Current Electricity · Question-type atlas

Power &
heating

Three algebraically identical formulas that behave in opposite ways depending on the circuit. Twenty-three question types, sorted A to F and ranked — plus the three results where the intuitive answer is the wrong one.

The counter-intuitive one

Which bulb actually glows brighter?

Two bulbs, both rated for the supply voltage. Switch between series and parallel and watch which one lights up — the answer reverses, and most students guess it wrong in series.

rated 40 W

20.4 W actual

R = 1210 Ω

rated 100 W

8.2 W actual

R = 484 Ω

Total power drawn
28.6 W
Current in the circuit
0.130 A
Combined-rating formula
P₁P₂/(P₁+P₂)

In series the current is common, so P = I²R and the higher-resistance bulb wins. The 40 W bulb has the higher resistance, so it burns brighter than the 100 W one.

One fork governs the topic

Before reaching for any formula, ask which quantity the two components share. That single decision fixes whether power rises or falls with resistance, and almost every question in Groups A to C turns on it.

P = VI = I²R = V²/R SERIES current I is common P = I²R P ∝ R — bigger R wins PARALLEL voltage V is common P = V²/R P ∝ 1/R — smaller R wins for a rated bulb R = V²/P, so a LOWER wattage bulb has HIGHER resistance series ⟹ low-wattage brighter · parallel ⟹ high-wattage brighter

The three that catch people

Each of these has an intuitive answer and a correct answer, and they are not the same. Examiners know it, which is why all three recur.

You expect: the 100 W bulb is brighter

In series, the 40 W bulb wins

Same current through both, so P = I²R — and the low-wattage bulb is the high-resistance one. A 40 W bulb in series with a 100 W bulb dissipates roughly 20 W against 8 W.

You expect: maximum power means maximum efficiency

At Pmax, efficiency is only 50%

Maximum power transfer needs R = r, and then η = R/(R+r) = ½. Half the energy is burnt inside the cell. Maximum efficiency instead wants R as large as possible — a different condition entirely.

You expect: half the voltage gives half the power

Half the voltage gives a quarter

At fixed resistance P ∝ V², so P′ = P(V′/V)². A 100 W, 220 V bulb on a 110 V supply draws 25 W, not 50 W.

Formula sheet

Everything above reduces to these.

The three forms
P = VI = I²R = V²/RUse I²R when current is shared, V²/R when voltage is shared.
Joule heating
H = I²Rt = Pt
H(cal) = I²Rt / 4.2
1 kWh = 3.6 × 10⁶ J
Rated device
R = V² / PLower wattage means higher resistance.
Non-rated voltage
P′ = P (V′ / V)²
Combined ratings
series: P = P₁P₂ / (P₁ + P₂)
parallel: P = P₁ + P₂The series form has the reciprocal structure — opposite to resistances.
Maximum power transfer
R = r ⟹ Pmax = ε² / 4rEfficiency there is exactly 50%.
Efficiency of a cell
η = R / (R + r) = V / ε
Heating a liquid
P t = m c ΔT ⟹ t = m c ΔT / P
Heater combinations
series: t = t₁ + t₂
parallel: t = t₁t₂ / (t₁ + t₂)Heating times behave exactly like resistances.