Electric Circuits

Eight topics on charge in motion through wires and what a battery, a resistor and a capacitor each do to it. Electric current as a drift far slower than the signal that starts it, simple circuits, where current is not used up and a battery fixes voltage rather than current, resistance and resistivity as geometry times material with Ohm's law a property some devices have, electric power computed from each element's own V and I, compound circuits reduced by how things are connected rather than how they are drawn, Kirchhoff's loop rule with signs read from the traversal, Kirchhoff's junction rule, where a negative solved current is an answer and not an error, and RC circuits, where the capacitor's voltage is continuous, tau = RC uses the resistance the capacitor actually sees, and the exponential never finishes.

AP exam 15-25%8 topics

Topics

11.1

Electric Current

Simple Circuits

Resistance, Resistivity, and Ohm's Law

Electric Power

Compound Direct Current Circuits

Kirchhoff's Loop Rule

Kirchhoff's Junction Rule

Resistor-Capacitor (RC) Circuits

Lesson

Diagnostic

after diagnostic

Key forms For every problem in this unit

Current

I = dQ/dt, in amperes (C/s). Conventional direction is the way POSITIVE charge would move, from high potential to low outside the source

Drift speed

I = n q A v(d). Carriers drift at mm/s; the field that starts them arrives at nearly c. At fixed I, a FATTER wire means SLOWER drift

Current density

J = I/A = n q v(d), and J = σE = E/ρ inside the conductor

Resistance from geometry

R = ρL/A. Longer is MORE, thicker is LESS. ρ is the material's, R is this piece's

Ohm's law

V = IR defines R for any element; an OHMIC one keeps R fixed as V changes. Its I-V graph is a straight line through the origin with slope 1/R

Ideal wires

zero resistance, zero voltage drop: every point on one unbroken wire is the SAME node at the same potential

Power

P = IV = I²R = V²/R, using THIS element's own I and V. Never a battery voltage on a resistor that does not have it across it

Energy vs power

energy = P × t, joules. A kilowatt-hour is 3.6 × 10⁶ J of ENERGY, not a power

Bulb brightness

rank by each bulb's own power, recomputed after every change. Counting components ranks nothing

EMF and internal resistance

terminal voltage V = ε − Ir. Sags under load, equals ε only at zero current. Power to the load peaks when R = r

Series

same current through each; R(eq) = R₁ + R₂ + ...; voltages add and split in PROPORTION to R, never evenly

Parallel

same voltage across each; 1/R(eq) = 1/R₁ + 1/R₂ + ...; R(eq) is SMALLER than the smallest branch. Two: R₁R₂/(R₁ + R₂)

Adding a branch

a new parallel path LOWERS the total resistance and RAISES the total current. Never reduce across a junction as if series

Topology

series = one path, no junction between; parallel = both ends share nodes. Decided by the CONNECTIONS, not by the drawing

Loop rule

Σ ΔV = 0 around any closed loop. Resistor: −IR traversed WITH the current, +IR against it. Battery: + crossing − to +, whatever the current does

Junction rule

Σ I(in) = Σ I(out) at every node. A NEGATIVE solved current is a correct answer running the other way: keep the arrow, keep the sign

Meters

ammeter in SERIES, near-zero R; voltmeter in PARALLEL, near-infinite R. Neither may disturb what it measures

Capacitor endpoints

V(C) is continuous: t = 0 uncharged means a WIRE, long time means an OPEN branch with I = 0. Re-solve the circuit at each endpoint

RC time constant

τ = RC, using the resistance the capacitor actually SEES from its own terminals, not the nearest resistor

Charging

Q = Cε(1 − e^(−t/RC)), I = (ε/R) e^(−t/RC). At t = τ it is 63% of the way, at 5τ about 99%. It never FINISHES

Discharging

Q = Q₀ e^(−t/RC), I = (Q₀/RC) e^(−t/RC). Each τ multiplies what is left by 1/e; the initial slope extended reaches zero at exactly τ

Energy in charging

battery supplies Cε²; capacitor banks Cε²/2; the other half is heat in R, whatever R is

Unit 11 tools

Challenge bank

1 / 60

60 open-ended problems.

Read the question, work it out, then flip the card to compare your reasoning to the worked solution. Mark each card so you can return to the ones that still bite.

Question

Tap card to reveal explanation

Worked solution

Tap card to return to question

Cumulative assessment

Test the unit.

Twenty mixed items drawn from across all 8 topics, with guaranteed misconception-code coverage. Identifies which misconceptions still bite when you cannot see which topic the question came from.

Begin assessment →

Check what stuck.

Units 8 through 11, drawn evenly so earlier units get the same share as this one. Twenty questions or a full 42-question section, your choice. Even coverage means this is a retention check rather than a score estimate.

20 or 42questions

21topics

76codes covered

Begin assessment →