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Archived What is the voltage drop?

  • Thread starter Jarfi
  • Start date
363
12
1. Homework Statement
The light bulb is 60 watts at 230 volts, how much does the voltage need to drop for the lightbulb to be 40 watts'?


2. Homework Equations

V=IA, P=IV, P=IR^2, V=IR

3. The Attempt at a Solution

I tried 60w/230v=0,26A..... 40w=0,26A*XV<->x=154V so voltage drop is 230-154=76.. I know the current drops with the voltage but this is all I can think of.


nevermind, I found out.... 40w=v^2/885ohms->> v^2=35400(ACCIDENTALLY DEVIDED) so V=188.
 
Last edited:

CWatters

Science Advisor
Homework Helper
Gold Member
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2,292
The key to understanding this problem is to realise that the resistance is the same for both situations. After all its the same bulb.

Using the relevant equations...

P=I*V
and
V=I*R
You can write..
P=(V^2)/R
Then rearrange to give...
R=(V^2)/P

Then plug in the numbers giving...
R=(230^2)/60=882Ohms

Then using...
P=(V^2)/R
Rearrange to give..
V=SQRT(R*P)

Substitute the new numbers...
V=SQRT(882*40)
=188V

However the question asks how much does the voltage need to drop and that is....

230-188=42V
 

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