Circuits Part 1: Current & Resistance

Circuits Part 1: Current & Resistance

Part 1 of 4 — Charge, Current & Voltage

Where this fits. This lesson builds DC circuits from the ground up: what current and voltage are, how resistance limits current, how to reduce any resistor network, and how power decides which bulb glows brightest. Its companion, the Circuits Part 2: EMF & Capacitors lesson, picks up with real batteries (EMF and internal resistance), meters, capacitors and bioelectricity. Here every source is ideal and every element is a resistor.

Charge and Current

Charge comes in whole multiples of the elementary charge, $e = 1.6 \times 10^{-19}\;\text{C}$: a proton carries $+e$, an electron $-e$. One coulomb is about $6.25 \times 10^{18}$ elementary charges, which is why even a small current involves enormous numbers of carriers.

Current is the rate at which charge passes a point:

$I = \frac{\Delta Q}{\Delta t}$

measured in amperes: $1\;\text{A} = 1\;\text{C/s}$. Current is a RATE, not an amount. A "dead" battery has run out of stored chemical energy, not out of charge.

Two rearrangements do most of the work on the exam:

• Charge moved: $\Delta Q = I\Delta t$
• Number of carriers: $N = \Delta Q / q_{carrier}$, where $q_{carrier} = e$ for an electron, $\text{Na}^+$ or $\text{K}^+$, but $2e$ for $\text{Ca}^{2+}$

Conventional Current vs. Electron Drift

• Conventional current is defined as the direction POSITIVE charge would flow: out of the source's $+$ terminal, through the external circuit, and back into the $-$ terminal. Every circuit rule on the MCAT uses this direction.
• In a metal wire the mobile carriers are electrons, so they drift the OPPOSITE way, from $-$ through the circuit toward $+$.
• A negative charge moving left is the same current as a positive charge moving right. Only the net flow of charge matters.
• Electrons drift slowly, but the electric field that pushes them spreads through the wire at a large fraction of the speed of light, so every carrier in the loop starts moving almost at once. A lamp lights instantly because the wire is already full of carriers, not because electrons race from the switch to the bulb.

• Typical drift speeds in household wiring are only about $10^{-4}\;\text{m/s}$, and the sign convention itself is a historical accident: "positive" was named long before the electron was discovered.

Metallic vs. Electrolytic Conduction

• Conductivity $\sigma$ measures how readily a material carries current. It is the reciprocal of resistivity (resistivity is defined in Part 2 of this lesson): $\sigma = 1/\rho$, in siemens per metre, where $1\;\text{S} = 1\;\Omega^{-1}$.
• Pure water conducts poorly because it has very few ions. Dissolve a strong electrolyte such as NaCl and conductivity jumps; a weak electrolyte such as acetic acid dissociates only partly and conducts weakly.
• In a solution, both ion types contribute to current in the SAME direction. A cation moving right and an anion moving left both carry net positive charge to the right, so their contributions ADD; they never cancel.
• The body is an electrolytic conductor. Blood, cytosol and interstitial fluid carry current by moving $\text{Na}^+$, $\text{K}^+$, $\text{Cl}^-$ and $\text{Ca}^{2+}$, never free electrons. A nerve impulse is ions crossing membranes, and an ECG electrode needs conductive gel to bridge the skin to the metal lead.

Practice: Charge, Current & Carriers 🎯

Voltage — Energy per Charge

The potential difference (voltage) between two points is the work an external agent does (against the field) per unit charge to move a charge between them:

$\Delta V = \frac{W}{q} \qquad \Delta U = q\Delta V$

measured in volts: $1\;\text{V} = 1\;\text{J/C}$. A $9\;\text{V}$ battery gives every coulomb that passes through it $9\;\text{J}$ of electrical potential energy; the resistors in the circuit turn that energy into heat and light.

• Positive charges move spontaneously from HIGH to LOW potential, losing potential energy.
• Negative charges (electrons, $\text{Cl}^-$) move spontaneously from LOW to HIGH potential, and they ALSO lose potential energy, because $q$ is negative in $\Delta U = q\Delta V$.
• The electron-volt is the energy one elementary charge gains or loses crossing $1\;\text{V}$: $1\;\text{eV} = 1.6 \times 10^{-19}\;\text{J}$. It is the natural unit for single electrons and ions.

Ground and Reference Points

Voltage is always a DIFFERENCE between two points. Ground is the point we declare to be $0\;\text{V}$, usually the battery's negative terminal or literally the Earth. Choosing a different reference shifts every potential by the same amount and changes no difference, no current and no energy. On a circuit figure, all points joined by ideal (zero-resistance) wire are at the same potential: wires are equipotential, and voltage drops happen only across elements.

Why Current Flows — and the Water Analogy

A battery maintains a potential difference between its terminals, so charge sent around an external path gains energy inside the battery and spends it in the circuit. (What a real battery does inside, its EMF and internal resistance, is covered in the Circuits Part 2: EMF & Capacitors lesson.)

Where the analogy breaks down:

• Charge is not used up. The same current that enters a bulb leaves it; the bulb converts potential ENERGY into heat and light. Picturing current "draining away" bulb by bulb is the root of most series-circuit errors.
• The pipe is always full. Wires already contain carriers everywhere, which is why the response is instant.
• A break stops everything. Cut a water pipe and water spills out; cut a wire and charge simply stops moving.

Open vs. Closed Circuits

Current needs a complete conducting loop, a closed circuit. Open a switch anywhere in a single loop and current stops EVERYWHERE in that loop at once. With no current, no resistor has any voltage across it ($V = IR = 0$), so the ENTIRE source voltage appears across the open gap. A broken wire, a burned-out filament and an open switch all behave the same way.

Worked Example — Counting Charge, Carriers and Energy

Problem 1. A $9\;\text{V}$ battery delivers a steady $0.50\;\text{A}$ to a small pump for $1$ minute.

Step 1 — Charge moved. $\Delta Q = I\Delta t = (0.50\;\text{A})(60\;\text{s}) = 30\;\text{C}$

Step 2 — Number of electrons. $N = \dfrac{30}{1.6 \times 10^{-19}} \approx 1.9 \times 10^{20}$. (Estimate: $30/1.6 \approx 19$.)

Step 3 — Energy delivered. Each coulomb gains $9\;\text{J}$: $W = qV = (30)(9) = 270\;\text{J}$.

Step 4 — Energy per electron. $9\;\text{eV} = 9(1.6 \times 10^{-19}) \approx 1.4 \times 10^{-18}\;\text{J}$. Check: $(1.9 \times 10^{20})(1.4 \times 10^{-18}) \approx 270\;\text{J}$. The two routes agree.

Problem 2 — A single ion channel. Recordings show one open potassium channel passing about $2\;\text{pA}$ ($2 \times 10^{-12}\;\text{A}$). How many ions cross in $1\;\text{ms}$?

Step 1 — Charge. $\Delta Q = (2 \times 10^{-12})(1 \times 10^{-3}) = 2 \times 10^{-15}\;\text{C}$

Step 2 — Ions. Each $\text{K}^+$ carries $+e$: $N = \dfrac{2 \times 10^{-15}}{1.6 \times 10^{-19}} \approx 1.25 \times 10^{4}$, about $12{,}500$ ions.

Step 3 — The divalent trap. If the same $2\;\text{pA}$ were carried by $\text{Ca}^{2+}$, each ion would carry $2e$, so only HALF as many ions (about $6{,}250$) would be needed. Always divide by the charge per carrier, not automatically by $e$.

Step 4 — Direction. $\text{K}^+$ leaving the cell is an outward current. If $\text{Cl}^-$ ions entered the cell instead, that would ALSO be an outward current, because negative charge moving in is equivalent to positive charge moving out.

Practice: Voltage, Energy & Closed Loops 🎯

Key Takeaways

• $I = \Delta Q/\Delta t$, with $1\;\text{A} = 1\;\text{C/s}$; count carriers with $N = \Delta Q/q_{carrier}$ ($2e$ for $\text{Ca}^{2+}$)
• Conventional current is the direction positive charge flows, $+$ to $-$ through the external circuit; electrons in metals drift the opposite way
• Metals conduct with electrons (conductivity falls when heated); electrolytes and body fluids conduct with ions moving both ways (conductivity rises with temperature and ion concentration); $\sigma = 1/\rho$
• Voltage is energy per charge: $\Delta U = q\Delta V$, $1\;\text{V} = 1\;\text{J/C}$, $1\;\text{eV} = 1.6 \times 10^{-19}\;\text{J}$
• Positive charges fall to lower potential, negative charges to higher; both lose potential energy
• Only differences matter: ground is a chosen $0\;\text{V}$, and ideal wires are equipotential
• Open loop: zero current everywhere, and the full source voltage across the gap

Common traps

• Drawing current along the electron motion
• Thinking cation and anion flows cancel in a solution
• Believing bulbs "use up" current
• Dividing by $e$ for a divalent ion
• Forgetting to convert millivolts before computing energy

• Low-yield extras: the actual drift-speed magnitude; the historical origin of the sign convention.

Circuits Part 1: Current & Resistance

Part 2 of 4 — Resistance, Ohm's Law & Series vs. Parallel

Resistance and Ohm's Law

Resistance $R$ measures how strongly an element opposes current. It is the ratio of the voltage across the element to the current through it, measured in ohms ($1\;\Omega = 1\;\text{V/A}$):

$V = IR$

Its reciprocal is the conductance, $G = 1/R$, measured in siemens ($1\;\text{S} = 1\;\Omega^{-1}$): the more conductance, the more current per volt.

Read Ohm's law three ways, depending on what the circuit holds fixed:

• Fixed $R$: current is proportional to voltage. Double $V$, double $I$.
• Fixed $V$ (an outlet, or a branch wired straight across a battery): $I \propto 1/R$. Halve the resistance, double the current.
• Fixed $I$ (elements in one series path): $V \propto R$. The bigger resistor takes the bigger share of the voltage.

What sets $R$: material and geometry.

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

• $\rho$ is the resistivity, in $\Omega \cdot \text{m}$. It belongs to the MATERIAL, not the object: very low for metals, intermediate for saline and tissue, enormous for glass, rubber and lipid. Its reciprocal is the conductivity introduced earlier in this lesson.
• Longer means more resistance ($R \propto L$): carriers must cross more obstacles.
• Fatter means less resistance ($R \propto 1/A$): more side-by-side lanes. For a round wire $A = \pi r^2$, so doubling the DIAMETER quadruples $A$ and cuts $R$ to one-quarter.
• Stretching at constant volume: if $L$ rises by a factor $n$, $A$ falls by $n$, so $R$ rises by $n^2$.
• Biology link: an axon's cytoplasm is a resistor. Its internal resistance per unit length goes as $1/r^2$, so a wider axon lets current spread farther down its length and conducts faster.
• Fluid trap: pipe flow resistance goes as $1/r^4$ (Poiseuille), but a wire's electrical resistance goes as $1/r^2$. Do not mix them up.

Temperature dependence. For metals, resistivity INCREASES with temperature: a hotter lattice scatters electrons more often. A glowing tungsten filament has many times the resistance it has when cold. (Electrolytes go the other way, as Part 1 of this lesson showed.)

• Semiconductors lose resistance as they warm, because heating frees more carriers; certain materials become superconductors, with exactly zero resistance, below a critical temperature.

Ohmic vs. non-ohmic. An ohmic element keeps the same $R$ whatever the voltage. Its graph of $I$ against $V$ is a straight line through the origin with slope $1/R$ (a graph of $V$ against $I$ has slope $R$). A non-ohmic element curves. On an $I$-vs-$V$ graph, a filament bulb's curve flattens as the filament heats (the slope $1/R$ shrinks): its resistance climbs, so doubling the voltage less than doubles the current. $V = IR$ still DEFINES $R$ at each instant; "ohmic" simply means that $R$ stays constant.

• A diode is an extreme non-ohmic device: it passes current easily in one direction and almost none in the other.

Practice: Resistance & Ohm's Law 🎯

Series vs. Parallel Resistors

Identify the arrangement by current paths and shared connection points (nodes), never by how the figure is drawn:

• Series: elements sit one after another on a single path, with no junction between them. They carry the SAME current.
• Parallel: elements each connect the SAME two nodes. They have the SAME voltage across them.

Why the rules look the way they do. Series resistors are like added LENGTH, so resistances add. Parallel resistors are like added cross-sectional AREA, so it is the conductances ($G = 1/R$) that add.

Shortcuts worth memorizing:

• Two in parallel: $R_{eq} = \dfrac{R_1 R_2}{R_1 + R_2}$ ("product over sum").
• $n$ identical resistors: $nR$ in series, $R/n$ in parallel.
• Bounds: a parallel $R_{eq}$ is always SMALLER than the smallest branch (every new path lets more current through); a series $R_{eq}$ is always LARGER than the largest element.
• Lopsided pairs: a huge resistor in parallel with a small one barely changes the small one's value; the small path dominates.

Short circuits. A wire of negligible resistance connected across an element holds both of its ends at the same potential. Then $V = 0$ across the element and $I = V/R = 0$ through it: the element is bypassed and drops out of the circuit, and all the current takes the wire. Shorting a source directly makes $R_{eq}$ nearly zero, which drives a dangerously large current and heats the wiring.

Open branches. A broken branch carries no current and can be erased from the drawing. If an ideal source sits directly across a parallel bank, the remaining branches keep their voltage and current. (Parts 3 and 4 of this lesson show what changes when a series resistor sits in front of the bank.)

Worked Example — Series String, Parallel Bank, Identical Branches

A — Series string. A $24\;\text{V}$ ideal battery drives three resistors in series: $2\;\Omega$, $4\;\Omega$ and $6\;\Omega$.

• Equivalent resistance: $R_{eq} = 2 + 4 + 6 = 12\;\Omega$
• Current, the same everywhere: $I = 24/12 = 2.0\;\text{A}$
• Voltage drops: $V_2 = (2.0)(2) = 4\;\text{V}$, $V_4 = 8\;\text{V}$, $V_6 = 12\;\text{V}$
• Check: $4 + 8 + 12 = 24\;\text{V}$. The drops are in the ratio $1 : 2 : 3$, the same as the resistances.

B — Parallel bank. A $12\;\text{V}$ ideal battery sits across a $4\;\Omega$ and a $12\;\Omega$ resistor in parallel.

• Product over sum: $R_{eq} = \dfrac{(4)(12)}{4 + 12} = \dfrac{48}{16} = 3\;\Omega$, below $4\;\Omega$ as it must be
• Each branch sees the full $12\;\text{V}$: $I_4 = 12/4 = 3\;\text{A}$ and $I_{12} = 12/12 = 1\;\text{A}$
• Junction check: $3 + 1 = 4\;\text{A}$, and $V/R_{eq} = 12/3 = 4\;\text{A}$
• The currents are in the ratio $3 : 1$, the INVERSE of the resistance ratio $1 : 3$.

C — Identical branches. Four identical $20\;\Omega$ heating elements are connected to a $10\;\text{V}$ source.

• In parallel: $R_{eq} = 20/4 = 5\;\Omega$, so the source supplies $10/5 = 2\;\text{A}$, or $0.5\;\text{A}$ per element.
• In series: $R_{eq} = 4 \times 20 = 80\;\Omega$, so $I = 10/80 = 0.125\;\text{A}$.
• The parallel bank draws $2/0.125 = 16$ times as much current, that is $n^2$ for $n = 4$: the resistance fell by $n$ in one arrangement and rose by $n$ in the other.

Practice: Series, Parallel & Shorts 🎯

Key Takeaways

• Ohm's law $V = IR$: fixed $R$ gives $I \propto V$; fixed $V$ gives $I \propto 1/R$; fixed $I$ gives $V \propto R$
• $R = \rho L/A$: long and thin means high resistance; doubling the diameter quarters $R$; stretching by $n$ at constant volume multiplies $R$ by $n^2$
• Metal resistivity rises with temperature; an ohmic element has a straight $I$–$V$ line (slope $1/R$), a filament bulb does not
• Series: same current, voltages add, $R_{eq}$ grows above the largest element, largest $R$ takes the most voltage
• Parallel: same voltage, currents add, $R_{eq}$ shrinks below the smallest branch, smallest $R$ takes the most current
• Product over sum for two; $R/n$ for $n$ identical in parallel
• A shorted element carries no current and drops out; an open branch can be erased

Common traps

• Using diameter instead of area (or the fluid $r^4$ law) for wire resistance
• Swapping the series and parallel formulas
• Forgetting to invert $1/R_{eq}$ at the end
• Accepting a parallel $R_{eq}$ larger than a branch

• Low-yield extras: semiconductors and superconductors versus temperature; diodes as non-ohmic devices.

Circuits Part 1: Current & Resistance

Part 3 of 4 — Kirchhoff's Rules & Reducing Networks

Kirchhoff's Rules

Every circuit calculation rests on two conservation laws.

Junction rule (conservation of charge). At any node, the total current flowing in equals the total current flowing out:

$\sum I_{in} = \sum I_{out}$

Charge cannot pile up in a wire, so whatever arrives at a junction must leave it.

Loop rule (conservation of energy). Around any closed loop, the changes in potential add to zero:

$\sum \Delta V = 0$

A charge that travels around a loop and returns to its starting point is back at the same potential, so every volt gained in a source is dropped somewhere along the way.

Sign conventions for walking a loop. Choose a walking direction (either one works) and a direction for each current (guess if you must; a negative answer just means the current runs the other way). Then record:

A single loop, walked. A $24\;\text{V}$ ideal battery drives $2\;\Omega$, $4\;\Omega$ and $6\;\Omega$ resistors in series. Walking with the current: $+24 - 2I - 4I - 6I = 0$, so $I = 2\;\text{A}$. The drops are $4\;\text{V}$, $8\;\text{V}$ and $12\;\text{V}$, and they add back to $24\;\text{V}$. Now ground the battery's negative terminal and track potentials: $24\;\text{V}$ just past the positive terminal, $20\;\text{V}$ after the $2\;\Omega$ resistor, $12\;\text{V}$ after the $4\;\Omega$, and $0\;\text{V}$ after the $6\;\Omega$, right back at ground. The loop closes.

Two sources in one loop. Sources that push the same way add; sources that oppose subtract. The net source voltage drives the current through the total resistance. For example, a $12\;\text{V}$ and a $3\;\text{V}$ battery opposing each other across $3\;\Omega$: $+12 - 3 - 3I = 0$, so $I = 3\;\text{A}$.

Most MCAT circuits are single loops or networks that collapse into one. Expect to apply the two rules as reasoning tools, not to solve systems of equations.

• When sources sit in more than one branch, the network does not collapse: write one junction equation and enough loop equations, then solve them simultaneously. This is rarely required on the MCAT.

Practice: Kirchhoff's Rules 🎯

Collapsing Networks Inside-Out

MCAT networks hide simple rules inside layers. Use the same routine every time:

1. Trace nodes, not drawings. Treat every set of points joined by bare wire as a single node. Two elements between the same two nodes are in parallel, however the figure stretches them. Two elements are in series only if no junction lies between them.
2. Collapse the innermost cluster. Find a pure series or pure parallel group, replace it with one equivalent resistor, and redraw. Repeat until a single $R_{eq}$ remains.
3. Find the total current. $I = V/R_{eq}$.
4. Walk back out. Through series pieces the current is shared, so split the VOLTAGE; across parallel pieces the voltage is shared, so split the CURRENT. Finish with a junction check and a loop check.

Never try to write every current at once. One layer at a time keeps each step a one-line Ohm's law calculation.

Voltage Dividers and Current Dividers

Voltage divider. Two resistors in series across a voltage $V$ share it in proportion to resistance:

$V_1 = V\,\frac{R_1}{R_1 + R_2}$

Current divider. Two resistors in parallel carrying a total current $I$ split it in INVERSE proportion to resistance:

$I_1 = I\,\frac{R_2}{R_1 + R_2}$

Notice the OTHER resistor in the numerator: current favours the smaller resistance. A $1\;\Omega$ and a $3\;\Omega$ branch in parallel split the current $3 : 1$, so three-quarters goes through the $1\;\Omega$ path.

Biology link. Current passing between two electrodes on the body splits among parallel tissue paths in inverse proportion to their resistances, so most of it travels through low-resistance blood and muscle rather than fat or bone.

Worked Example — A Two-Step Network

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

Step 1 — Collapse the parallel pair.

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

Step 2 — Total resistance and current.

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

Step 3 — Walk back out. The series resistor drops $V = (3)(2) = 6\;\text{V}$, leaving $12 - 6 = 6\;\text{V}$ across the pair:

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

Junction check: $1 + 2 = 3\;\text{A}$. Loop check: $12 - 6 - 6 = 0$. The $3\;\Omega$ branch carries twice the current of the $6\;\Omega$ branch, inverse to resistance, exactly as the current divider predicts.

Step 4 — What if the $6\;\Omega$ branch breaks open? Now the $2\;\Omega$ and $3\;\Omega$ resistors are simply in series: $R_{eq} = 5\;\Omega$ and $I = 12/5 = 2.4\;\text{A}$. The $2\;\Omega$ resistor drops $4.8\;\text{V}$ and the $3\;\Omega$ resistor $7.2\;\text{V}$. The total current FELL (3 A to 2.4 A), yet the current in the surviving $3\;\Omega$ branch ROSE (2 A to 2.4 A), because the series resistor now takes less of the voltage.

Step 5 — What if instead the $2\;\Omega$ resistor is shorted? It drops out, so $R_{eq} = 2\;\Omega$ and $I = 6\;\text{A}$. The pair now sees the full $12\;\text{V}$: $2\;\text{A}$ through the $6\;\Omega$ and $4\;\text{A}$ through the $3\;\Omega$, both doubled.

Practice: Dividers & Networks 🎯

Key Takeaways

• Junction rule = conservation of charge: current in equals current out at every node
• Loop rule = conservation of energy: potential changes around any closed loop sum to zero
• Walking a loop: resistor with the current $-IR$, against $+IR$; source $-$ to $+$ is a rise
• Ground is a reference choice; it shifts potentials but never differences
• Collapse networks inside-out, find the total current, then walk back out splitting voltage (series) and current (parallel)
• Voltage divider: share proportional to $R$. Current divider: share inversely proportional to $R$ (the other resistor goes on top)

Common traps

• Trusting the drawing instead of the nodes
• Adding opposing sources instead of subtracting
• Assuming total current stays fixed after the network changes

• Low-yield extras: simultaneous-equation solutions for multi-source networks.

Circuits Part 1: Current & Resistance

Part 4 of 4 — Circuit Changes, Power & Bulb Brightness

What Changes When the Circuit Changes

Part 3 of this lesson collapsed networks that stay put. Now predict how a network responds when something changes. Reason in a fixed order: the change, then $R_{eq}$, then the total current, then the voltage across any fixed series resistor, then whatever voltage is left for the parallel part.

• Add a branch in parallel: $R_{eq}$ falls and the total current rises. If the bank sits directly across an ideal source, the existing branches are unaffected. If a series resistor sits in front of the bank, the larger total current drops MORE voltage across that resistor, leaving LESS for the bank, so the existing branch currents FALL.
• Remove (open) a branch: the reverse. $R_{eq}$ rises, the total current falls, the series resistor drops less voltage, and the remaining branches get MORE voltage and more current.
• Short one branch of a parallel bank: that zero-resistance branch shorts the WHOLE bank, because every branch shares the same two nodes; all of them now have $0\;\text{V}$ across them and carry no current.
• Short a resistor: its voltage and current drop to zero, $R_{eq}$ falls, and everything left in series with it carries more current.
• Add a resistor in series: $R_{eq}$ rises and the current through every element in that path falls.

Bulbs, later in this part, follow exactly the same logic: a brighter bulb is simply one dissipating more power.

Electrical Power

Each coulomb that moves through a potential drop $\Delta V$ gives up energy $q\Delta V$, and the current $I$ is the number of coulombs per second. So the power delivered to an element (energy per time) is:

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

measured in watts: $1\;\text{W} = 1\;\text{J/s} = 1\;\text{V} \cdot \text{A}$. All three forms are the same law combined with $V = IR$. Choose the form that matches what the circuit holds constant:

• Same current (elements in series): use $P = I^2R$. The LARGER resistance dissipates more power.
• Same voltage (parallel branches, anything plugged into an outlet): use $P = V^2/R$. The SMALLER resistance dissipates more power.
• The same physical resistor obeys both forms. The apparent paradox, "does more resistance mean more power or less?", disappears once you ask which quantity is fixed.

Proportional reasoning. At fixed $R$, doubling the current quadruples the heating ($P \propto I^2$). At fixed $V$, halving $R$ DOUBLES the power, because the current doubles too; plugging into $I^2R$ while wrongly holding $I$ fixed would predict that power halves.

Energy bookkeeping. For an ideal source, the power it delivers, $V_{source} I_{total}$, equals the sum of the powers dissipated in all the resistors. Energy in equals energy out.

Energy, the Kilowatt-Hour and Joule Heating

Energy is power multiplied by time: $E = Pt$. Utilities bill in kilowatt-hours, a unit of ENERGY, not power:

$1\;\text{kWh} = (1000\;\text{W})(3600\;\text{s}) = 3.6 \times 10^6\;\text{J}$

Joule heating. In a resistor, all the electrical energy becomes thermal energy at the rate $P = I^2R$. That is the working principle of heaters, toasters and incandescent filaments, and the reason wires warm under load. It also explains high-voltage power lines: delivering the same power $P = IV$ at a higher voltage needs a smaller current, and the line's heating loss, $I^2R_{line}$, falls with the square of that current.

Practice: Changing Circuits, Power & Energy 🎯

Bulb Brightness in Series and Parallel

Brightness tracks the POWER a bulb dissipates. Unless told otherwise, treat bulbs as ohmic resistors (an MCAT simplification; real filaments are non-ohmic, as Part 2 of this lesson noted). For identical bulbs of resistance $R$ on an ideal source $V$, call one bulb's power alone $P_0 = V^2/R$:

Two identical bulbs in series each glow at one-QUARTER of their solo power; in parallel each glows at full power.

Different bulbs. A rating such as "60 W at 120 V" gives the resistance: $R = V^2/P$, so a LOWER wattage rating means a HIGHER resistance. In parallel (normal household use) each bulb gets its rated voltage and the 100 W bulb is brighter. In series the order SWAPS, because the shared current makes the higher-resistance bulb dissipate more.

When a Bulb Burns Out or One Is Added

A burned-out bulb is an open circuit.

• Series string: one bulb burns out and they ALL go dark.
• Parallel bank across an ideal source: the others are unaffected; the source just supplies less total current.
• Mixed circuit (bulb A in series with a parallel pair B and C): if C burns out, $R_{eq}$ rises, the total current falls, and A DIMS. With less voltage across A, more is left for B, so B BRIGHTENS.
• Adding a bulb in series dims every bulb in that path. Adding one in parallel directly across the source leaves the others unchanged but raises the total power drawn.

Household wiring is parallel. Every outlet gets the full line voltage, and appliances run independently. Each appliance switched on adds a branch, lowering $R_{eq}$ and raising the total current in the supply wire. A fuse or circuit breaker, in SERIES with that branch circuit, opens when the total current gets too large, which is why too many appliances on one circuit "trips the breaker".

Electricity and the Body

Current, not voltage, does the harm. What disrupts nerve and heart tissue, and heats it, is the current through it, $I = V/R_{body}$. Voltage matters only because it drives that current. Most of the body's resistance is in dry skin; the interior is salty, ionic fluid and conducts well (Part 1 of this lesson).

• Dry skin gives a body path on the order of $10^5\;\Omega$; wet or broken skin drops it to the order of $10^3\;\Omega$. At the same voltage, wet skin can pass about $100$ times the current.
• The path matters: hand-to-hand or hand-to-foot current crosses the chest and can disrupt the heart's rhythm.
• A static spark can be thousands of volts yet harmless, because it delivers very little charge; a modest voltage across wet skin can be lethal.

• Approximate body-current thresholds: about $1\;\text{mA}$ is perceptible, roughly $10$ to $20\;\text{mA}$ prevents letting go, and about $100\;\text{mA}$ across the chest can trigger ventricular fibrillation. The values vary with path, duration and frequency, and passages supply them.

MCAT Integration — Rank Without Computing

Passage questions reward ratios over arithmetic:

1. Find what is held constant for each element: current in series, voltage in parallel.
2. Pick the power form containing that constant and compare with the remaining variable.
3. Multiply factor changes. If $V$ triples and $R$ doubles, $P = V^2/R$ changes by $3^2/2 = 9/2$.
4. Sanity-check the extremes. A short ($R \to 0$) across a fixed voltage draws enormous power; an open branch ($R \to \infty$) draws none.

Most "which element dissipates the most power" questions fall to steps 1 and 2 in ten seconds.

Worked Example — Which Bulb Is Brighter, and How Big Is the Shock?

A — Rated bulbs in series. A $60\;\text{W}$ bulb and a $100\;\text{W}$ bulb, both rated at $120\;\text{V}$, are wired in series across a $120\;\text{V}$ outlet. Which glows brighter?

• Resistances from the ratings: $R_{60} = \dfrac{120^2}{60} = \dfrac{14{,}400}{60} = 240\;\Omega$ and $R_{100} = \dfrac{14{,}400}{100} = 144\;\Omega$.
• Series means the same current, so $P = I^2R$ and $P \propto R$: $\dfrac{P_{60}}{P_{100}} = \dfrac{240}{144} = \dfrac{5}{3}$.
• Conclusion: the 60 W bulb is brighter, the reverse of the normal parallel case. Both glow well below their ratings, because each gets less than $120\;\text{V}$.

B — A shock through dry and wet skin. A person grips a $120\;\text{V}$ conductor with one hand while touching a grounded pipe with the other.

• Dry skin, body path $1 \times 10^5\;\Omega$: $I = \dfrac{120}{10^5} = 1.2 \times 10^{-3}\;\text{A} = 1.2\;\text{mA}$, a tingle.
• Wet skin, body path $1 \times 10^3\;\Omega$: $I = \dfrac{120}{10^3} = 0.12\;\text{A} = 120\;\text{mA}$, a hundred times larger.
• Heating in the body: $P = V^2/R$ gives $0.144\;\text{W}$ dry but $14.4\;\text{W}$ wet. Same voltage, $100$ times the current and $100$ times the power.
• A passage that gives a fibrillation threshold near $0.1\;\text{A}$ would flag the wet case as potentially lethal: the voltage never changed, only the resistance.

Practice: Brightness, Safety & Ratios 🎯

Key Takeaways

• When a circuit changes, reason in order: $R_{eq}$, total current, the drop across any series resistor, then what is left for the parallel bank; shorting one branch of a bank shorts the whole bank
• $P = IV = I^2R = V^2/R$: use $I^2R$ when current is shared (series, larger $R$ wins) and $V^2/R$ when voltage is shared (parallel, smaller $R$ wins)
• Energy $E = Pt$; $1\;\text{kWh} = 3.6 \times 10^6\;\text{J}$ is energy, not power
• Joule heating converts all of a resistor's electrical energy to heat; high-voltage transmission cuts $I^2R$ losses
• Brightness = power. $n$ identical bulbs in series each get $P_0/n^2$; in parallel each gets $P_0$
• Lower wattage rating means higher resistance; rated bulbs swap brightness order in series
• Series burnout darkens everything; parallel burnout across an ideal source changes nothing else; in mixed circuits, track $R_{eq}$, then total current, then the voltage split
• Household wiring is parallel; breakers sit in series and trip on total current
• Current through the body does the harm: wet skin can pass about 100 times the current at the same voltage
• Rank without computing: identify the constant, pick the form, multiply factors

Common traps

• Assuming parallel branches never change (true only directly across an ideal source)
• Using $I^2R$ at fixed voltage while holding $I$ constant
• Calling the kilowatt-hour a unit of power
• Assuming the higher-wattage bulb is brighter in every circuit
• Saying voltage, rather than current, causes shock injury

• Low-yield extras: specific body-current thresholds (perception, let-go, fibrillation).

Next up: the Circuits Part 2: EMF & Capacitors lesson — real batteries with internal resistance, meters, capacitors and dielectrics, RC behaviour, and the membrane as a circuit.