Advertisements
Advertisements
Question
An electric bulb of resistance 500 Ω draws current 0.4 A from the source. Calculate:
- the power of bulb and
- the potential difference at its end.
Solution
- Resistance of electric bulb (R) = 500Ω
current drawn from the source (I) = 0.4 A
Power of the bulb (P) = VI
V = I × R
V = 0.4 × 500
V = 200 V - The potential difference at its end is 200 V.
Hence,
Power (P) = VI
P = 200 × 0.4
P = 80 W
The power of the bulb is 80 Watt.
APPEARS IN
RELATED QUESTIONS
A 2 kWh heater, a 200 W TV and three 100 W lamps are all switched on from 6 p.m. to 10 p.m. What is the total cost at Rs 5.50 per kWh?
At a given time, a house is supplied with 100 A at 220 V. How many 75 W, 220 V light bulb could be switched on in the house at the same time (if they are all connected in parallel)?
State whether an electric heater will consume more electrical energy or less electrical energy per second when the length of its heating element is reduced. Give reasons for your answer.
A bulb marked 12 V, 24 W operated on a 12 volt battery for 20 minutes. Calculate:
(i) the current flowing through it, and
(ii) the energy consumed
What would happen and why, to an electric bulb when it is connected across a supply of voltage (i) lower (ii) higher than its proper rating?
Three 250 W heaters are connected in parallel to a 100 V supply, Calculate the resistance of each heater.
The resistance of filament of an electric heater is 500 Ω It is operated at 200V for 1 hour daily. Calculate the current drawn by the heater and the energy consumed in kWh, by the heater in a month of 30 days.
Unit of electric power may also be expressed as:
