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"A 60 W bulb and a 100 W bulb are connected in series to the mains. Which glows brighter?" Most students say the 100 W bulb, and most students are wrong. The trick is that a power rating describes the bulb under one condition only. This post shows how to turn ratings into resistances, then into actual power, with worked numbers.
What does a power rating really tell you?
A bulb marked 100 W, 220 V uses 100 W only when it has 220 V across it. Using P = V²/R from NCERT Class 12 Chapter 3, Current Electricity, the rating fixes the resistance:
R = V²/P.
The same chapter notes that this I²R power is what heats a bulb's coil until it glows. Brightness follows the power the bulb actually dissipates in the circuit, not the number printed on it. The JEE (Main) 2026 syllabus lists electrical energy and power and series and parallel combinations of resistors, and the JEE (Advanced) 2026 brochure lists "Heating effect of current". Check your year's syllabus.
Worked example 1: 60 W and 100 W in parallel
Both bulbs are rated at 220 V.
R(60 W) = 220²/60 = 48 400/60 ≈ 807 Ω.
R(100 W) = 220²/100 = 484 Ω.
In parallel across 220 V, each bulb gets its rated voltage, so each uses its rated power: 60 W and 100 W. The 100 W bulb is brighter. This is how bulbs in a house are connected.
Worked example 2: the same bulbs in series
Now connect them in series across 220 V.
Total resistance: 807 + 484 ≈ 1291 Ω.
Current: I = 220/1291 ≈ 0.170 A, the same in both.
Power in the 60 W bulb: I²R = 0.170² × 807 ≈ 23.4 W.
Power in the 100 W bulb: I²R = 0.170² × 484 ≈ 14.1 W.
The 60 W bulb is now brighter. In series, the same current flows through both, so P = I²R favours the larger resistance, and the lower-rated bulb has the larger resistance. Both are dimmer than in example 1, and the total, about 37.5 W, is less than either rating.
Worked example 3: a bulb on a lower voltage
A bulb rated 100 W, 220 V is run on 110 V. Its resistance is 484 Ω, so P = 110²/484 = 12 100/484 = 25 W. Halving the voltage quarters the power, because P depends on V². Two identical 100 W bulbs in series across 220 V give the same result: each gets 110 V and uses 25 W.
How accurate are these numbers?
These examples assume each bulb keeps the resistance it has at full brightness. A real filament is cooler when it runs dim, and NCERT section 3.8 shows that a metal's resistivity falls as it cools. So a dim bulb has a lower resistance than R = V²/P suggests, and the actual powers differ somewhat from these figures. The comparison of which bulb glows brighter still holds in typical exam problems, which make this assumption. Our post on the limitations of Ohm's law explains why a lamp is non-ohmic.
Why do students get this wrong?
Engelhardt and Beichner (2004) developed a diagnostic test on direct-current resistive circuits for high school and university students. They found that students tend to hold several misconceptions even after instruction. The idea that a battery is a constant source of current was the one used most often in interviews, and students tended to focus on current and confuse its properties with those of voltage and resistance. "The 100 W bulb always uses 100 W" is the same kind of error: treating a rating as fixed instead of working out what the circuit gives the bulb.
Which mistakes cost marks?
Assuming the rated power applies in every circuit. It applies only at the rated voltage.
Using P = V²/R in series with the supply voltage. Each bulb gets only part of the voltage; use I²R with the shared current.
Rounding resistance too early. Keep 807 Ω, not 800 Ω, until the end.
Forgetting which bulb has more resistance. Lower rating at the same voltage means higher resistance.
A practice routine
Find R for 25 W, 40 W, 60 W and 100 W bulbs rated at 220 V.
Put each pair in series across 220 V and decide which glows brighter before calculating.
Check with I²R, then repeat in parallel.
Then try mixed networks using our guide to combination of resistors.
If series and parallel still blur together, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning rather than printing an answer.
Next step: predict, then calculate, which glows brighter when a 40 W and a 100 W bulb, both rated at 220 V, are in series.
