How Do You Calculate the Voltage Drop Across a 100 Watt Light Bulb?

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To calculate the voltage drop across a 100-watt light bulb, it is essential to understand the relationship between power, voltage, and current. A 100-watt bulb dissipates 100 joules of energy per coulomb of charge, leading to the conclusion that the voltage drop is 100 volts. This is derived from the equation where power (in watts) equals voltage (in volts) multiplied by current (in amperes). Therefore, if the bulb dissipates 100 watts, the current is 1 ampere, confirming that the voltage drop is indeed 100 volts. Understanding these fundamental electrical principles is crucial for solving related problems in electrical engineering.
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Right now I'm a first year electrical engineering student at Purdue University. I have a problem that I have to do for a homework assignment I read the entire chapter and cannot find out how to do this problem. Any help would be appreciated, if this isn't the right place to ask this question just let me know and if not I would appreciate any information on how I could find help.

A 100 watt light bulb dissipates 100 joules of energy per coulomb of charge. Find its voltage drop.

Thanks in advance.
 
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call me dense but I don't see the relationship there, I don't see how those equations relate to the problem given. Most of those deal with current and resistance, please explain. sorry I'm kinda slow when it comes to learning stuff and I want to make sure I fully understand this.

edit: So what you're saying is that 1Watt = J/C? so the equation would read 100w = 100J/1C = 100V?
 
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I was showing that 1 V = 1 J/C and so 100 V = 100 J/C.

Now, 1 W = 1 J/s, and if 100 W dissipates 100 J/C, that implies (100 W)/(100 J/C) = (100 J/s)/(100 J/C) = 1 C/s = 1 A.

So for a 100 W light bulb to dissipate 100 J/C, the voltage drop is 100 V, and the current is 1 A.


See also - http://science.howstuffworks.com/question501.htm
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