The power rating of an CFL is 11 W when operated at 220 V. What is the maximum value of safe current flowing through it?
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Answers
Explanation:
At least two numbers are needed to answer the question because both the watts and voltage are ratings and not actual measurements.
The formula is:
Current = watts / volts
Therefore, if the light bulb is connected to a 220 volt source the equation is 60 watts / 220 volts = 0.2727 amperes (current).
However, since the actual voltage is not specified, we can look at any other voltage so that there are an infinite number of answers.
But, just to make it interesting, let's try a source voltage of 120 NOW it would seem like 60W / 120V = 0.5A. But, if you understand how light bulbs are rated, you will realize that that the bulb will not be producing 60 watts. Instead, it will be producing much less wattage. In order to calculate the actual current it is now necessary to find the electrical resistance of the light bulb.
Now the formula is: Resistance = voltage squared / watts, or 220V x 220V / 60W = 807 ohms (about)
And, the new current calculation result is: A=V/R or 120V/807ohms = 0.1847 amperes. It is also only emitting 120Vx0.1487A= about 18 watts!
Does this make sense?
The abbreviations above are:
V=VOLTS
A=AMPERES (current)
R=RESISTANCE (ohms)
W=WATTS (power in light and mostly heat)
❥A᭄ɴsᴡᴇʀ࿐:
GIVEN ,
POWER=P=11W,
POTENTIAL DIFFERNCE=V=220V
WE KNOW THAT
P=VI {HERE , I STANDS FOR CURRENT }
I=P/V
I=11/220
I=1/20
I=0.05A
max value of Current is 0.05 {say, 1} Ampere