Rating of a bulb is 12 W, 6V. How it can be run on a 12 V supply line?
(Calculate the value of resistance to be connected with the bulb)
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The answer to the first question is to drop the voltage so that the lamp gets 6V across it rather than the full 12V.
12W/6V=2A, so we need a resistor value that drops 6V and will result in 2A of current. 6V/2A=3 Ohm.
However, in practice, you'll need a very beefy resistor, as the power dissipation would be 2^2A*3Ohm=12W. That's a lot. It's better to invest in a 6V battery or a buck converter
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