Three formulas that are all the same formula, and one question type — which bulb glows brighter — that catches more students than anything else in the chapter.
NCERT §3.9 · fully in syllabus · frequent
Part 1 · The idea, told simply
Why does a wire get hot when current flows through it? Where does the energy come from, and where does it go?
Story track
Picture children sliding down a long slide. At the top they have height — potential energy. At the bottom they have none. If the slide were perfectly smooth and frictionless, all that height would turn into speed: they would arrive at the bottom moving very fast.
But a real slide is rough. The children keep bumping and rubbing along the way, so they arrive at the bottom moving at a steady modest speed, not faster and faster. The energy did not vanish — it went into warming the slide.
A resistor is the rough slide. Electrons fall from high potential to low potential, and the energy they lose does not turn into ever-increasing speed, because they keep colliding with the metal ions. Every collision hands energy to the ions, which vibrate harder. Vibrating harder is being hotter. That is Joule heating.
Maths track · deriving the power formula
NCERT §3.9 sets it out in three lines. A charge ΔQ = IΔt falls through a potential difference V, so it loses potential energy:
ΔU = ΔQ · V = I V Δt
If there were no collisions, this would all become kinetic energy and the electrons would accelerate. But we know from Topic 3 that the drift speed is steady, so the kinetic energy is not increasing. The energy is instead handed to the atoms at every collision, and the conductor heats up:
ΔW = I V Δt P = ΔW/Δt = V I
Now bring in Ohm's law V = IR, and the one formula becomes three:
P = VI = I²R = V²/R
These three are algebraically identical. Choosing the right one is not about correctness — it is about which quantity you already know and which one is common to the components you are comparing. That choice is the entire skill of this topic.
Animation 1 · Where the energy goes
Electrons fall through the resistor. Each collision hands energy to an ion — and the ions get visibly hotter.
Part 2 · The bulb question
Almost every power question in NEET is a disguised version of this: two bulbs of different wattage are connected together — which is brighter? The answer flips depending on how they are connected, and that flip is the whole trap.
The two-step method — use it every single time
Step 1. Convert every rating into a resistance. A bulb marked "100 W, 220 V" does not always consume 100 W. That is only what it consumes at exactly 220 V. What the bulb actually is, permanently, is a resistance:
R = V²_rated / P_rated Higher wattage ⇒ LOWER resistance
Step 2. Ask what is common.
SERIES → current is common → use P = I²R → P ∝ R → the HIGHER-resistance bulb glows brighter → so the LOWER-wattage bulb wins
PARALLEL → voltage is common → use P = V²/R → P ∝ 1/R → the LOWER-resistance bulb glows brighter → so the HIGHER-wattage bulb wins
In parallel, the rating tells the truth: a 100 W bulb does beat a 60 W bulb. In series, the rating lies: the 60 W bulb glows brighter. Every household appliance is wired in parallel, which is exactly why our intuition is trained the wrong way for series questions.
Animation 2 · The same two bulbs, wired two ways
A 100 W and a 60 W bulb on a 220 V supply. Switch between series and parallel and watch which one brightens.
Why the low-wattage bulb has the high resistance
This feels backwards until you see the numbers. On a 220 V supply:
Bulb rating
Resistance R = V²/P
In series (P ∝ R)
In parallel (P ∝ 1/R)
100 W, 220 V
484 Ω
14.1 W — dimmer
100 W — brighter
60 W, 220 V
806.7 Ω
23.4 W — brighter
60 W — dimmer
A high-wattage bulb is designed to draw a lot of current at its rated voltage, and drawing a lot of current means offering little resistance. So high power rating means thick, low-resistance filament. Put it in series with a thinner filament and it is the thin one that takes the punishment.
Running a bulb at the wrong voltage
The resistance is fixed by the bulb. So if the supply voltage changes, the power follows the square of the voltage:
P_actual = P_rated × (V_actual / V_rated)²
Halve the voltage and the power drops to a quarter — not a half. A 100 W, 220 V bulb on a 110 V supply consumes only 25 W.
Animation 3 · Power follows the square of the voltage
Slide the supply voltage. The bulb's resistance never changes — but its brightness collapses fast.
Combining wattages directly
There is a shortcut worth having. For bulbs all rated at the same voltage, the ratings themselves combine — but note that they combine the opposite way round from resistances, because P is inversely proportional to R:
So a 100 W and a 60 W bulb in series consume 37.5 W between them — less than either would alone. In parallel they consume 160 W.
Part 3 · Heat, energy and cost
Heat produced: H = I²Rt = VIt = (V²/R)t in joules
Energy in kWh: E = P(kW) × t(hours)
1 kWh = 1000 W × 3600 s = 3.6 × 10⁶ J
Cost = number of units (kWh) × rate per unit
One "unit" on an electricity bill is one kilowatt-hour. This is a unit of energy, not power — a distinction NEET tests directly.
Why power is transmitted at high voltage
Maths track · NCERT §3.9
A power P must be delivered to a distant device through cables of resistance R_c. The current needed is I = P/V, and the power wasted heating the cables is
P_c = I²R_c = (P/V)² R_c = P²R_c / V²
The waste is inversely proportional to the square of the transmission voltage. Deliver 100 kW through cables of 10 Ω at 200 V and you lose 2.5 MW — impossible, far more than you are sending. Do it at 20 000 V and you lose only 250 W.
That is why transmission lines carry the high-voltage danger signs seen outside towns, and why a transformer at the far end steps the voltage back down to something safe to use.
Animation 4 · Transmission loss against voltage
Same power delivered, same cables. Only the transmission voltage changes.
Animation 5 · The energy meter
Power is a rate; energy is the total. Watch the units accumulate and the bill grow.
Trap · the six that cost marks
Assuming a 100 W bulb always consumes 100 W. It does so only at its rated voltage.
Saying the higher-wattage bulb is brighter in series. It is the dimmer one there.
Using P = V²/R for series components. In series the current is common, so use I²R.
Halving the supply voltage and halving the power. Power goes as V², so it quarters.
Adding wattages for a series combination. In series the reciprocals add.
Treating the kWh as a unit of power. It is energy — power multiplied by time.
Part 4 · Formula sheet
Power — the three faces of one formula
P = VI (always true)
P = I²R (use when CURRENT is common → series)
P = V²/R (use when VOLTAGE is common → parallel)
Total power from a cell: P = εI
Power lost inside a cell: P = I²r
Bulb ratings
Resistance from a rating: R = V²_rated / P_rated
Current at rated voltage: I = P_rated / V_rated
Higher wattage ⇒ LOWER resistance ⇒ thicker filament
At a different supply: P_actual = P_rated (V_actual/V_rated)²
Half the rated voltage → one quarter of the rated power
Which bulb glows brighter
SERIES (I common): P ∝ R → higher R wins → LOWER-wattage bulb is brighter
PARALLEL (V common): P ∝ 1/R → lower R wins → HIGHER-wattage bulb is brighter
In series, the bulb with the SMALLEST rating glows brightest
In parallel, the bulb with the LARGEST rating glows brightest
Combining ratings (same rated voltage)
Parallel: P_total = P₁ + P₂ + …
Series: 1/P_total = 1/P₁ + 1/P₂ + …
n identical bulbs of rating P, in series across rated V:
total = P/n, each bulb gets P/n²
n identical bulbs of rating P, in parallel across rated V:
total = nP, each bulb gets its full P
Heat and energy
H = I²Rt = VIt = (V²/R)t (joule)
1 kWh = 3.6 × 10⁶ J
Units consumed = P(kW) × t(hours)
Cost = units × rate
Heating water: Q = mcΔT, time t = mcΔT/P
Transmission
P_c = I²R_c = P²R_c/V²
Loss ∝ 1/V² → raise V tenfold, cut the loss a hundredfold
This is why long-distance lines run at very high voltage
Sanity checks
Series combination consumes LESS power than either bulb alone
Parallel combination consumes MORE than either alone
Adding a bulb in parallel raises the total power drawn
Adding a bulb in series lowers it
Power at half voltage = a quarter, never a half
Part 5 · 50 NEET-pattern questions with full solutions
Includes 4 graph-based questions, 3 assertion–reason questions, and 12 questions tagged Bulb — the dominant sub-family. Year labels are not attached: the patterns are authentic, the wording is mine.