: series favors big $R$ ($I^2R$), parallel favors small $R$ ($V^2/R$)
• Membrane = RC circuit: bilayer capacitor, channel resistors, $\tau = R_mC_m$; myelin cuts $C_m$
• Defibrillator: energy $\frac{1}{2}CV^2$ stored slowly, released fast — huge power from modest energy
• Full capacitor in DC = open branch; empty capacitor = momentary wire