Welcome to EDAboard.com

Welcome to our site! EDAboard.com is an international Electronic Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.

Physics capacitor question...

Status
Not open for further replies.

Xiah

Newbie level 3
Joined
May 6, 2011
Messages
4
Helped
0
Reputation
0
Reaction score
0
Trophy points
1,281
Location
Newcastle
Activity points
1,308
Hi guys i'm pretty new here and i'm just looking for abit of help really. I'm attempting my physics assignment and i can't for the world of me figure out how to work this one out and searching online for equations is just making it more confusing for me! The question is below.

A 4700 uF capacitor is charged to a p.d. of 10v. When the capacitor is discharged through a small electric motor, the motor lifts a weight of 0.1N through a height of 0.12m.
I need to work out the energy stored in the charged capacitor - I've done this one I think! E = 1/2 CV^2 which is 1/2 * 4700 * 100
The gravitational potential energy gained by the weight
The efficiency of the energy transfer - This one i think i can do!

Any help at all would be greatly appreciated so thanks in advance!!
 
Last edited:

ferdem

Full Member level 2
Joined
Oct 18, 2008
Messages
121
Helped
29
Reputation
58
Reaction score
29
Trophy points
1,308
Location
Türkiye
Activity points
2,412
You have overlooked "uF". Stored energy in the cap= 0.5*4700e-6*100 =235 mJ (millijoule)
0.1N does 0.1*0.12=12mJ much work.
e=12/235=0.05 (!)
 
  • Like
Reactions: Xiah

    Xiah

    Points: 2
    Helpful Answer Positive Rating

ServoAmp

Newbie level 5
Joined
May 6, 2011
Messages
9
Helped
4
Reputation
8
Reaction score
4
Trophy points
1,283
Location
Austin, TX
Activity points
1,393
Wtransfer is the energy transferred = Work physical
Wloss = energy lost in the transfer
Total energy = Wloss + Wtransfer = Work Capacitor
Efficiency = Wtransfer/total energy

Work (physical) = .1N*.12M =.012Nm
Work Capacitor = 0.5*4700E-6*100 = 0.235 Joules = .235 Nm
Wloss = Wcapacitor=Wphysical = .235-.012 = .223
Efficiency = .012/.235 = .051 = 5%

Work (physical) = F*d
http://en.wikipedia.org/wiki/Work_(physics)
Work Capacitor = 0.5CV^2
Capacitance - Wikipedia, the free encyclopedia
I Joule = 1N*1m
Joule - Wikipedia, the free encyclopedia
 
  • Like
Reactions: Xiah

    Xiah

    Points: 2
    Helpful Answer Positive Rating

Xiah

Newbie level 3
Joined
May 6, 2011
Messages
4
Helped
0
Reputation
0
Reaction score
0
Trophy points
1,281
Location
Newcastle
Activity points
1,308
I thought the gravitational potential energy gained by the weight would be = mgh 0.1 * 9.8 * 0.12 = 0.1176?

Thank you both so much for your help by the way!
 

_Eduardo_

Full Member level 5
Joined
Aug 31, 2009
Messages
295
Helped
118
Reputation
238
Reaction score
103
Trophy points
1,323
Location
Argentina
Activity points
2,843
It´s a weight of 0.1N not a mass of 0.1Kg
Already was multiplied by 9.8 ( weight = mg )
 

Status
Not open for further replies.

Part and Inventory Search

Welcome to EDABoard.com

Sponsor

Top