Bulbs and Heaters: Power Rating and Series-Series Problems
Physics · Current Electricity · NEET
A rating like "100 W, 220 V" means the bulb gives 100 W ONLY at 220 V. Its true fixed property is resistance: R = V^2/P = 220^2/100 = 484 ohm. Memory hook: "Rating gives R; R never changes, power does." Once you find R from the rating, every series/parallel or voltage-drop problem is just P = I^2 R or P = V^2/R with that fixed R.
The same two bulbs swap brightness. In series (common current) power is I^2 R, so the higher-resistance 60 W bulb is brighter. In parallel (common voltage) power is V^2/R, so the lower-resistance 100 W bulb is brighter.
Your doubts, answered
Does a bulb always give exactly its rated power?
No. The rating (say 100 W, 220 V) is only true at the stated voltage 220 V. The bulb's real, fixed property is its resistance R = V^2/P. If you run it at a different voltage or put it in series with something, the current changes, so the actual power = I^2 R changes. The label is a promise at one voltage, not a constant.
Why does a 100 W bulb glow dimmer than a 60 W bulb when they are in series?
Higher wattage means LOWER resistance: R = V^2/P, so R_100 = 220^2/100 = 484 ohm and R_60 = 220^2/60 = 807 ohm. In series both carry the SAME current I. Power = I^2 R, so the bulb with the bigger R (the 60 W one) dissipates more power and glows brighter. This flips the usual expectation, and NEET loves it.
In series, which bulb is brighter: the one with high wattage or low wattage?
The LOW-wattage bulb. In series, current is common, so P = I^2 R and the higher-resistance (lower-wattage) bulb wins. In parallel the voltage is common, so P = V^2/R and the LOW-resistance (higher-wattage) bulb glows brighter. Remember: series -> low watt bright; parallel -> high watt bright.
Two heaters, one strong one weak, joined in series to boil water fastest?
Neither individually. In series the current is small (resistances add), so total power V^2/(R1+R2) is LESS than either heater alone. Heaters give maximum combined heat in PARALLEL, where each still gets full mains voltage. Series always reduces power below the smallest rated value.
If mains voltage drops, does the heater still consume rated power?
No. Resistance R stays fixed (P is proportional to V^2). If a 400 W, 220 V heater runs at 200 V, new power = (200/220)^2 x 400 = 331 W. Lower voltage means less power and less heat. This exact case appeared in NEET 2026.
⚠️ The NEET trap ✗ Students assume a 100 W bulb always gives 100 W, and that in series the 100 W bulb is brighter because it is 'more powerful'. ✓ Convert the rating to a FIXED resistance R = V^2/P first. In series, current is common, so P = I^2 R and the higher-R (lower-wattage) bulb is brighter. The 100 W bulb (lower R) actually glows dimmer in series. 🧠 Series -> low watt is bright; Parallel -> high watt is bright. The rating only holds at its own voltage.
Real NEET questions
NEET 2016
A filament bulb (500 W, 100 V) is to be used in a 230 V supply. A resistance R is connected in series so it works perfectly at 500 W. The value of R is:
A · 230 ohm
B · 46 ohm
C · 26 ohm ✓
D · 13 ohm
Solution: Step 1: For the bulb to run at 500 W it must get 100 V and its rated current. Rated current I = P/V = 500/100 = 5 A. Step 2: The extra voltage the series resistor must drop = 230 - 100 = 130 V. Step 3: Same current flows through R, so R = V_R/I = 130/5 = 26 ohm. Answer: 26 ohm.
NEET 2024
Two heaters A and B rated 1 kW and 2 kW are connected first in series and then in parallel to a fixed source. The ratio of power outputs (series : parallel) is:
A · 2 : 9 ✓
B · 1 : 2
C · 2 : 3
D · 1 : 1
Solution: Step 1: R is proportional to 1/P (R = V^2/P). Take R1 = V^2/1 and R2 = V^2/2, so R1 = 1 and R2 = 1/2 in units of V^2. Step 2 (series): R_s = R1 + R2 = 3/2, so P_s = V^2/R_s = 2/3. Step 3 (parallel): 1/R_p = 1/R1 + 1/R2 = 1 + 2 = 3, so P_p = V^2/R_p = 3. Step 4: Ratio = (2/3) : 3 = 2 : 9. Answer: 2 : 9.
NEET 2026
A room heater is rated 400 W, 220 V. If the supply voltage drops to 200 V, the power consumed (approximately) is:
A · 200 W
B · 400 W
C · 331 W ✓
D · 121 W
Solution: Step 1: Resistance is fixed, so with P = V^2/R, power is proportional to V^2. Step 2: New power = (V_new/V_rated)^2 x P_rated = (200/220)^2 x 400. Step 3: (200/220)^2 = (0.909)^2 = 0.826, so P = 0.826 x 400 = 331 W. Answer: about 331 W.
Solved Current Electricity NEET PYQs
Try the real previous-year questions from this chapter — each with the answer and a full solution.
What is the fixed quantity in a bulb: power or resistance?
Resistance. Find it once from the rating using R = V^2/P (for example 484 ohm for a 100 W, 220 V bulb), then use that R in every problem. Power and current change with the circuit; R does not (ignoring temperature effects).
What is the formula to find a bulb's resistance from its rating?
R = V^2/P, where V is the rated voltage and P is the rated power. Example: a 60 W, 220 V bulb has R = 220^2/60 = 807 ohm.
In series, does the low-power or high-power bulb glow brighter?
The low-power (high-resistance) bulb. Current is common in series, so P = I^2 R makes the higher-R bulb brighter.
Do heaters give more heat in series or parallel?
Parallel. In parallel each heater gets full mains voltage, so total power V^2/R_p is maximum. In series the resistances add, current drops, and total power falls below even the weaker heater.
How does power change if the supply voltage falls?
Since R is fixed, P is proportional to V^2. If voltage drops to a fraction f of rated, power becomes f^2 times rated. A 20/220 drop (about 9%) cuts power by roughly 17%, as in the NEET 2026 heater question.
