Unit 3

Electric Circuits

Every circuit problem is two conservation laws in disguise: charge is conserved, so current that flows into a junction flows out; energy is conserved, so the voltage a battery provides is all used up around any loop. Series rules, parallel rules, Kirchhoff, even capacitors charging — all of it is bookkeeping for those two sentences.

0 · Foundations: charge, energy, and the circuit diagram

Circuits are where Unit 2’s potential difference stops being abstract. Learn to read the diagram, then hold on to what current and voltage each mean — almost every conceptual question tests the meaning, not the algebra.

Circuit symbols

What current and voltage mean

Those two prepositions are the whole foundation. A battery does not supply current; it maintains a potential difference, and the rest of the circuit decides how much current that difference drives. Resistance links the two, and power is their product:

The two conservation laws

The series and parallel rules of Topic 3 are just these two laws applied to the simplest arrangements: series elements share a current (nothing branches off), and parallel elements share a voltage (each is its own loop with the same two endpoints).

1 · Current, resistance & Ohm’s law

A circuit is a closed loop that charge flows around. Current says how much charge passes per second; resistance says how hard the material makes that flow; and Ohm’s law ties both to the potential difference that drives it. Every later topic in the unit is built from these three quantities.

What current is

Pick any slice through a wire and count the charge crossing it each second — that is the current. Because charge is conserved and does not pile up anywhere in a steady circuit, the current is the same through every slice of a single, unbranched wire, before a bulb and after it. A bulb converts the charges’ energy; it does not use up the charges.

Conventional current vs. electron flow

The direction of current was defined (by Franklin, long before the electron) as the direction positive charge would move: out of the + terminal of a battery, through the external circuit, into the − terminal. In metal wires the charges that actually move are electrons, and they drift the opposite way.

The drift picture

AP wants this qualitatively. Free electrons in a metal are already moving very fast — around a thousand kilometres per second (~10⁶ m/s) — but in random directions, so with no field there is no net flow. Connect a battery and a field appears in the wire. Each electron now picks up a small bias opposite to the field between collisions with the vibrating ion lattice. That bias is the drift velocity, and it is tiny: for ordinary currents in copper, well under a millimetre per second.

• Why do lights come on instantly, then? The field is set up throughout the circuit at a large fraction of the speed of light, so electrons everywhere in the circuit start drifting at once. No electron has to travel from the switch to the bulb.
• What is resistance, microscopically? Collisions. Each collision hands the electron’s gained kinetic energy to the lattice, which is why a current-carrying resistor gets warm.
• What sets the drift speed? More current, a thinner wire, or fewer free electrons per cubic metre all mean faster drift. The same current in a thin section of wire drifts faster than in a thick section.

Resistance & resistivity

Resistance R is a property of a particular object — this length of this wire. Resistivity ρ is a property of the material. The two are linked by the geometry:

Temperature and resistance

Resistivity is quoted “at 20 °C” because it changes with temperature. AP asks for the direction and the reason, not a formula.

The everyday consequence: an incandescent bulb’s tungsten filament runs at roughly 2500–3000 K, and its hot resistance is on the order of ten times its cold resistance. That is why the filament is the textbook non-ohmic device, and why bulbs tend to fail at the moment they are switched on — the cold filament draws a large surge of current.

Ohm’s law & I–V graphs

Resistance is defined for any device at any operating point as

Ohm’s law is the stronger, experimental claim that for some materials this ratio is constant — independent of the current. Those materials are ohmic. Anything whose ratio changes as you turn the voltage up is non-ohmic. Ohm’s law is a property of particular materials, not a law of nature like charge conservation.

Worked examples

1. Counting electrons

A phone charger cable carries a steady 0.40 A for 2.0 minutes. How much charge passes through it, and how many electrons is that?

ΔQ = IΔt = (0.40 A)(120 s) = 48 C Convert minutes to seconds first: 2.0 min = 120 s.

N = ΔQe = 481.60×10⁻¹⁹ = 3.0×10²⁰ electrons

An enormous number of electrons, each drifting very slowly — which is exactly the drift picture: big current from a huge crowd moving at a crawl.

2. Resistance of a copper wire

An extension cord conductor is 20 m of copper wire with a diameter of 1.0 mm. Find its resistance.

Area first. r = 0.50 mm = 5.0×10⁻⁴ m.

A = πr² = π(5.0×10⁻⁴)² = 7.85×10⁻⁷ m²

R = ρLA = (1.68×10⁻⁸)(20)7.85×10⁻⁷ = 0.43 Ω

Small but not zero: carrying 10 A, this wire would have ΔV = IR ≈ 4.3 V across it — which is why long, thin extension cords get warm under heavy loads.

3. Stretching a wire (proportional reasoning)

A wire of resistance R₀ is drawn out until it is twice as long. Its volume (and material) stays the same. What is its new resistance?

Constant volume means LA = constant, so doubling L halves A. Both changes raise R:

R_new = ρ(2L)A/2 = 4ρLA = 4R₀

Contrast a different question: same material, new wire with twice the length and twice the diameter. Then A is 4× larger and R = ρ(2L)/(4A) = R₀/2. Always write the ratio out; never guess.

4. Resistance from a curved I–V graph

The bulb in the figure above carries 0.10 A at 1.0 V and 0.25 A at 6.0 V. Find its resistance at each point. A student instead computes (6.0 − 1.0)/(0.25 − 0.10). What does that number mean?

R₁ = 1.00.10 = 10 Ω      R₂ = 6.00.25 = 24 Ω

The resistance more than doubles as the filament heats up — the non-ohmic signature. The student’s value, 5.0/0.15 = 33 Ω, is the inverse slope of the chord between the points. It is not the resistance at either point (the ratio ΔV/I is), and quoting it would lose the point.

5. Designing a nichrome heater element

A small heater needs a 12 Ω element made of nichrome wire of diameter 0.40 mm. What length of wire is required?

Area. r = 2.0×10⁻⁴ m, so A = π(2.0×10⁻⁴)² = 1.26×10⁻⁷ m².

L = RAρ = (12)(1.26×10⁻⁷)1.10×10⁻⁶ = 1.4 m

About 1.4 m, which is why heater elements are coiled. The same 12 Ω in copper of the same diameter would need about 90 m — nichrome’s resistivity is 65 times larger, and that, plus its stability when hot, is why it is used.

Checklist

• I can compute current from charge and time, and count electrons with N = ΔQ/e.
• I draw conventional current from + to − through the external circuit, and know electrons drift the other way.
• I can explain in words why drift is slow but bulbs light instantly, and why resistors heat up.
• I can use R = ρL/A with correct units and predict ratios when L or d change.
• I know metals’ resistance rises with temperature and can explain why using collisions.
• I can tell ohmic from non-ohmic on an I–V graph and find R at a point as ΔV/I.

2 · Electric power & energy

A battery gives every coulomb that passes through it some energy; a resistor takes that energy back and turns it into heat and light. Power is how fast that hand-off happens. Get the three forms of the power equation straight, and bulb-brightness questions — a guaranteed AP favourite — become one-line arguments.

Where electric power comes from

Potential difference is energy per unit charge: a charge q moving through ΔV changes its electric potential energy by qΔV. Current is charge per unit time. Multiply them and the coulombs cancel:

The battery delivers power P = Iε to the charges; the circuit elements dissipate it. Energy conservation says those two totals match — which is the loop rule you will meet in Kirchhoff’s rules, seen as a statement about power.

Three forms, and when to use which

For a resistor, substitute ΔV = IR into P = IΔV one way or the other:

Energy, kilowatt-hours & cost

Electricity companies bill in kilowatt-hours, a unit of energy (not power, despite the name):

Brightness is power

For identical-type bulbs, AP treats brightness as determined by the power dissipated in the bulb — not by its current alone, not by its voltage alone, and certainly not by its “wattage” label. The ranking procedure:

1. Find what the bulbs have in common (same current? same voltage?).
2. Pick the power form that keeps the common quantity and puts the difference in R.
3. Rank by power. If two bulbs are identical (same R), whichever has the larger current — equivalently the larger voltage — is brighter.

Rated power at rated voltage

A bulb stamped “60 W, 120 V” is making a conditional promise: if you put exactly 120 V across it, it will dissipate 60 W. The label is really telling you the bulb’s resistance:

Notice the direction: at the same rated voltage, the higher-wattage bulb has the lower resistance. That surprises people, and it is the key to the next section. Put any other voltage across the bulb and (treating R as fixed) the power scales with the square of the voltage:

The classic: 60 W and 100 W in series

Question. A 60 W and a 100 W bulb, both rated for 120 V, are connected in series to a 120 V supply. Which is brighter? Assume constant resistances.

Most people say the 100 W bulb — it is the “brighter bulb,” after all. It is the dimmer one here.

Resistances from the ratings (previous section): 240 Ω and 144 Ω.

Current (series, so the same in both):

Power in each — same I, so use I²R:

The 60 W bulb is brighter, and both are far dimmer than their labels, because each now gets only part of the 120 V. Connected in parallel instead, each bulb gets the full 120 V, both run at their ratings, and the 100 W bulb wins.

Worked examples

1. An electric kettle

A 1500 W kettle runs on 120 V. Find the current it draws, its resistance, and the energy it uses in 4.0 minutes, in joules and in kWh.

I = PΔV = 1500120 = 12.5 A      R = ΔVI = 12012.5 = 9.6 Ω Cross-check with the rated-power formula: 120²/1500 = 14 400/1500 = 9.6 Ω.

E = PΔt = (1500)(240 s) = 3.6×10⁵ J = 3.6×10⁵3.6×10⁶ kWh = 0.10 kWh Or directly: 1.5 kW × (4.0/60) h = 0.10 kWh.

12.5 A is a large current for a household circuit (typically fused at 15 A), which is why running a kettle and a toaster on the same outlet trips the breaker.

2. What a light bulb costs

A 10 W LED bulb replaces a 60 W incandescent bulb. Each is on for 6.0 hours a day for 30 days, and electricity costs $0.15 per kWh. Find the monthly cost of each.

Total time: (6.0 h/day)(30 days) = 180 h.

LED: (0.010 kW)(180 h) = 1.8 kWh → (1.8)($0.15) = $0.27

Incandescent: (0.060 kW)(180 h) = 10.8 kWh → (10.8)($0.15) = $1.62

The LED saves $1.35 a month per bulb — the same light for one-sixth of the energy, because the incandescent bulb turns most of its power into heat, not light.

3. Three forms, one answer

A 50 Ω resistor carries 0.20 A. Find the power three ways, and the energy dissipated in 5.0 minutes.

First the voltage: ΔV = IR = (0.20)(50) = 10 V.

IΔV = (0.20)(10) = 2.0 W     I²R = (0.040)(50) = 2.0 W     (ΔV)²/R = 100/50 = 2.0 W

E = (2.0 W)(300 s) = 600 J

If the three ever disagree on a problem, you have used a current or voltage that belongs to a different element of the circuit.

4. A bulb on half its rated voltage

A 60 W, 120 V bulb is connected to 60 V. Assuming constant resistance, find the current and the power.

Resistance from the rating: R = 120²/60 = 240 Ω.

I = 60240 = 0.25 A      P = 60²240 = 3600240 = 15 W Half the voltage gives half the current (0.25 A vs the rated 0.50 A), and power = IΔV drops by both halves: one quarter of 60 W.

A real filament would run a little above 15 W, because the cooler filament has less resistance than 240 Ω. The answer “30 W” is the trap.

5. The ranking flips: series vs parallel

A 2.0 Ω and a 4.0 Ω resistor are connected to a 12 V battery (negligible internal resistance), first in parallel and then in series. In each case, which dissipates more power?

Parallel — each has the full 12 V across it:

P₂ = 12²2.0 = 72 W      P₄ = 12²4.0 = 36 W

Series — same current, I = 12/(2.0 + 4.0) = 2.0 A:

P₂ = (2.0)²(2.0) = 8.0 W      P₄ = (2.0)²(4.0) = 16 W

The smaller resistor wins in parallel; the larger wins in series. Also notice the totals: 108 W in parallel against 24 W in series. Parallel connections draw more total power from the same battery, because the equivalent resistance is smaller.

Checklist

• I can derive P = IΔV from “volts are joules per coulomb, amps are coulombs per second.”
• I pick I²R for elements sharing a current and (ΔV)²/R for elements sharing a voltage, and say so in my justification.
• I can convert between joules and kWh and compute an electricity bill.
• I rank bulb brightness by power dissipated, not by current, voltage, or label alone.
• I can get a bulb’s resistance from its rating and know the higher-wattage bulb has lower R.
• I can explain, with no numbers, why a 60 W bulb outshines a 100 W bulb in series.

3 · Series & parallel circuits

Every rule on this page comes from two conservation laws. Charge cannot pile up in a wire, so current splits and rejoins without loss. Energy per charge depends only on position, so every path between the same two points drops the same voltage. Hold those two sentences and you can rebuild series and parallel from scratch in the middle of an exam.

Series: one path

Resistors are in series when there is exactly one path through them, with no junction in between. Two consequences follow, one from each conservation law.

• Same current through each (charge conservation). At steady state no charge can accumulate anywhere along a single path, so whatever flows into R₁ per second must flow out of it and into R₂. There is nowhere else for it to go.
• Voltages add (energy conservation). A charge q that crosses the whole chain loses qΔV₁ in the first resistor, qΔV₂ in the second, and so on. The total it loses must equal the energy the battery gave it, qΔV_total.

Put Ohm’s law into the energy statement and the common I cancels:

Parallel: same two nodes

Resistors are in parallel when both ends of each connect to the same two nodes. Again one consequence per conservation law:

• Same ΔV across each (energy conservation). Electric potential is a property of a location. Node a has one potential and node b has one potential, so every branch running from a to b has the same ΔV_ab, whatever is in it.
• Currents add (charge conservation). The current arriving at node a has to leave it, split among the branches: I = I₁ + I₂ + …

Write each current with Ohm’s law and the common ΔV cancels:

Voltage and current division

Two results fall straight out of the rules above, and they save a lot of arithmetic.

Series: the voltage divider

The current is shared, so voltage is handed out in proportion to resistance. For two resistors across a total ΔV:

Parallel: the current divider

The voltage is shared, so current goes in inverse proportion to resistance — most of it takes the easy road. For two branches carrying a total I:

Reducing a compound circuit

Real exam circuits mix both. The method is to collapse the circuit from the inside out — find a group that is purely series or purely parallel, replace it with one equivalent resistor, redraw, repeat — then rebuild outward to get each individual current and voltage.

The circuit: a 24 V ideal battery, R₁ = 4 Ω in the main line, then two branches — R₂ = 6 Ω alone, and R₃ = 2 Ω in series with R₄ = 10 Ω.

Now rebuild outward. The 3 A flows through R₁ and through the 4 Ω parallel block. Expand the block back into its branches using the shared voltage:

1. ΔV₁ = (3)(4) = 12 V. The parallel block gets the rest: ΔV_p = (3)(4) = 12 V. (Check: 12 + 12 = 24 ✓)
2. Branch currents: I₂ = 12/6 = 2 A; I₃₄ = 12/12 = 1 A. (Check: 2 + 1 = 3 ✓)
3. Inside the 12 Ω branch: ΔV₃ = (1)(2) = 2 V, ΔV₄ = (1)(10) = 10 V. (Check: 2 + 10 = 12 ✓)

Bulbs: removed, added, shorted

The most common qualitative circuit question on the AP exam. Bulbs are treated as identical resistors R, and brightness tracks power, P = I²R = (ΔV)²/R. For identical bulbs, the one with more current (equivalently more ΔV) is brighter.

Now change one thing at a time. The table gives each bulb’s power in units of ε²/R (ideal battery), but notice that every row can be predicted without numbers by asking what happened to the total resistance, and so to the current through A?

Summary table

Worked examples

1. Three resistors in parallel

Resistors of 2.0 Ω, 3.0 Ω, and 6.0 Ω are connected in parallel across a 6.0 V ideal battery. Find R_eq, the battery current, and the current in each resistor.

1R_eq = 12 + 13 + 16 = 3 + 2 + 16 = 1 Ω⁻¹  ⇒  R_eq = 1.0 Ω Smaller than the smallest (2.0 Ω) ✓.

Battery current: I = 6.0/1.0 = 6.0 A. Each resistor has the full 6.0 V across it: I = 6/2 = 3.0 A, 6/3 = 2.0 A, 6/6 = 1.0 A. Check: 3 + 2 + 1 = 6 A ✓.

2. Voltage divider

A 9.0 V battery is connected across a 1.0 kΩ and a 2.0 kΩ resistor in series. Find the current and the potential difference across each.

I = 9.03.0×10³ = 3.0×10⁻³ A = 3.0 mA

ΔV_1k = (3.0 mA)(1.0 kΩ) = 3.0 V; ΔV_2k = (3.0 mA)(2.0 kΩ) = 6.0 V. Or by division directly: 9.0 × 2/3 = 6.0 V. The larger resistor takes the larger share, and 3.0 + 6.0 = 9.0 V ✓.

3. Current divider

A current of 3.0 A arrives at a junction and splits between a 4.0 Ω and a 12 Ω resistor in parallel. How much goes through each?

I₄ = 3.0 · 124 + 12 = 2.25 A     I₁₂ = 3.0 · 44 + 12 = 0.75 A

Check via the shared voltage: R_p = 48/16 = 3.0 Ω, so ΔV = (3.0)(3.0) = 9.0 V; then 9.0/4 = 2.25 A and 9.0/12 = 0.75 A ✓. Three times the resistance, one third the current.

4. Rank the bulbs, then unscrew one

Three identical bulbs and an ideal battery ε: bulb A alone on one branch; bulbs B and C in series on a second branch; both branches directly across the battery. Rank the brightnesses. Then bulb B is unscrewed — what happens to A and C?

Each branch has the full ε across it. A gets all of it: P_A = ε²/R. B and C split it equally, ε/2 each: P = (ε/2)²/R = ε²/4R. A > B = C, with A four times as bright.

Unscrew B: its branch is now open, so C goes out. A still has ε across it and is unchanged. The battery current drops from ε/R + ε/2R = 1.5ε/R to ε/R, but that does not affect A. (With a real battery that has internal resistance, A would get very slightly brighter — less current means less voltage lost inside the battery. See emf & internal resistance.)

5. Shorting a bulb

Two identical bulbs, 1 and 2, are in series with an ideal battery. A switch connected directly across bulb 2 is then closed. Describe both bulbs and compare the power of bulb 1 before and after.

Before: R_eq = 2R, each bulb has ε/2: P₁ = ε²/4R.

After: the closed switch is a zero-resistance path in parallel with bulb 2. The potential difference across a zero-resistance wire is I · 0 = 0, and bulb 2 is connected across the same two nodes, so it has 0 V too — bulb 2 goes out. All the current takes the wire. Bulb 1 now has the full ε:

P_1,after = ε²R = 4 × P_1,before Double the voltage on a fixed resistance quadruples the power.

Checklist

• I can derive the series and parallel rules from charge and energy conservation, not just recite them.
• I identify series and parallel by tracing nodes, not by how the drawing looks.
• I know a parallel combination is smaller than its smallest resistor.
• I can reduce a compound circuit step by step and then rebuild it to find every current and voltage.
• I check my answers: branch currents add up, series voltages add up, total power equals εI.
• I can predict brightness changes when a bulb is removed, added, or shorted by asking what happened to R_eq.

4 · Kirchhoff’s rules

Series and parallel shortcuts run out the moment a circuit has two batteries in different branches. Kirchhoff’s rules never run out. They are the same two conservation laws you already used — charge and energy — written so that they apply to any circuit, one junction and one loop at a time.

The junction rule

This is conservation of charge. In a steady-state circuit, charge does not build up at a point in a wire: every coulomb per second that arrives at a junction must leave it. If it did not, charge would pile up there, its field would push back on the incoming current, and the circuit would not be steady.

You get one useful junction equation for each junction except one — with n junctions, the last equation is always a combination of the others. In a two-loop circuit with two junctions, that means exactly one junction equation.

The loop rule

This is conservation of energy. Electric potential is energy per unit charge, and it is a property of position. Carry a test charge around a closed path and back to the start: it ends at the same potential, so its electric potential energy is unchanged. Every volt gained in a battery must be spent in the resistors along that loop. That is why the rule is sometimes stated as “Σε = ΣIR around a loop.”

Walking a loop: sign conventions

Before writing any loop equation, draw and label a direction for every branch current. Guess if you have to — a wrong guess just produces a negative answer, meaning the current really flows the other way. Then pick a direction to walk around the loop and record each change in potential as you cross each element:

A full two-loop circuit

Three branches join a top node to a bottom node. Left: an 18 V battery and a 2.0 Ω resistor. Middle: a 6.0 Ω resistor. Right: an 8.0 V battery and a 4.0 Ω resistor. Both batteries have their + terminal toward the top. Find all three currents.

Step 1 — junction rule (top node)

Step 2 — loop 1 (left loop, clockwise from the bottom-left corner)

Up through ε₁, − to +: +18. Up through R₁, with I₁: −2I₁. Across the top, then down through R₃, with I₃: −6I₃.

Step 3 — loop 2 (right loop, clockwise from the bottom of the middle branch)

Up through R₃, against I₃: +6I₃. Across the top, then down through R₂, against the assumed I₂: +4I₂. Down through ε₂, + to −: −8.

Step 4 — solve

From loop 1: I₁ = 9 − 3I₃. From loop 2: I₂ = 2 − 1.5I₃. Substitute both into the junction equation:

Then I₁ = 9 − 3(2) = 3.0 A and I₂ = 2 − 1.5(2) = −1.0 A.

Step 5 — check by substituting back into every loop

emf & internal resistance

A real battery is an ideal emf ε in series with a small internal resistance r — the resistance of its own chemistry and electrodes. You can only connect to the two terminals outside, so you never get the full ε when current flows: part of it is used up pushing current through r.

Walk from b to a through the battery: +ε across the emf, then −Ir across the internal resistance (with the current):

Worked example

A battery with ε = 9.0 V and r = 0.50 Ω drives a 4.0 Ω load.

• Current: I = 9.0/(4.0 + 0.50) = 9.0/4.5 = 2.0 A.
• Terminal voltage: ΔV = 9.0 − (2.0)(0.50) = 8.0 V. Check: IR = (2.0)(4.0) = 8.0 V ✓.
• Power to the load: (2.0)²(4.0) = 16 W. Wasted inside the battery: (2.0)²(0.50) = 2.0 W. Total from the emf: εI = 18 W = 16 + 2 ✓.
• If the terminals were shorted (R = 0): I = ε/r = 18 A, terminal voltage 0, and all 162 W heat the battery — why shorting a battery is dangerous.

Ammeters and voltmeters

A quick ammeter illustration: a 10 V ideal battery and a 4.0 Ω resistor carry 10/4.0 = 2.5 A. Insert an ammeter with 1.0 Ω of resistance and it reads 10/5.0 = 2.0 A — a 20% error, because the meter’s resistance is not small compared with the circuit’s. Example 5 below does the voltmeter version.

Worked examples

1. Junction rule at a node

At a junction, 5.0 A flows in along one wire and 2.0 A and 1.5 A flow out along two others. A fourth wire also connects. Find its current and direction.

5.0 = 2.0 + 1.5 + I₄  ⇒  I₄ = 1.5 A, out Had the answer come out negative, the fourth current would flow in.

2. One loop, two opposing batteries

A single loop contains a 12 V battery, a 2.0 Ω resistor, a 4.0 V battery oriented to oppose the first, and a 6.0 Ω resistor. Find the current, and say what each battery is doing.

Assume the current flows the way the 12 V battery pushes it, and walk the loop in that direction: +12 − 2.0I − 4.0 − 6.0I = 0.

I = 12 − 4.02.0 + 6.0 = 1.0 A Positive, so the guessed direction is right. Opposing emfs subtract.

Power check: the 12 V battery supplies (12)(1.0) = 12 W. The resistors take (1.0)²(2.0) + (1.0)²(6.0) = 8.0 W. The 4.0 V battery has current forced into its + terminal, so it absorbs (4.0)(1.0) = 4.0 W — it is being charged. 12 = 8 + 4 ✓.

3. Potential difference between two points

In the two-loop circuit above, what is the potential of the top node relative to the bottom node? Show that the answer does not depend on the path.

Start at the bottom (call it 0 V) and walk up each branch, using the actual currents:

• Left: +18 across ε₁, then −(3.0)(2.0) = −6 across R₁ (with the current): 18 − 6 = 12 V.
• Middle: up through R₃, against the 2.0 A flowing down: +(2.0)(6.0) = 12 V.
• Right: +8 across ε₂, then up through R₂ against the 1.0 A flowing down: +(1.0)(4.0) = 4. Total 8 + 4 = 12 V.

Three paths, one answer. That path-independence is the loop rule — potential belongs to the point, not to the route.

4. Finding ε and r from two readings

With one load, a battery’s terminal voltage is 8.0 V while it delivers 2.0 A. With a different load it reads 7.0 V at 4.0 A. Find ε and r.

Write ΔV = ε − Ir twice:

8.0 = ε − 2.0r     7.0 = ε − 4.0r Subtract: 1.0 = 2.0r, so r = 0.50 Ω. Then ε = 8.0 + (2.0)(0.50) = 9.0 V.

Check the second reading: 9.0 − (4.0)(0.50) = 7.0 V ✓. Graphically, this is reading the slope and intercept of the ΔV-vs-I line above — the same analysis you would do with lab data.

5. A voltmeter that is not ideal

Two 10 kΩ resistors are in series with a 12 V ideal battery. A voltmeter with 10 kΩ internal resistance is connected across one of them. What does it read, and what “should” it read?

Without the meter: equal resistors split 12 V evenly — 6.0 V each.

With the meter: the meter and the resistor it touches form a parallel pair, (10)(10)/(10 + 10) = 5.0 kΩ. The circuit is now 10 + 5.0 = 15 kΩ:

I = 1215×10³ = 0.80 mA  ⇒  ΔV_meter = (0.80 mA)(5.0 kΩ) = 4.0 V Check: the other resistor has (0.80 mA)(10 kΩ) = 8.0 V, and 8.0 + 4.0 = 12 ✓.

It reads 4.0 V instead of 6.0 V. A voltmeter needs a resistance much larger than the element it measures — real digital meters are around 10 MΩ, a thousand times this one, and the error becomes negligible.

Checklist

• I can say which conservation law each Kirchhoff rule expresses, and why.
• I label a direction for every branch current before writing any equation.
• I get resistor signs from the current direction and battery signs from the terminal I meet first.
• I can set up and solve a two-loop circuit, and I interpret a negative current correctly.
• I check answers with an unused loop and with power balance.
• I can use ΔV = ε − Ir, and explain why terminal voltage drops as current rises.
• I know ammeters go in series (ideal: zero resistance) and voltmeters in parallel (ideal: infinite resistance), and what a real meter does to its reading.

5 · Capacitors in circuits (RC)

Everything else in this unit is steady: switch on, and the currents are instantly what they will always be. A capacitor adds time. It starts out letting current through as if it were a wire, fills up, and ends up blocking current completely. The whole topic is knowing those two endpoints and the shape of the curve between them.

Capacitors in series and parallel

Recall C = Q/ΔV: capacitance is charge stored per volt. The combination rules come from the same two conservation laws as for resistors — but they come out the other way round.

Parallel: same ΔV, charges add

Both capacitors connect the same two nodes, so each has the full ΔV (energy). The total charge the battery delivered is the sum of what each holds (charge):

Series: same Q, voltages add

The battery puts +Q on the left plate of C₁, which holds −Q on its partner. That partner, the wire, and the left plate of C₂ form an isolated conductor that started neutral and has no way to gain charge — so it carries +Q on C₂’s plate (charge). Every series capacitor holds the same Q, and the voltages add up to the battery’s (energy):

The two limits: wire and open circuit

Put a resistor R, an uncharged capacitor C, a battery ε, and a switch in one loop, and close the switch. Two moments are all the AP exam really needs:

Charging and the time constant

Between the limits, the reason for the curve shape is a feedback loop. As charge builds, ΔV_C rises, which leaves less voltage for the resistor, so the current falls, so charge arrives more slowly, so ΔV_C rises more slowly… The capacitor approaches full charge ever more gently and never quite gets there. That is an exponential.

Discharging

Now take a capacitor charged to ΔV₀ and connect it across a resistor with no battery. It acts as a temporary battery whose emf runs down as it empties. The charge, the voltage, and the current all decay together, with the same time constant τ = RC:

A capacitor inside a resistor network

The most common free-response setup: a capacitor buried in an otherwise ordinary resistor circuit, and questions about “immediately after the switch closes” and “a long time later.” Here ε = 12 V (ideal), R₁ = 4.0 Ω, R₂ = 8.0 Ω, and a branch with R₃ = 8.0 Ω in series with an initially uncharged C = 6.0 μF. A switch in the main line (not drawn) closes at t = 0.

Immediately after the switch closes

C is uncharged ⇒ wire. Then R₃ is simply in parallel with R₂: (8)(8)/(8 + 8) = 4.0 Ω, and the total is 4.0 + 4.0 = 8.0 Ω.

• Battery current: 12/8.0 = 1.5 A; ΔV₁ = (1.5)(4.0) = 6.0 V.
• The parallel part has 12 − 6.0 = 6.0 V, so R₂ and R₃ each carry 6.0/8.0 = 0.75 A. The capacitor’s initial charging current is 0.75 A.

A long time later

C is full ⇒ open circuit. No current in the R₃–C branch at all, which leaves R₁ and R₂ in series: 12.0 Ω.

• Battery current: 12/12 = 1.0 A, through R₁ and R₂; ΔV₂ = (1.0)(8.0) = 8.0 V.
• R₃ carries no current, so it has no potential difference across it. The capacitor branch is in parallel with R₂, so ΔV_C = 8.0 V — not 12 V.
• Charge: Q = CΔV = (6.0 μF)(8.0 V) = 48 μC. Energy: ½(6.0×10⁻⁶)(8.0)² = 1.9×10⁻⁴ J (192 μJ).

Energy stored

A capacitor stores energy in the electric field between its plates. Charging it from 0 to Q means moving each new bit of charge against a voltage that grows linearly from 0 to ΔV, so the average voltage is ΔV/2 — the energy is the triangular area under a Q-vs-ΔV graph:

Worked examples

1. Combining capacitors

C₁ = 3.0 μF and C₂ = 6.0 μF are in series across a 12 V battery. Find the charge on each and the voltage across each. Then C₃ = 4.0 μF is connected in parallel with the pair; find the new total charge drawn from the battery.

1C_s = 13.0 + 16.0 = 12.0  ⇒  C_s = 2.0 μF

Series ⇒ same charge: Q = (2.0 μF)(12 V) = 24 μC on each. Voltages: 24/3.0 = 8.0 V on C₁, 24/6.0 = 4.0 V on C₂. Check: 8.0 + 4.0 = 12 ✓. The smaller capacitor takes the larger voltage.

Adding C₃ in parallel: C_eq = 2.0 + 4.0 = 6.0 μF, Q_total = (6.0)(12) = 72 μC. C₃ holds (4.0)(12) = 48 μC; the series pair still holds 24 μC; 48 + 24 = 72 ✓.

2. Time constant and values at t = τ

An uncharged 20 μF capacitor is charged through a 50 kΩ resistor by a 9.0 V battery. Find τ, the initial current, and the capacitor voltage and current one time constant after the switch closes.

τ = RC = (50×10³)(20×10⁻⁶) = 1.0 s     I₀ = 9.050×10³ = 0.18 mA

At t = τ: ΔV_C = 0.632 × 9.0 = 5.7 V and I = 0.368 × 0.18 mA = 0.066 mA. Check with the loop rule: the resistor has (0.066 mA)(50 kΩ) = 3.3 V, and 5.7 + 3.3 = 9.0 V ✓. The capacitor is essentially full (>99%) after 5τ = 5.0 s.

3. Bulbs and a capacitor — what you see

Identical bulbs A and B are in series with an ideal battery ε. An uncharged capacitor is connected directly across bulb B. The switch closes. Describe both bulbs immediately after, and a long time after.

Immediately after: the capacitor acts as a wire across B, so B is shorted — B is dark. A has the full ε: P_A = ε²/R, bright.

Long after: the capacitor is an open circuit, so it is as if it were not there: A and B in series, each with ε/2 and power ε²/4R. A dims to a quarter of its initial power; B brightens from nothing until the two are equal. In between, both change smoothly with time constant set by the circuit.

Contrast: a single bulb in series with the capacitor does the opposite of B — it flashes bright at the instant the switch closes and fades to dark as the capacitor fills.

4. Energy bookkeeping while charging

A 100 μF capacitor is charged fully by a 12 V battery through a resistor. Find the energy stored, the work done by the battery, and the thermal energy produced in the resistor.

Final charge: Q = Cε = (100×10⁻⁶)(12) = 1.2×10⁻³ C.

U_C = ½Qε = ½(1.2×10⁻³)(12) = 7.2×10⁻³ J Same as ½Cε² = ½(100×10⁻⁶)(144) = 7.2 mJ ✓.

Battery work: Qε = (1.2×10⁻³)(12) = 14.4 mJ. Thermal energy in the resistor: 14.4 − 7.2 = 7.2 mJ — exactly half, for any value of R.

5. Discharging a capacitor

A 470 μF capacitor charged to 10 V is connected across a 2.0 kΩ resistor. Find τ, the initial current, and the voltage, current, and stored energy after one time constant.

τ = (2.0×10³)(470×10⁻⁶) = 0.94 s     I₀ = 102.0×10³ = 5.0 mA

At t = τ, both have fallen to 37%: ΔV_C = 3.7 V, I = 1.8 mA (check: 3.68 V / 2.0 kΩ = 1.84 mA ✓).

Energy: initially ½(470×10⁻⁶)(10)² = 23.5 mJ. At t = τ, ½(470×10⁻⁶)(3.68)² = 3.2 mJ — only 14% remains, not 37%, because energy goes as (ΔV)²: (0.368)² = 0.135. The other 20.3 mJ has become thermal energy in the resistor.

Checklist

• I can derive the series and parallel capacitor rules and say why they are reversed relative to resistors.
• I treat an uncharged capacitor as a wire at t = 0 and a charged one as an open circuit at steady state.
• I can sketch ΔV_C, Q, and I against time for both charging and discharging.
• I know τ = RC, its units, and how changing R or C changes the curves.
• I can find the steady-state charge on a capacitor buried in a resistor network.
• I can compute stored energy three ways and know that charging through a resistor wastes half the battery’s work.

6 · Formula sheet & constants

Everything the unit uses, with the condition each one needs to be legal.

Constants worth memorizing

7 · Unit-wide traps

• “Current gets used up.” It does not. The current leaving a bulb equals the current entering it; what the bulb uses is energy. See Section 1.
• A battery is not a constant-current source. It sets a potential difference (its emf, minus Ir); the circuit decides the current. Add a parallel branch and the battery’s current goes up.
• Choosing the wrong power form. “Bigger R, more power” is true in series and false in parallel. Name the shared quantity first — Section 2.
• Swapping the resistor and capacitor rules. Capacitors in parallel add; in series they combine by reciprocals — the opposite of resistors.
• Meters in the wrong place. An ammeter goes in series (and has nearly zero resistance); a voltmeter goes in parallel (and has nearly infinite resistance). An ammeter placed in parallel is a short circuit.
• Reading the wrong slope. On an I-vs-ΔV graph the slope is 1/R; on a ΔV-vs-I graph it is R.

8 · Labs for this unit

Every lab for this unit lives in Interactive Laboratories, each one paired with the section above that explains it.

Interactive lab Ohm’s law & resistivity Open it in Interactive Laboratories → Interactive lab Series vs parallel bulbs Open it in Interactive Laboratories → Interactive lab Two-loop circuit solver Open it in Interactive Laboratories → Interactive lab RC charging & discharging Open it in Interactive Laboratories → Bench lab Resistivity of a wire Open it in Interactive Laboratories →