246
A “75watt” lightbulb uses an average power of 75 W when connected to an rms voltage of 120 V.
(a)
What is the resistance of the lightbulb?
(b)
What is the maximum current in the bulb?
(c)
What is the
maximum power used by the bulb at any given instant of time?
Picture the Problem
: A light bulb dissipates power as the voltage oscillates across its filament resistance.
Strategy:
Calculate the resistance from the average power and the rms voltage using equation 216. Then,
from the resistance and
rms voltage, solve for the
rms current using Ohm’s Law (equation 212). Convert
the
rms current to maximum current by multiplying it by the square root of two. Finally, use the resistance
and maximum current to calculate the peak power dissipation.
Solution:
1. (a)
Solve equation 216 for
R
:
2
2
rms
av
120 V
190
75 W
V
R
P
2.
(b)
Use Ohm’s Law to calculate
rms
:
I
rms
rms
120 V
0.625 A
192
V
I
R
3.
Convert to peak current:
max
rms
2
2 0.625 A
0.88 A
II
4.
(c)
Calculate maximum power:
2
2
max
max
0.884 A
192
150 W
0.15 kW
PI
R
Insight:
The peak power dissipated is equal to twice the average power dissipated.
2479
A 4.40 F
and an 880
.F
capacitor are connected in
parallel
to a 60.0Hz generator operating with an
rms voltage of 115 V. What is the rms current supplied by the generator?
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview.
Sign up
to
access the rest of the document.
 Fall '06
 Alexandrakis
 Physics, Current, Resistance, Power, Light, Resistor, Inductor, Electrical resistance

Click to edit the document details