I Calculating Watt Hours for a time interval less than 1 second

AI Thread Summary
Using an online physics calculator can yield confusing results, especially when inputting time values less than one second for watt-hour calculations. The initial error message regarding negative values is misleading, as the calculator ultimately provides a result. The discussion highlights uncertainty about the unit "J watts hours" and the accuracy of the output, specifically the value of 0.1. A proper calculation shows that multiplying power (10W) by time (0.01s) and converting from joules to watt-hours results in approximately 27.8 micro watt-hours. Understanding the conversion process is crucial for accurate energy calculations.
pete94857
Messages
99
Reaction score
9
TL;DR Summary
Is it OK to put less than 1 second into the equation ?
I use an online physics calculator a lot to help me for various reasons. When I input less than 1 second into a watt hour calculation it initially tells me it can not be a negative value although that does disappear and it does do the calculation.

Example power 10 w time 0.01 seconds 0.1 J watts hours.
 

Attachments

  • Screenshot_20241111_232524_Chrome.jpg
    Screenshot_20241111_232524_Chrome.jpg
    20.3 KB · Views: 70
  • Screenshot_20241111_232347_Chrome.jpg
    Screenshot_20241111_232347_Chrome.jpg
    17.5 KB · Views: 46
Physics news on Phys.org
pete94857 said:
Example power 10 w time 0.01 seconds 0.1 J watts hours.
I have no idea what a "J watts hours" is as a unit, and the number 0.1 looks incorrect to me.

Can you show us by hand how you would calculate 10W*0.01s with the results in Watt*Hours?
 
  • Like
Likes russ_watters and pete94857
berkeman said:
I have no idea what a "J watts hours" is as a unit, and the number 0.1 looks incorrect to me.

Can you show us by hand how you would calculate 10W*0.01s with the results in Watt*Hours?
It is an ambiguous interface. The answer is in joules.
 
  • Informative
  • Like
Likes pete94857 and berkeman
pete94857 said:
TL;DR Summary: Is it OK to put less than 1 second into the equation ?

I use an online physics calculator a lot to help me for various reasons. When I input less than 1 second into a watt hour calculation it initially tells me it can not be a negative value although that does disappear and it does do the calculation.

Example power 10 w time 0.01 seconds 0.1 J watts hours.
It doesn’t like zero, but greatery than zero seems ok. Watt-hours should be Energy.
 
berkeman said:
I have no idea what a "J watts hours" is as a unit, and the number 0.1 looks incorrect to me.

Can you show us by hand how you would calculate 10W*0.01s with the results in Watt*Hours?
Power x time = Energy. Power is the required energy to power something, time is the amount of time needed and energy is the energy needed to perform the task at that time. Or reverse E/t=w
 
pete94857 said:
Power x time = Energy. Power is the required energy to power something, time is the amount of time needed and energy is the energy needed to perform the task at that time. Or reverse E/t=w
Well, that didn't really answer my question, but I'll show you how to do it since I don't think this is a schoolwork-related question.

So to calculate the Energy based on power and time and convert units from W*s to W*hr:

$$E = 10W ~0.01s ~ \frac{1hr}{3600s} = 27.8\mu Whr $$

Note how I converted from W*s (Joules) to W*hr by multiplying by "1" in this form: ##1=\frac{1hr}{3600s}## to get the right cancellations of units from the initial answer to what I wanted. :wink:
 
Thread 'Gauss' law seems to imply instantaneous electric field'
Imagine a charged sphere at the origin connected through an open switch to a vertical grounded wire. We wish to find an expression for the horizontal component of the electric field at a distance ##\mathbf{r}## from the sphere as it discharges. By using the Lorenz gauge condition: $$\nabla \cdot \mathbf{A} + \frac{1}{c^2}\frac{\partial \phi}{\partial t}=0\tag{1}$$ we find the following retarded solutions to the Maxwell equations If we assume that...
Hello! Let's say I have a cavity resonant at 10 GHz with a Q factor of 1000. Given the Lorentzian shape of the cavity, I can also drive the cavity at, say 100 MHz. Of course the response will be very very weak, but non-zero given that the Loretzian shape never really reaches zero. I am trying to understand how are the magnetic and electric field distributions of the field at 100 MHz relative to the ones at 10 GHz? In particular, if inside the cavity I have some structure, such as 2 plates...
Back
Top