**PLEASE HELP! 2 questions**
1) In a lightning bolt, 30C of charge moves through a potential difference of 10^8V in 0.002 seconds. A) What is the current in the bolt? B) What is the total energy of the bolt? **for A, I got an answer of (1.5x10^3)A .. Wondering if the answer to B will just be (30x10^8)J ???
2) An iron, designed for use in the United States where outlets provide 120V and 5A. The iron is taken to Europe and connected to an outlet that supplies 240V. A) What is the current it will draw there? B) What effect will this have on the iron?
* No clue how to do #2.

Question

**PLEASE HELP! 2 questions**
1) In a lightning bolt, 30C of charge moves through a potential difference of 10^8V in 0.002 seconds. A) What is the current in the bolt? B) What is the total energy of the bolt?  **for A, I got an answer of (1.5x10^3)A .. Wondering if the answer to B will just be (30x10^8)J ???
2) An iron, designed for use in the United States where outlets provide 120V and 5A. The iron is taken to Europe and connected to an outlet that supplies 240V. A) What is the current it will draw there? B) What effect will this have on the iron?
* No clue how to do #2.

anonymous · Answer

1)Current= Rate of flow of charge = charge/time.
Voltage= Work done to move the charge in the given potential difference/ charge. 
So you gotta think, we do the work but the where does the work go? 
It is stored as energy in the bolt.
That means, Energy of the bolt = Voltage * Charge. 

2) With the given outlet values, you can find the resistance, which will remain constant no matter where you take it. 
So, for A, you'll know the resistance and voltage. Find the current using ohm's law i.e., r=V/i 
B, I'm not very sure. You should check this answer. The power drawn from the iron will be more as Power(joule's heating, precisely) = i^2 * r. As i increases, power must also increase. Hence, more heat will be generated in the iron. 
 Hope this helps :)

anonymous · Answer

In 1) I said "we do the work", but in a lightning bolt, "we" technically are not doing any work. But it still works the same way :)

anonymous · Answer

Okay, I understand 1a and 2a & b... But for 1b, the amount of charge is 30C and the potential difference is 10^8V. It doesn't make sense that the total energy would be equal to (30x10^8) J??? Am I understanding the problem incorrectly? @Nivi

anonymous · Answer

Do you understand integral calculus? I can probably help you understand this better that way.

anonymous · Answer

Yes ma'am!! :) @Nivi

anonymous · Answer

Lol,alright!
You know that Power=current * Voltage. 
Power is rate at which energy is transferred. Integrate power relative to time and you get energy!
|dw:1392867272463:dw|
substitute the values of current and voltage. You'll get this. 
|dw:1392867343440:dw|
Solve the definite integral. 
And you'll get 3 * 10^9 J
That's how you actually get the value. 
It's okay if it's such a large value. It's a lightning bolt travelling through a large potential difference in a very short time. It's perfectly okay for it to take this value.