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PhysicsCurrent Electricity
A room heater is rated 400 W, 220 V. If the supply voltage drops to 200 V, what will be the power consumed (approximately)?
Options
1
200 W
2
400 W
3
331 W
4
121 W
Correct Answer
Option 3 : 331 W
Step-by-Step Solution
1

Find resistance of heater:

Using rated values: P = V²/R

R = V²/P = (220)²/400 = 48400/400 = 121 Ω

2

New power at V' = 200V:

P' = V'²/R = (200)²/121 = 40000/121

P' = 330.6 W ≈ 331 W ✓

✓

Quick ratio method:

P' = P × (V'/V)² = 400 × (200/220)² = 400 × (10/11)²

= 400 × 100/121 = 40000/121 ≈ 331 W ✓

Theory: Electrical Power & Appliance Ratings
1. Three Forms of Electrical Power

P = VI

P = V²/R (use when V and R known)

P = I²R (use when I and R known)

All three are equivalent — derived from Ohm's law (V = IR). For a fixed resistor, P ∝ V² and P ∝ I². This problem uses P = V²/R because the heater's resistance is fixed and we're changing voltage.

2. Appliance Ratings

Every electrical appliance has a rated power (P_rated) and rated voltage (V_rated). These are the values at which the appliance is designed to operate optimally. From the rated values, we can calculate the resistance: R = V²_rated / P_rated. This resistance is assumed constant (good approximation for small voltage changes).

📌 R = V²_rated / P_rated (resistance from rating)

📌 P_actual = V²_actual / R (power at actual voltage)

📌 P_actual = P_rated × (V_actual/V_rated)²

📌 P ∝ V² for fixed resistance — key relationship!

3. Effect of Voltage Change on Power

Since P = V²/R, power scales with the square of voltage. A 10% drop in voltage causes a 19% drop in power (since (0.9)² = 0.81). In this problem: voltage drops by 200/220 = 10/11 ≈ 9.1%. Power drops by (10/11)² = 100/121 ≈ 17.4%. New power = 400 × 0.826 ≈ 331 W.

4. Heating Effect of Current (Joule's Law)

Heat generated by a current-carrying conductor: H = I²Rt = P×t = V²t/R. This is the basis of electric heaters, incandescent bulbs, electric irons, and fuses. Joule's law states that heat produced is proportional to: square of current (I²), resistance (R), and time (t).

H = I²Rt = V²t/R = Pt (Joule's Law)

5. Appliances in Series vs Parallel

📌 Parallel (household wiring): Each appliance gets full voltage. Higher-rated appliance draws more current.

📌 Series: All appliances share the same current. Higher resistance appliance dissipates more power (P = I²R).

📌 In series: Lower-rated (higher R) bulb glows brighter — counterintuitive!

📌 In parallel: Higher-rated (lower R) bulb glows brighter — normal expectation.

6. Fuse Rating

A fuse is a thin wire that melts when current exceeds the safe limit. Fuse rating = maximum safe current. For a 400W, 220V heater: I = P/V = 400/220 = 1.82A. Use a fuse rated 3A (next standard size above 1.82A) for safety. Fuses protect against short circuits and overloading.

Frequently Asked Questions
1. Why is P = V²/R and not P = V/R? ⌄
From Ohm's law: I = V/R. Power P = VI = V × (V/R) = V²/R. It comes from substituting Ohm's law into P = VI. Similarly P = I²R comes from P = VI = (IR) × I = I²R.
2. Why does resistance stay constant when voltage changes? ⌄
For ohmic conductors (metals at roughly constant temperature), resistance is constant regardless of voltage or current. For a heater with small voltage change (220→200V), assuming constant R is a good approximation. This assumption gives the "approximately" in the question.
3. What is the ratio method for this problem? ⌄
P'/P = (V'/V)². So P' = P × (V'/V)² = 400 × (200/220)² = 400 × (100/121) = 40000/121 ≈ 331 W. This ratio method is faster than finding R first — use it in NEET to save time.
4. What is the resistance of this heater? ⌄
R = V²/P = 220²/400 = 48400/400 = 121 Ω. This is a clean answer (121 = 11²) because 220 = 11 × 20 and 400 = 20². In NEET, clean numbers often signal you're on the right track.
5. What is the current drawn at rated voltage? ⌄
I = P/V = 400/220 = 20/11 ≈ 1.82 A. Or I = V/R = 220/121 = 20/11 ≈ 1.82 A. At 200V: I' = 200/121 ≈ 1.65 A. The current also decreases when voltage drops.
6. Which bulb is brighter when 60W and 100W bulbs are in series? ⌄
60W bulb! In series, same current flows through both. R_60W = V²/P = 220²/60 ≈ 807Ω (higher R). R_100W = 220²/100 = 484Ω. P = I²R → higher R → more power dissipated. 60W bulb glows brighter in series — counterintuitive but correct!
7. What is Joule's heating effect? ⌄
When current I flows through resistance R for time t, heat generated H = I²Rt joules. This is used in heaters, irons, electric stoves. It is also the cause of power loss in transmission lines (P_loss = I²R_line), which is why high voltage (low current) transmission is preferred.
8. Why is power transmission done at high voltage? ⌄
P_loss = I²R_line. To transmit same power P = VI at high voltage, current I = P/V is small. Smaller current → much smaller I²R losses. Doubling transmission voltage → halves current → reduces power loss by factor 4. This is why power grids use 11kV, 33kV, 220kV lines.
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