Find out How Long 9.5 W Battery Can Run Camera

[Travian]

Homework Statement

How long the camera can be
operated in the 9.5 Watts recording mode using an initially fully charged
battery.

P = 9.5 W,
V = 7.2 V,
I = 1.3194 A

Homework Equations

(Cb) = I x t

The Attempt at a Solution

(Cb) = I x t = 1.3194 x 3600 = 4749,84 minutes (79+ hours)?

 

[gneill]

What are the characteristics (specifications) for the battery?

 

[Travian]

“BP-927” DC rechargeable lithium battery pack.

Damn i thought it doesn't matter.

 

[gneill]

“BP-927” DC rechargeable lithium battery pack.

Damn i thought it doesn't matter.

Sure it matters! There are many different battery types available with different energy storage capabilities. There are a number of "replacement" batteries for the original BP-927 that have various ratings, but a quick web search seems to indicate that the original battery pack had specs of 7.2V and 2000 mAh. So 2000 mAh is the amount of charge that the battery can deliver before it is exhausted and requires recharging.

The device power consumption of 9.5W at 7.2V can be converted to a value for the operating current. Current x time is the amp-hours.

 

[Travian]

Im having some trouble. What should I do next? Calculate current x time?

 

[gneill]

Im having some trouble. What should I do next? Calculate current x time?

More or less, yes; It's the time that you need to solve for. First calculate the current drawn from the battery pack when it's providing 9.5W to the load. Then current x time is the amp-hours. Solve for time.

 

[Travian]

Current:

I = P / V

I = 9.5W / 7.2V = 1.319A

Current x time:

Cb = I x t

Cb = 1.319A x 3600s = 4748.4s = 79.14minutes

 

[Travian]

If this is correct then what does 200mAh have to do with this?

 

[gneill]

Time is what you want to solve for. Why are you plugging it a value?

The battery holds 2000 mAh of charge. At 9.5W it's delivering current at 1.319A. How many seconds at 1.319A will make 2000 mAh of charge?

 

[Travian]

mm so it's 2000/1,319 = 1516s = 25 min?

 

[gneill]

You've got the right idea, but you need to keep track of all the units:

(2000 mA*hr)/(1319 mA) = ?

 

[Travian]

2000 mA x 3600 / 1319 mA = 5458,68s = 90,978 min.

 

[Travian]

Thank you for your help!

There is more to this exercise, should i post it here, or create a new thread?

 

[gneill]

Thank you for your help!

There is more to this exercise, should i post it here, or create a new thread?

It's your choice.

 

[Travian]

Homework Statement

Calculate the total energy stored in the BP-927 battery when it is fully charged.
P = 9.5 W,
V = 7.2 V,
I = 1.3194 A

Homework Equations

E = V x Ah

The Attempt at a Solution

E = 7.2V x 2Ah = 14.4 Joules..?

 

[Quinzio]

14 J ?
With this energy you can use your camera for some seconds.

 

[gneill]

E = 7.2V x 2Ah = 14.4 Joules..?

Once again, be careful of the units. 2Ah is 2 amps x 1hr. An hour is 3600 seconds.

 

[Travian]

Oh right.
E = 7.2V x 2A x 3600s = 7.2V x 7200Ah = 51840 J

However what bothers me is...

It was 2 Ah before. After i multiply 2 Ah by 3600s, is it again Ah?

 

[gneill]

Oh right.
E = 7.2V x 2A x 3600s = 7.2V x 7200Ah = 51840 J

However what bothers me is...

It was 2 Ah before. After i multiply 2 Ah by 3600s, is it again Ah?

Sorry, I don't understand the question. Can you point to a particular calculation where this was the case?

 

[Travian]

I figured it out.

Thank you for your help.

I'll keep in mind that Ah = A x 3600s :]