Chapter 18
Electric Circuits
Electric Power
Suppose some charge Δq emerges from a battery in time Δt and the
potential difference between the battery terminals is V. The battery is
putting out power to the circuit since it is outputting energy per time,
I
>
energy
V +|
Δq, Δt
(
Δq )V
P=
Δt
Δq
=
V = IV
Δt
power
time
Electric Power
ELECTRIC POWER
When there is current in a circuit as a result of a voltage, the electric
power delivered to the circuit is:
P = IV
SI Unit of Power: watt (W)
Many electrical devices are essentially resistors, so the power
dissipated by them can be calculated using their resistance, R
P = I (IR ) = I 2 R
V2
&V #
P = $ !V =
R
%R"
Electric Power
Example: The Power and Energy Used in a
Flashlight
In the flashlight, the current is 0.40A and the voltage
is 3.0 V. Find (a) the power delivered to the bulb and
(b) the energy dissipated in the bulb in 5.5 minutes
of operation.
Electric Power
(a)
(b)
P = IV = (0.40 A )(3.0 V ) = 1.2 W
E = Pt = (1.2 W )(330 s ) = 4.0 ×10 2 J
Alternating Current
In an AC circuit, the charge flow reverses direction periodically.
Alternating Current
V = Vo sin ( 2π ft ) = V0 sin (ω t )
V0 = peak voltage
f = frequency
ω = 2πf = angular frequency
Alternating Current
In circuits that contain only resistance, the current reverses direction each time
the polarity of the generator reverses.
V Vo
I = = sin (2π ft ) = I o sin (2π ft )
R R
peak current
Alternating Current
I = I o sin (2π ft )
V = Vo sin (2π ft )
P = IV = I oVo sin 2 (2π ft )
Alternating Current
I oVo ⎛ I o ⎞⎛ Vo ⎞
P=
= ⎜
⎟⎜
⎟ = I rmsVrms
2
⎝ 2 ⎠⎝ 2 ⎠
Irms = root mean square current
Vrms = root mean square voltage
Alternating Current
AC (rms) versions of Ohm’s Law and Power
Vrms = I rms R
P = Vrms I rms
2
P = I rms
R
2
Vrms
P=
R
Example: What is the resistance of the tungsten filament in
a 100 W light bulb meant to be used in a lamp which is to be
plugged into an AC outlet with peak voltage V0 = 170 V?
2
rms
2
rms
V
V
P=
→R=
R
P
V0 170
Vrms =
=
= 120 V
2
2
2
Vrms
120 2
R=
=
= 144 Ω
P
100
(US AC outlet
voltage)
Alternating Current
Example: Electrical Power Sent to a
Loudspeaker
A stereo receiver applies a peak voltage of
34 V to a speaker. The speaker behaves
approximately as if it had a resistance of 8.0 Ω.
Determine (a) the rms voltage, (b) the rms
current, and (c) the average power for this
circuit.
Alternating Current
(a)
(b)
(c)
Vrms
I rms
Vo 34 V
=
=
= 24 V
2
2
Vrms 24 V
=
=
= 3.0 A
R
8.0 Ω
P = I rmsVrms = (3.0 A )(24 V ) = 72 W
Alternating Current
Conceptual Example: Extension Cords and a Potential Fire Hazard
During the winter, many people use portable electric space heaters to keep
warm. Sometimes, however, the heater must be located far from a 120-V wall
receptacle, so an extension cord must be used. However, manufacturers often
warn against using an extension cord. If one must be used, they recommend
a certain wire gauge, or smaller. Why the warning, and why are smaller-gauge
wires better then larger-gauge wires?