Archived Therefore the voltage needs to drop by 42V for the light bulb to be 40 watts.

[Jarfi]

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'?

Homework Equations

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

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.

 

[CWatters]

The key to understanding this problem is to realize 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