Current electricity / concept toolkit
A formula becomes useful when we know its conditions. Use this reference, attempt the six questions, then check and retry.
| Relationship | When and how to use it |
|---|---|
| V = IR | For an ohmic resistor with resistance R under the stated conditions. |
| P = VI = I²R = V²/R | Power dissipated in a resistor. At fixed V, power decreases as R increases. At fixed I, power increases as R increases. |
| Rs = R₁ + R₂ + … | Series: the same current flows through every resistor. |
| 1/Rp = 1/R₁ + 1/R₂ + … | Parallel: the same potential difference acts across every branch. |
| R = ρL/A | Uniform wire. If volume AL and resistivity ρ stay constant, R is proportional to L². |
| Rrated = Vrated²/Prated | A lamp’s rated resistance. Treating it as constant is an approximation for incandescent filaments, not a model for arbitrary LED lamps. |
V: potential difference (V) · I: current (A) · R: resistance (Ω) · P: power (W)
ρ: resistivity (Ω·m) · L: wire length (m) · A: cross-sectional area (m²)
Identify the source. Work out Req. State what stays constant. Decide whether the question asks for total power or one component’s power.
An ideal battery keeps its voltage fixed. An ideal current source keeps its total current fixed. They do not lead to the same power comparison.
Use the pack: questions on pages 2–3 · worked solutions on page 4 · error review and retry record on page 5. These are original practice questions, not claimed past-paper questions.
Practice / fixed voltage
Show your reasoning and units. Use an ideal voltage source and constant resistor values in questions 1–3.
Two identical resistors consume 60 W in total in series across an ideal battery. They are reconnected in parallel across the same battery. Find the new total power and the power in each resistor.
What stays constant? _________ Rs = _________ Rp = _________
Three identical resistors consume 30 W in total when all are in series across an ideal battery. Find the total power when all three are in parallel across that same battery.
Derive the power ratio before using the given 30 W.
A 3 Ω and a 6 Ω resistor are connected first in series and then in parallel across an ideal 12 V battery. Find total power in each connection. Is the parallel-to-series power ratio four?
Calculate both equivalent resistances. State why the equal-resistor shortcut does or does not apply.
Practice / changing the conditions
The source, the material and the connection tell us what can be held constant.
A uniform wire is stretched uniformly to twice its length without changing its volume or resistivity. It originally dissipates 20 W across an ideal constant-voltage source. Express the new resistance in terms of the original R, and find its new power.
Start with volume AL. What happens to the cross-sectional area?
Two incandescent lamps rated 100 W, 220 V and 60 W, 220 V are connected in series across 220 V. Use the constant rated-resistance approximation. Which lamp dissipates more power? Calculate both powers and state one limitation of the model.
Find each rated resistance first. Which quantity is shared in series?
Two identical resistors dissipate 60 W in total in series across an ideal current source. They are reconnected in parallel to the same source, which maintains the same total current. The resistor values stay constant. Find the new total power.
Compare your reasoning with question 1. Is voltage still fixed?
Worked solutions / check after attempting
Compare each step with your own working. Record the first point where your reasoning differed.
Rs = 2R and Rp = R/2. At the same V, P = V²/Req.
Pp/Ps = (2R)/(R/2) = 4. Total power = 240 W.
The equal parallel branches share total power equally: 120 W each.
Rs = 3R; Rp = R/3. Power ratio = (3R)/(R/3) = 9.
New total power = 9 × 30 = 270 W. For n identical resistors at the same ideal voltage, the ratio is n².
Rs = 3 + 6 = 9 Ω; Rp = (3 × 6)/(3 + 6) = 2 Ω.
Ps = 12²/9 = 16 W; Pp = 12²/2 = 72 W.
Ratio = 4.5. The factor of four requires two equal resistors.
AL is unchanged. If L′ = 2L, then A′ = A/2.
R′ = ρ(2L)/(A/2) = 4R. At constant V, P′ = P/4.
New power = 5 W. Holding current fixed would give a different result.
R100 = 220²/100 = 484 Ω; R60 = 220²/60 ≈ 806.67 Ω.
Series current I = 220/(484 + 806.67) ≈ 0.17045 A.
P100 = I²R100 ≈ 14.06 W; P60 ≈ 23.44 W.
The 60 W-rated lamp dissipates more power in this model. Real filament resistance changes with temperature, so these numerical powers are approximate.
The source keeps I fixed; its terminal voltage changes. Use P = I²Req.
Pp/Ps = (R/2)/(2R) = 1/4.
New total power = 60/4 = 15 W. The fixed-voltage shortcut from question 1 does not apply.
Close the loop / record and retry
Choose questions that exposed a gap. Cover the solutions and retry them on another day.
| Question / first attempt | First incorrect step Concept · condition · calculation |
Correction in our own words | Retry date / result |
|---|---|---|---|
On __________, at __________,
we will retry question(s) __________
with the solutions covered.
What can we now explain independently?
What still needs another attempt?
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