{ "format": "recallweave-deck/1", "title": "DC circuits: current, connections and power", "attribution": "Original classroom scenarios, questions, explanations and diagrams authored for RecallWeave (2026). Physics background: OpenStax, University Physics Volume 2, sections 9.1, 9.4, 9.5 and 10.2. Worked solutions and source links accompany this course.", "license": "Original course content created for this repository; no textbook passages, figures or exercises are reproduced. The cited OpenStax reference book has its own CC BY-NC-SA 4.0 terms. No new repository-wide license is declared.", "concepts": [ "Current and resistance", "Series connections", "Parallel connections", "Energy and power" ], "items": [ { "id": "dc-current-1", "concept": "Current and resistance", "prerequisites": [], "prompt": "An ideal source maintains 15 V across a fixed 5 Ω ohmic resistor. What steady current passes through the resistor?", "options": [ "75 A", "3 A", "0.33 A", "10 A" ], "answer": 1, "explanation": "For the stated fixed ohmic resistor, I = V/R = 15/5 = 3 A. Multiplying voltage by resistance gives 75 with the wrong units for current. Reversing the quotient gives about 0.33, and subtracting the two values does not express the current–voltage relationship.", "transfer": "Keep the resistor fixed and change the source to 10 V. Predict the current and state which assumption lets you use the same resistance." }, { "id": "dc-current-2", "concept": "Current and resistance", "prerequisites": [], "prompt": "A resistor remains ohmic with the same resistance. Its applied voltage changes from 4 V to 8 V. What happens to the steady current through it?", "options": [ "It halves.", "It stays fixed.", "It doubles.", "It quadruples." ], "answer": 2, "explanation": "With R unchanged, I = V/R is proportional to V. Doubling the applied voltage doubles the current. This conclusion depends on the fixed-resistance assumption; it does not describe every electrical component or a resistor whose temperature changes its resistance.", "transfer": "If both the voltage and the resistance double, how does the current change? Explain using their ratio." }, { "id": "dc-current-3", "concept": "Current and resistance", "prerequisites": [], "prompt": "A constant current of 0.75 A passes one cross-section of a wire for 8 s. What net amount of charge passes that cross-section during this interval?", "options": [ "6 C", "0.094 C", "10.7 C", "8.75 C" ], "answer": 0, "explanation": "One ampere is one coulomb per second. With constant current, Q = It = 0.75 × 8 = 6 C. Dividing current by time or time by current reverses the required relationship; adding values with different units does not give charge. This counts charge passing one location, not charge created by the source.", "transfer": "At the same constant current, how long would 3 C take to pass? Distinguish charge from the energy transferred as charge moves through a voltage difference." }, { "id": "dc-series-1", "concept": "Series connections", "prerequisites": [ "Current and resistance" ], "prompt": "A 4 Ω resistor R1 and an 8 Ω resistor R2 form a single series loop with an ideal 12 V source and zero-resistance wires. What is the voltage drop across R2 in steady state?", "options": [ "4 V", "6 V", "12 V", "8 V" ], "answer": 3, "explanation": "The series resistance is 4 + 8 = 12 Ω, so the loop current is 12/12 = 1 A. Therefore V2 = IR2 = 1 × 8 = 8 V. The other resistor drops 4 V and the drops sum to 12 V. Equal division would require equal resistances; both resistors cannot each have the full source voltage in this series loop.", "transfer": "Swap the physical positions of R1 and R2 without changing the series connection. Which resistor still has the larger voltage drop, and what happens to the total current?" }, { "id": "dc-series-2", "concept": "Series connections", "prerequisites": [ "Current and resistance" ], "prompt": "Two resistors, 3 Ω and 9 Ω, are in one unbranched loop carrying a steady current. The current immediately before the first resistor is 0.5 A. What is the current in the wire between them?", "options": [ "0.125 A", "1.5 A", "0.5 A", "0.375 A" ], "answer": 2, "explanation": "In a steady unbranched circuit, the same charge per second passes each cross-section: the current between the resistors is 0.5 A. Resistors transfer electrical energy, but they do not consume charge. Unequal resistances produce unequal voltage drops at this shared current, not different steady currents along the loop.", "transfer": "For this current, calculate each resistor's voltage drop. Why can the energy transferred per coulomb differ while the charge per second remains the same?" }, { "id": "dc-series-3", "concept": "Series connections", "prerequisites": [ "Current and resistance" ], "prompt": "An ideal 18 V source drives fixed 3 Ω and 6 Ω resistors in series. Replace only the 6 Ω resistor with a 15 Ω resistor, leaving an unbranched loop and ideal wires. What is the new steady source current?", "options": [ "1 A", "2 A", "6 A", "1.2 A" ], "answer": 0, "explanation": "The new series resistance is 3 + 15 = 18 Ω. The new current is 18 V / 18 Ω = 1 A. The original current was 18/(3 + 6) = 2 A, so it falls throughout the loop. Dividing by either resistor alone overlooks the other resistor's voltage drop.", "transfer": "At the same 18 V, replace the 15 Ω resistor with 9 Ω instead. Find the current and check that the two voltage drops add to the source voltage." }, { "id": "dc-parallel-1", "concept": "Parallel connections", "prerequisites": [ "Current and resistance" ], "prompt": "Fixed 6 Ω and 3 Ω resistors are connected in parallel directly across an ideal 12 V source. What is the total steady current supplied to the two branches?", "options": [ "1.33 A", "2 A", "4 A", "6 A" ], "answer": 3, "explanation": "Each branch has 12 V across it. The currents are 12/6 = 2 A and 12/3 = 4 A; the source supplies their sum, 6 A. Adding the resistances as 9 Ω would describe a series connection. Either branch current alone leaves out the other branch.", "transfer": "Find the single resistance that would draw the same total current from 12 V. Explain why it is smaller than either branch resistance." }, { "id": "dc-parallel-2", "concept": "Parallel connections", "prerequisites": [ "Current and resistance" ], "prompt": "A 9 Ω resistor and an 18 Ω resistor are connected between the same two ideal wires across an ideal 18 V source. What is the voltage across the 18 Ω resistor?", "options": [ "6 V", "18 V", "9 V", "12 V" ], "answer": 1, "explanation": "Both resistors connect to the same two nodes, so each has the source's 18 V potential difference. Their currents differ: 2 A through 9 Ω and 1 A through 18 Ω. A voltage divider describes a series arrangement, not these shared parallel nodes.", "transfer": "Redraw the two resistors in different positions while keeping both connected to the same two nodes. What determines their parallel relationship: their position on the page or their electrical connections?" }, { "id": "dc-parallel-3", "concept": "Parallel connections", "prerequisites": [ "Current and resistance" ], "prompt": "An ideal 10 V source keeps 10 V across two parallel branches: R1 = 5 Ω and R2 = 10 Ω. Replace R2 with 20 Ω while leaving R1 and the ideal source unchanged. What happens to R1's steady current?", "options": [ "It stays at 2 A.", "It falls to 1 A.", "It rises to 4 A.", "It rises to 3 A." ], "answer": 0, "explanation": "R1 still has 10 V across 5 Ω, so its current stays 2 A. R2's current falls from 1 A to 0.5 A and the total source current falls from 3 A to 2.5 A. The ideal source maintains voltage; it does not maintain a fixed total current that must be redistributed between branches.", "transfer": "Would this conclusion automatically hold if the source's terminal voltage fell as its load changed? Identify the assumption you would need to revisit." }, { "id": "dc-power-1", "concept": "Energy and power", "prerequisites": [ "Current and resistance" ], "prompt": "A fixed 12 Ω ohmic resistor carries a steady current of 0.5 A. At what rate does it convert electrical energy to thermal energy in this ideal resistor model?", "options": [ "6 W", "24 W", "3 W", "0.25 W" ], "answer": 2, "explanation": "The resistor's voltage is IR = 0.5 × 12 = 6 V. Its power is VI = 6 × 0.5 = 3 W, equivalently I²R = 0.25 × 12 = 3 W. The value 6 is the voltage, not the power. Power is energy per second; it is different from the amount of charge passing per second.", "transfer": "If the current doubles while resistance stays fixed, by what factor does the resistor's power change? Explain the square in I²R." }, { "id": "dc-power-2", "concept": "Energy and power", "prerequisites": [ "Series connections", "Parallel connections" ], "prompt": "Two identical fixed 8 Ω resistors are first connected in series, then reconnected in parallel to the same ideal 12 V source. By what factor does the total steady power supplied to the pair change (parallel power divided by series power)?", "options": [ "× 1/2", "× 4", "× 2", "× 1" ], "answer": 1, "explanation": "In series, the pair is 16 Ω and draws P = V²/R = 144/16 = 9 W. In parallel, the equivalent resistance is 4 Ω and the total power is 144/4 = 36 W. The ratio is 36/9 = 4. The source voltage is held fixed in both arrangements; a fixed-current source would be a different comparison.", "transfer": "For two identical resistors of any fixed resistance R at the same nonzero ideal voltage V, derive the power ratio symbolically before choosing numbers." }, { "id": "dc-power-3", "concept": "Energy and power", "prerequisites": [ "Parallel connections" ], "prompt": "Fixed 3 Ω and 6 Ω resistors are connected in parallel across an ideal 6 V source. Ideal wires dissipate no power. What total steady power must the source supply?", "options": [ "6 W", "12 W", "3 W", "18 W" ], "answer": 3, "explanation": "The branch currents are 6/3 = 2 A and 6/6 = 1 A. Their powers are 6 × 2 = 12 W and 6 × 1 = 6 W. The source supplies 18 W in total, also found from 6 V × (2 + 1) A. In this steady model, that supplied electrical power equals the combined resistor dissipation.", "transfer": "If the same values persisted for 5 s, how much energy would the source transfer? Check that the energies dissipated in the two resistors add to the same total." } ] }