# In a typical lightning strike, 2.5 C flows from cloud to ground in 0.20 ms.

1. Feb 23, 2010

### csimon1

1. The problem statement, all variables and given/known data

In a typical lightning strike, 2.5 c flows from cloud to ground in 0.20 ms. What is the current during the strike?

2. Relevant equations

I = Q/t

3. The attempt at a solution

The reason I keep getting this wrong is because it says I am rounding something off wrong. I am very confused. What I keep getting is 2.5 C/0.0002 s = 12500 A. Anyone know what I'm doing wrong?

2. Feb 23, 2010

### sylas

It looks correct to me; although if you are given data with accuracy of one significant figure, then you can't justify all the accuracy of your answer. Does the question want to you to give the answer with an appropriate precision?

3. Feb 23, 2010

### csimon1

It wants me to give the answer in A. This is the feedback I got:
12500 or 1.25×10e4

Not quite. Check through your calculations; you may have made a rounding error or used the wrong number of significant figures.

4. Feb 23, 2010

### sylas

Is this online?

5. Feb 23, 2010

### csimon1

yes it's using mastering physics.

6. Feb 23, 2010

### Coto

Well double check your significant figures. (i.e. 2.5 C is only 2 significant digits)

7. Feb 23, 2010

### csimon1

im sorry, i dont really understand what you mean? 2.5 C was given to me in the problem. I don't know what else to use.

8. Feb 23, 2010

### sylas

Looks interesting. This is a product by Pearson. It is possible that there might be a glitch with the problem, but try giving the answer to a more appropriate number of figures accuracy, and see if that is accepted. You've obtained the answer 12500. But this has three figures of accuracy. What would be that value, with less precision?

9. Feb 23, 2010

### Coto

See: http://en.wikipedia.org/wiki/Significant_figures: [Broken]
spurious digits introduced, for example, by calculations carried out to greater accuracy than that of the original data, or measurements reported to a greater precision than the equipment supports.

In this case you only have 2 relevant digits. You would have to round to 13,000 since:
leading and trailing zeros where they serve merely as placeholders to indicate the scale of the number.

Last edited by a moderator: May 4, 2017
10. Feb 23, 2010

### csimon1

ohh thank you very much i would have never gotten it otherwise!