An incandescent electric bulb converts 20 % of its power consumption into light, and the remaining power is dissipated as heat. The bulb’s filament has a resistance of 200 Ω and 2 A current flows through it. If the bulb remains ON for 10 h and the rate of electricity charge is ₹5 per unit, then which among the following is the correct amount for the money spent on producing light?
- (a)₹ 5
- (b)₹ 6
- (c)₹ 7
- (d)₹ 8
Correct — D, ₹8. Power drawn by the filament P = I²R = (2)² × 200 = 800 W = 0.8 kW. Energy used in 10 h = 0.8 kW × 10 h = 8 kWh = 8 units. Total cost = 8 units × ₹5 = ₹40. Only 20% of this goes into light, so the money spent on producing light = 20% of ₹40 = ₹8.
- (a)₹ 5 — This is just the cost of 1 unit; it ignores the 8 kWh total energy and the 20% light fraction.
- (b)₹ 6 — Does not follow from P = 800 W, 8 kWh, a ₹40 total and a 20% light share (which give ₹8).
- (c)₹ 7 — No combination of the given data yields ₹7; the light cost is 20% of ₹40 = ₹8.
Electrical power dissipated in a resistor is P = I²R (equivalently VI or V²/R). Energy = power × time, billed in kilowatt-hours (1 unit = 1 kWh). Cost = energy (in units) × rate. An extra step here splits the energy: only 20% becomes light and the other 80% is wasted as heat, so the 'light' cost is 20% of the total bill.
Chain the steps: power (I²R) -> energy in kWh (× time) -> total cost (× rate) -> light's share (× 20%). Watch the units: 800 W = 0.8 kW, so 10 hours gives exactly 8 kWh. The 20%/80% split is the twist that separates 'money spent on light' from the full bill.
- Power in a resistor: P = I²R = 2² × 200 = 800 W = 0.8 kW.
- Energy = P × t = 0.8 kW × 10 h = 8 kWh (8 units).
- 1 unit of electricity = 1 kWh; total cost = 8 × ₹5 = ₹40.
- Only 20% is converted to light -> 0.20 × ₹40 = ₹8.
- Forgetting to convert watts to kilowatts (800 W = 0.8 kW) before multiplying by hours.
- Reporting the full bill (₹40) instead of the 20% spent specifically on light.
Asked as a numerical: compute power/energy/cost from I, R, time and tariff, sometimes with an efficiency fraction.
No directly related past PYQ was found.
- practice — not a real PYQ
An electric heater draws 5 A at 220 V for 2 hours. The energy consumed is
- (a)1.1 kWh
- (b)2.2 kWh
- (c)4.4 kWh
- (d)11 kWh
Answer(b) 2.2 kWh — P = 220 × 5 = 1100 W = 1.1 kW; energy = 1.1 × 2.
- practice — not a real PYQ
The commercial unit of electrical energy, the 'unit', is equal to
- (a)1 watt
- (b)1 joule
- (c)1 kilowatt-hour
- (d)1 volt
Answer(c) 1 kilowatt-hour — 1 unit = 1 kWh.