Transcript What the heck does that mean???

Homework 18.65
In a “brownout” situation, the voltage supplied by the electric
company falls. Assuming the percent drop is small, show that the
power output of a given appliance falls by approximately twice that
percent, assuming the resistance does not change. How much of a
voltage drop does it take for a 60 W light bulb to begin acting like a
50 W light bulb?
This problem has two parts.
(a) Show that the power output of a given appliance falls by
approximately twice the percent voltage drop, assuming the
resistance does not change.
What the heck does that mean???
(b) Calculate the voltage drop for 60 W bulb to act like 50 W bulb.
(a) Show that the power output of a given appliance falls by
approximately twice the percent voltage drop, assuming the
resistance does not change.
Let me do this first, and then maybe you’ll see what it means. I
don’t think I’ll give a problem like this on the exam!
Even though your home power is ac rather than dc, we will use our
dc equations, as we have done throughout chapter 18.
OSE:
P = V2 / R
Assume the voltage starts at some initial value Vi and decreases by
an amount V = Vf – Vi.
The fractional drop is V /Vi and the percentage drop is 100 V /Vi.
Using P = V2 / R, we get
Pi = Vi2 / R
Pf = Vf2 / R = (Vi - V)2 / R
The power ratio is
  Vi - ΔV 2 


R


Pf
=

2
Pi
 Vi 
R 
 
 Vi
Pf 
ΔV 
= 1 
Pi 
Vi 
- ΔV 
Vi2
2
2
assuming R does
not change
According to the binomial theorem (see page 1043), for x small,
1
 x
n
 1  nx
If V is small compared to Vi, then
Pf 
ΔV 
= 1 
Pi 
Vi 
2
1 - 2
ΔV
Vi
The change in power is approximately 2(V/Vi), which is what
Giancoli wanted you to prove (except I have expressed the result as
a ratio and he expressed it as a percent).
“Do you really expect me to remember this?”
Not for an exam. But I expect you to file this “trick” away in your
brain for future reference, because it is used by many disciplines in
many situations.
(b) Calculate the voltage drop for 60 W bulb to act like 50 W bulb.
If you do the calculation using our OSE’s, you won’t get Giancoli’s
approximate result. Instead, using our result of part (a):
P
ΔV
 -2
Pi
Vi
ΔV
1 P
 Vi
2 Pi
ΔV
1 50  60
 Vi
2 60
ΔV
1

 .083  8.3% .
Vi
12