Homework 18'65

1
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.
2
(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 youll see
what it means. I dont think Ill 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.
3
Using P V2 / R, we get
Pi Vi2 / R
Pf Vf2 / R (Vi - ?V)2 / R
The power ratio is
assuming R does not change
4
According to the binomial theorem (see page
1043), for x small,
If ?V is small compared to Vi, then
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.
5
(b) Calculate the voltage drop for 60 W bulb to
act like 50 W bulb.
If you do the calculation using our OSEs, you
wont get Giancolis approximate result.
Instead, using our result of part (a)