This week's CR4 Challenge Question:
A light bulb is connected to the power line by a coiled wire. You insert an iron cylinder inside the coil of the coiled wire. How the light of the light bulb will be affected? Now you repeat this experiment, but this time you power the light bulb with a very powerful battery. What happens to the light of the light bulb?
Disclaimer: This is a theoretical challenge for discussion only – DO NOT ATTEMPT THIS EXPERIMENT! It is potentially dangerous.
And The Answer is....
In the first case, the light bulb will dim. Because the light bulb is connected to an AC source (the line voltage) the iron cylinder will produce a variable magnetic field which will produce a a voltage drop in the coil. Because, the power source is constant (110 V rms), by having a voltage drop in the coil, the voltage across the light bulb will diminish.
In the second case, the power source is DC voltage. The dc current flowing through the coil will not produce a variable magnetic field. Without a variable magnetic field in the coil no voltage drop will develop across it. Therefore, in this case the brightness of the light bulb remains unchanged.
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