Circuits Part 2: Kirchhoff's Rules & Power

Circuits Part 2: Kirchhoff's Rules & Power

Part 1 of 2 — Kirchhoff's Rules & Reducing Networks

Where this fits. The Circuits Part 1: Current & Resistance lesson gave you current, voltage, Ohm's law and the series and parallel rules. This lesson applies them to whole circuits: Kirchhoff's junction and loop rules, network reduction and dividers, then how a circuit responds to changes and how power sets bulb brightness and electrical safety. Next comes the Circuits Part 3: EMF & Capacitance lesson, which makes the battery real and adds the capacitor.

Kirchhoff's Rules

Every circuit calculation rests on two conservation laws. Most MCAT circuits are single loops or networks that collapse into one, so expect to apply the rules as reasoning tools, not to solve systems of equations.

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:

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 $I = 3\;\text{A}$.

• 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.

Worked Example — A Single Loop, Walked

Problem. A $30\;\text{V}$ ideal battery drives $3\;\Omega$, $5\;\Omega$ and $7\;\Omega$ resistors in series. Walk the loop in the direction of the current.

• (a) What equation does the loop rule give, and what current does it predict?
• (b) What is the voltage drop across each resistor?
• (c) With the battery's negative terminal grounded, what is the potential after each element of the loop?

(a) Loop equation. Walking with the current: $+30 - 3I - 5I - 7I = 0$, so $I = 2\;\text{A}$.

(b) Drops. With $I = 2\;\text{A}$ from (a), the drops are $6\;\text{V}$, $10\;\text{V}$ and $14\;\text{V}$, and they add back to $30\;\text{V}$.

(c) Potentials. $30\;\text{V}$ just past the positive terminal, $24\;\text{V}$ after the $3\;\Omega$ resistor, $14\;\text{V}$ after the $5\;\Omega$, and $0\;\text{V}$ after the $7\;\Omega$, right back at ground. The loop closes.

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

Problem 1. A $12\;\text{V}$ ideal battery feeds a $2\;\Omega$ resistor in series with a parallel pair: $6\;\Omega$ and $3\;\Omega$.

• (a) What is the resistance of the parallel pair?
• (b) What is the total resistance, and what current leaves the battery?
• (c) What voltage is left across the parallel pair?
• (d) What current flows in each branch of the pair?
• (e) Do the junction and loop rules check out?

(a) Collapse the parallel pair. $R_{pair} = \dfrac{(6)(3)}{6 + 3} = \dfrac{18}{9} = 2\;\Omega$

(b) Total resistance and current. With $R_{pair} = 2\;\Omega$ from (a):

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

(c) Walk back out. With $I = 3\;\text{A}$ from (b), the series resistor drops $V = (3)(2) = 6\;\text{V}$, leaving $12 - 6 = 6\;\text{V}$ across the pair.

(d) Branch currents. With $6\;\text{V}$ across the pair from (c):

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

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

Problem 2. In the same circuit ($12\;\text{V}$, a $2\;\Omega$ resistor in series with $6\;\Omega$ parallel to $3\;\Omega$), the $6\;\Omega$ branch breaks open.

• (a) What are the new total resistance and current?
• (b) What voltage drops across each remaining resistor?
• (c) How did the total current and the current in the $3\;\Omega$ branch change from Problem 1?

(a) New total. Now the $2\;\Omega$ and $3\;\Omega$ resistors are simply in series: $R_{eq} = 5\;\Omega$ and $I = 12/5 = 2.4\;\text{A}$.

(b) New drops. With $I = 2.4\;\text{A}$ from (a), the $2\;\Omega$ resistor drops $4.8\;\text{V}$ and the $3\;\Omega$ resistor $7.2\;\text{V}$.

(c) Compare. 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.

Problem 3. Back in the original circuit, the $2\;\Omega$ series resistor is shorted instead, while the $6\;\Omega$ and $3\;\Omega$ pair stays on the $12\;\text{V}$ battery.

• (a) What are the new total resistance and current?
• (b) What current flows in each branch now, compared with Problem 1?

(a) New total. The shorted resistor drops out, so only the pair remains: $R_{eq} = 2\;\Omega$ and $I = 6\;\text{A}$.

(b) Branch currents. 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. Charge is conserved, so the current into every node equals the current out of it.
• Loop rule. Energy is conserved: around a closed loop, $-IR$ with the current and $+V$ from $-$ to $+$ sum to zero.
• Collapse inside-out. Reduce the innermost series or parallel group, find $I = V/R_{eq}$, then walk back out.
• Voltage and current dividers. Series voltage splits in proportion to $R$; parallel current splits inversely, favouring the smaller $R$.

Common traps

• Trusting the drawing. Elements between the same two nodes are in parallel, however the figure stretches them.
• Adding opposing sources. Sources that push against each other subtract; only sources pushing the same way add.
• Fixed total current. Changing a network changes $R_{eq}$, so the total current changes too; recompute it.
• Wrong divider numerator. In $I_1 = I\,R_2/(R_1 + R_2)$ the OTHER resistor goes on top, because current favours the smaller $R$.

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

Circuits Part 2: Kirchhoff's Rules & Power

Part 2 of 2 — Circuit Changes, Power & Bulb Brightness

What Changes When the Circuit Changes

Part 1 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. Bulbs, later in this part, follow exactly the same logic: a brighter bulb is simply one dissipating more power.

• 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.

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.

One resistor, both forms. 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). 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 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 salty, ionic interior conducts well.

• Dry skin gives a body path on the order of $10^5\;\Omega$; wet or broken skin drops it to about $10^3\;\Omega$, so at the same voltage wet skin passes 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 of thousands of volts is 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; most "which element dissipates the most power" questions fall to steps 1 and 2 in ten seconds:

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.

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

Problem 1. 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.

• (a) What is each bulb's resistance?
• (b) What is the ratio of their powers in series?
• (c) Which bulb glows brighter, and does either reach its rated power?

(a) Resistances from the ratings. $R_{60} = \dfrac{120^2}{60} = \dfrac{14{,}400}{60} = 240\;\Omega$ and $R_{100} = \dfrac{14{,}400}{100} = 144\;\Omega$.

(b) Power ratio. Series means the same current, so $P = I^2R$ and $P \propto R$. With the resistances from (a): $\dfrac{P_{60}}{P_{100}} = \dfrac{240}{144} = \dfrac{5}{3}$.

(c) 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 $120\;\text{V}$.

Problem 2. A person grips a $120\;\text{V}$ conductor with one hand while touching a grounded pipe with the other. The body path is $1 \times 10^5\;\Omega$ with dry skin and $1 \times 10^3\;\Omega$ with wet skin.

• (a) How much current flows through dry skin?
• (b) How much current flows through wet skin?
• (c) How much power is dissipated in the body in each case?
• (d) If a passage puts the fibrillation threshold near $0.1\;\text{A}$, which case is potentially lethal?

(a) Dry skin. $I = \dfrac{120}{10^5} = 1.2 \times 10^{-3}\;\text{A} = 1.2\;\text{mA}$, a tingle.

(b) Wet skin. $I = \dfrac{120}{10^3} = 0.12\;\text{A} = 120\;\text{mA}$, a hundred times larger than the $1.2\;\text{mA}$ from (a).

(c) 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.

(d) The danger. The wet case: its $0.12\;\text{A}$ from (b) is above the threshold. The voltage never changed, only the resistance.

Practice: Brightness, Safety & Ratios 🎯

Key Takeaways

• Reason in order. After any change: $R_{eq}$, then total current, then the series drop, then what is left for the bank.
• Pick the power form. $P = I^2R$ when current is shared (larger $R$ wins); $P = V^2/R$ when voltage is shared (smaller $R$ wins).
• Brightness is power. $n$ identical bulbs in series each get $P_0/n^2$; in parallel each keeps the full $P_0$.
• Current does the harm. $I = V/R_{body}$, so wet skin passes about 100 times the current at the same voltage.

Common traps

• Branches never change. A bank keeps its voltage only directly across an ideal source, not behind a series resistor.
• Fixing $I$ at fixed $V$. Halving $R$ at fixed voltage doubles $I$ and so doubles $P$; it does not halve it.
• The kWh as power. The kilowatt-hour is energy: $1\;\text{kWh} = 3.6 \times 10^6\;\text{J}$.
• Higher wattage always brighter. A lower rating means higher $R$, so in series the lower-wattage bulb glows brighter.

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

Next up: the Circuits Part 3: EMF & Capacitance lesson — real batteries with EMF and internal resistance, ammeters and voltmeters, then capacitance, dielectrics and stored energy.