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Four identical bulbs, each of same rating (100 W, 220 V) are connected across an ideal battery of emf
550 volts. Which of the 4 bulbs will have a voltage across it, which is greater than voltage rating. (i.e.
which of them will fuse)
Razvoltage division
Answers
P= W/t
W = P x t. The energy supplied to the circuit by the source in time t is
E = P x t.
The S.I unit of electrical energy is joule and we know that for commercial purposes we use a bigger unit of electrical energy which is called “kilowatt-hour”. One kilowatt-hour is the amount of electrical energy consumed when an electrical appliance has a power rating of 1 kilowatt and is used for 1 hour. 1 kilowatt-hour = 1000 w x 3600 sec = 3600000 joules = 3.6 x 106 J
Power – Voltage Rating of Electrical Appliances: If we look at a particular bulb in our home it may have the figures 220 V, 100 W written on it. Now 220 V means that this bulb is to be used on a voltage of 220 volts and 100 W which means it has a power consumption of 100 watts or 100 joules per second.