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CED objectives

Compound Direct Current (DC) Circuits

▶︎  Watch it animatedinteractive step-through · ~3 min · optional ⚙︎  Open the appletCompound Lab · two boxes drawn side by side in either wiring, coloured node dots, a loaded battery and a meter you can put in the wrong place

Elements are series or parallel according to the nodes they connect, not the way they are drawn: parallel means both ends touch the same pair of nodes, series means one shared node that nothing else touches. Series resistances add and parallel resistances combine through a sum of reciprocals whose last step is a flip, so a parallel combination always lands below its smallest branch. A parallel addition therefore lowers $R_{\text{eq}}$ and raises the battery current. A series potential difference divides in proportion to the resistances, and a real battery's terminal value is $\varepsilon - Ir$, matching the printed emf only at zero current.

Six errors dominate. Classifying elements from the layout rather than the nodes, so redrawing the circuit changes the answer. Adding parallel resistances directly, or doing the reciprocal arithmetic and forgetting the final inversion. Assuming any added resistor raises the total resistance, when a parallel addition opens a path and lowers it. Splitting a series voltage evenly regardless of the resistances, or handing the larger share to the smaller resistor. Using the printed emf as the terminal potential difference at any load. And clipping an ammeter across an element or wiring a voltmeter into a path, each of which destroys the circuit being measured.

one circuit, two layouts, the same node list R1 R2 R3 node b node a R1 R2 R3 node b node a parallel = both ends touch the SAME pair of nodes · series = one shared node nothing else touches “R2 is drawn above R3, so they are parallel” reads the page instead of the connections if redrawing changes your answer, the first reading came from the layout
The two panels are the same circuit. Labelling every stretch of plain wire as one node is what makes that visible.
one current prices both drops; and the source itself sags 2 Ω 10 Ω I = 1 A through both 2 V 10 V drops are in the SAME 1:5 ratio as the resistances 6 V each is right only when the resistances match V(term) I ε at I = 0 slope = −r V(term) = ε − Ir: the printed value appears only at zero current which is why more parallel bulbs each glow a little less brightly
Both panels are about the same current. It divides the series drops on the left, and it is what pulls the terminal reading down on the right.

The work

3 ways in · any order
Lesson
Compound Direct Current (DC) Circuits ›

Classifies elements by the nodes they touch, keeps the final flip in the parallel calculation, makes a parallel addition lower the total resistance, divides a series voltage by resistance, and places meters by what they sample.

Skill check · 10 scenarios
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Diagnostic
10-item topic check ›

Ten items spanning the failure modes of this topic: reading series and parallel off the drawing, summing parallel resistances straight or stopping before the flip, treating every addition as more opposition, splitting a series voltage evenly, using the printed emf under load, and swapping the two meters. Take it cold to find which one is yours, or after the lesson to confirm it is not.

Not started · 10 items · ~15 min
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Targeted Practice
Drill a single misconception ›

Pick one of the failure modes you missed and drill it on its own. The round is adaptive: two correct in a row clears it for now and moves you to the next. Two in a row is a checkpoint, not proof: if the error resurfaces later, the misconception comes back.

Take the diagnostic to identify your misconceptions
Diagnostic first