245:
Picture the Problem
: Power is dissipated at a rate proportional to the square of the voltage in a resistive AC
circuit.
Strategy:
Solve equation 246 for the average power dissipation and equation 216 for the peak power dissipation.
Solution:
1. (a)
Use equation 246 to find
av
:
P
2
2
max
av
3
141 V
11
2.99 W
22
3.33 10
V
P
R
2.
(b)
Calculate the peak power dissipation:
2
2
max
max
3
141 V
5.97 W
3.33 10
V
P
R
Insight:
For a sinusoidal voltage the peak power dissipation in a resistor is double the average power dissipation.
2416:
Picture the Problem
: An oscillating voltage drives an alternating current
through a capacitor as in the circuit depicted at the right.
Strategy:
Solve equation 248 for the capacitive reactance. Then insert the
reactance into equation 249 to solve for the capacitance. Combine
equations 248 and 249 to write the rms current as a function of the
frequency and solve for the current when the frequency is increased to 410
Hz.
Solution:
1. (a)
Insert
rms
rms
C
X
VI
into equation 249 and solve for
C
:
rms
rms
1
0.021 A
4.0 F
26
0
.
0
H
z1
4
V
C
I
C
XV
2.
(b)
The current in the capacitor is directly proportional to the frequency, so the current will increase if the
frequency increases.
4. (c)
Calculate the current at
f
= 410 Hz:
rms
rms
rms
rms
C
6
1
2
410 Hz
4.0 10 F 14 V
0.14 A
VV
IC
V
XC
Insight:
To increase the current to 0.50 A, the frequency should be increased to 1430 Hz.
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 Fall '06
 Alexandrakis
 Physics, Power, Frequency, Hertz, Doppler, Hz

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